Physics GRE Solutions Omnibus

Senior Editor: Taylor Faucett Editor-in-Chief: Taylor Faucett Associate Editor: Taylor Faucett Editorial Assistant: Taylor Faucett Art Studio: Taylor Faucett Art Director: Taylor Faucett Cover Design: Taylor Faucett

BiBTeX: @Book{FaucettOmnibus, author = {Taylor Faucett}, title = {Physics GRE Solutions Omnibus}, pages = {1-560}, year = {2010}, edition = {first}, }

c 2010 by Taylor Faucett, all rights reserved.

The “Physics GRE Solutions Omnibus” is in no way affiliated with ETS or the GRE. All media and information contained within is to be used strictly for educational and non-profit purposes. Any reproductions of this work, whether in part or in full, will be allowed provided that proper attribution is given. Additionally, contacting me about its use would be a nice gesture.

Contents 1 Introduction 13 1.1 About the Omnibus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2 PGRE8677 Solutions 2.1 PGRE8677 #1 . . 2.2 PGRE8677 #2 . . 2.3 PGRE8677 #3 . . 2.4 PGRE8677 #4 . . 2.5 PGRE8677 #5 . . 2.6 PGRE8677 #6 . . 2.7 PGRE8677 #7 . . 2.8 PGRE8677 #8 . . 2.9 PGRE8677 #9 . . 2.10 PGRE8677 #10 . . 2.11 PGRE8677 #11 . . 2.12 PGRE8677 #12 . . 2.13 PGRE8677 #13 . . 2.14 PGRE8677 #14 . . 2.15 PGRE8677 #15 . . 2.16 PGRE8677 #16 . . 2.17 PGRE8677 #17 . . 2.18 PGRE8677 #18 . . 2.19 PGRE8677 #19 . . 2.20 PGRE8677 #20 . . 2.21 PGRE8677 #21 . . 2.22 PGRE8677 #22 . . 2.23 PGRE8677 #23 . . 2.24 PGRE8677 #24 . . 2.25 PGRE8677 #25 . . 2.26 PGRE8677 #26 . . 2.27 PGRE8677 #27 . . 2.28 PGRE8677 #28 . . 2.29 PGRE8677 #29 . . 2.30 PGRE8677 #30 . . 2.31 PGRE8677 #31 . . 2.32 PGRE8677 #32 . . 2.33 PGRE8677 #33 . .

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15 16 18 19 21 22 24 26 28 30 31 32 33 34 35 36 37 38 39 41 42 44 45 46 47 48 49 51 53 54 55 56 58 59

CONTENTS

2.34 2.35 2.36 2.37 2.38 2.39 2.40 2.41 2.42 2.43 2.44 2.45 2.46 2.47 2.48 2.49 2.50 2.51 2.52 2.53 2.54 2.55 2.56 2.57 2.58 2.59 2.60 2.61 2.62 2.63 2.64 2.65 2.66 2.67 2.68 2.69 2.70 2.71 2.72 2.73 2.74 2.75 2.76 2.77 2.78 2.79 2.80 2.81

PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677 PGRE8677

CONTENTS

#34 . #35 . #36 . #37 . #38 . #39 . #40 . #41 . #42 . #43 . #44 . #45 . #46 . #47 . #48 . #49 . #50 . #51 . #52 . #53 . #54 . #55 . #56 . #57 . #58 . #59 . #60 . #61 . #62 . #63 . #64 . #65 . #66 . #67 . #68 . #69 . #70 . #71 . #72 . #73 . #74 . #75 . #76 . #77 . #78 . #79 . #80 . #81 .

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60 62 64 66 68 69 70 71 72 74 76 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 95 96 98 99 100 101 103 104 105 107 109 111 112 114 115 117 119 121 122 123 124

CONTENTS

2.82 PGRE8677 2.83 PGRE8677 2.84 PGRE8677 2.85 PGRE8677 2.86 PGRE8677 2.87 PGRE8677 2.88 PGRE8677 2.89 PGRE8677 2.90 PGRE8677 2.91 PGRE8677 2.92 PGRE8677 2.93 PGRE8677 2.94 PGRE8677 2.95 PGRE8677 2.96 PGRE8677 2.97 PGRE8677 2.98 PGRE8677 2.99 PGRE8677 2.100PGRE8677

CONTENTS

#82 . #83 . #84 . #85 . #86 . #87 . #88 . #89 . #90 . #91 . #92 . #93 . #94 . #95 . #96 . #97 . #98 . #99 . #100

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126 127 128 129 130 131 132 133 134 136 137 138 140 142 143 145 147 148 150

3 PGRE9277 Solutions 3.1 PGRE9277 #1 . . 3.2 PGRE9277 #2 . . 3.3 PGRE9277 #3 . . 3.4 PGRE9277 #4 . . 3.5 PGRE9277 #5 . . 3.6 PGRE9277 #6 . . 3.7 PGRE9277 #7 . . 3.8 PGRE9277 #8 . . 3.9 PGRE9277 #9 . . 3.10 PGRE9277 #10 . . 3.11 PGRE9277 #11 . . 3.12 PGRE9277 #12 . . 3.13 PGRE9277 #13 . . 3.14 PGRE9277 #14 . . 3.15 PGRE9277 #15 . . 3.16 PGRE9277 #16 . . 3.17 PGRE9277 #17 . . 3.18 PGRE9277 #18 . . 3.19 PGRE9277 #19 . . 3.20 PGRE9277 #20 . . 3.21 PGRE9277 #21 . . 3.22 PGRE9277 #22 . . 3.23 PGRE9277 #23 . . 3.24 PGRE9277 #24 . . 3.25 PGRE9277 #25 . . 3.26 PGRE9277 #26 . . 3.27 PGRE9277 #27 . .

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151 152 153 154 155 156 158 160 161 162 164 166 168 170 171 172 173 174 176 177 178 180 181 182 183 184 185 187

5

CONTENTS

3.28 3.29 3.30 3.31 3.32 3.33 3.34 3.35 3.36 3.37 3.38 3.39 3.40 3.41 3.42 3.43 3.44 3.45 3.46 3.47 3.48 3.49 3.50 3.51 3.52 3.53 3.54 3.55 3.56 3.57 3.58 3.59 3.60 3.61 3.62 3.63 3.64 3.65 3.66 3.67 3.68 3.69 3.70 3.71 3.72 3.73 3.74 3.75

PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277 PGRE9277

CONTENTS

#28 . #29 . #30 . #31 . #32 . #33 . #34 . #35 . #36 . #37 . #38 . #39 . #40 . #41 . #42 . #43 . #44 . #45 . #46 . #47 . #48 . #49 . #50 . #51 . #52 . #53 . #54 . #55 . #56 . #57 . #58 . #59 . #60 . #61 . #62 . #63 . #64 . #65 . #66 . #67 . #68 . #69 . #70 . #71 . #72 . #73 . #74 . #75 .

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188 189 190 191 192 193 194 195 196 197 198 200 201 202 203 204 205 206 207 209 211 212 213 214 216 218 220 222 224 225 227 228 229 230 232 233 234 235 238 240 241 243 244 245 247 249 251 253

CONTENTS

3.76 PGRE9277 3.77 PGRE9277 3.78 PGRE9277 3.79 PGRE9277 3.80 PGRE9277 3.81 PGRE9277 3.82 PGRE9277 3.83 PGRE9277 3.84 PGRE9277 3.85 PGRE9277 3.86 PGRE9277 3.87 PGRE9277 3.88 PGRE9277 3.89 PGRE9277 3.90 PGRE9277 3.91 PGRE9277 3.92 PGRE9277 3.93 PGRE9277 3.94 PGRE9277 3.95 PGRE9277 3.96 PGRE9277 3.97 PGRE9277 3.98 PGRE9277 3.99 PGRE9277 3.100PGRE9277

CONTENTS

#76 . #77 . #78 . #79 . #80 . #81 . #82 . #83 . #84 . #85 . #86 . #87 . #88 . #89 . #90 . #91 . #92 . #93 . #94 . #95 . #96 . #97 . #98 . #99 . #100

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4 PGRE9677 Solutions 4.1 PGRE9677 #1 . . 4.2 PGRE9677 #2 . . 4.3 PGRE9677 #3 . . 4.4 PGRE9677 #4 . . 4.5 PGRE9677 #5 . . 4.6 PGRE9677 #6 . . 4.7 PGRE9677 #7 . . 4.8 PGRE9677 #8 . . 4.9 PGRE9677 #9 . . 4.10 PGRE9677 #10 . . 4.11 PGRE9677 #11 . . 4.12 PGRE9677 #12 . . 4.13 PGRE9677 #13 . . 4.14 PGRE9677 #14 . . 4.15 PGRE9677 #15 . . 4.16 PGRE9677 #16 . . 4.17 PGRE9677 #17 . . 4.18 PGRE9677 #18 . . 4.19 PGRE9677 #19 . . 4.20 PGRE9677 #20 . . 4.21 PGRE9677 #21 . .

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254 255 256 257 258 259 261 263 265 266 268 269 271 272 273 274 275 276 277 278 279 281 282 283 284

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285 . 286 . 288 . 290 . 292 . 295 . 297 . 300 . 302 . 303 . 305 . 306 . 307 . 309 . 311 . 313 . 315 . 316 . 318 . 320 . 321 . 323

CONTENTS

4.22 4.23 4.24 4.25 4.26 4.27 4.28 4.29 4.30 4.31 4.32 4.33 4.34 4.35 4.36 4.37 4.38 4.39 4.40 4.41 4.42 4.43 4.44 4.45 4.46 4.47 4.48 4.49 4.50 4.51 4.52 4.53 4.54 4.55 4.56 4.57 4.58 4.59 4.60 4.61 4.62 4.63 4.64 4.65 4.66 4.67 4.68 4.69

PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677 PGRE9677

CONTENTS

#22 . #23 . #24 . #25 . #26 . #27 . #28 . #29 . #30 . #31 . #32 . #33 . #34 . #35 . #36 . #37 . #38 . #39 . #40 . #41 . #42 . #43 . #44 . #45 . #46 . #47 . #48 . #49 . #50 . #51 . #52 . #53 . #54 . #55 . #56 . #57 . #58 . #59 . #60 . #61 . #62 . #63 . #64 . #65 . #66 . #67 . #68 . #69 .

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324 325 326 327 329 330 331 333 335 337 338 339 340 341 342 343 344 345 346 347 348 349 351 352 354 356 358 360 361 362 364 365 366 368 370 371 372 373 374 376 377 378 379 380 382 383 384 385

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4.70 PGRE9677 4.71 PGRE9677 4.72 PGRE9677 4.73 PGRE9677 4.74 PGRE9677 4.75 PGRE9677 4.76 PGRE9677 4.77 PGRE9677 4.78 PGRE9677 4.79 PGRE9677 4.80 PGRE9677 4.81 PGRE9677 4.82 PGRE9677 4.83 PGRE9677 4.84 PGRE9677 4.85 PGRE9677 4.86 PGRE9677 4.87 PGRE9677 4.88 PGRE9677 4.89 PGRE9677 4.90 PGRE9677 4.91 PGRE9677 4.92 PGRE9677 4.93 PGRE9677 4.94 PGRE9677 4.95 PGRE9677 4.96 PGRE9677 4.97 PGRE9677 4.98 PGRE9677 4.99 PGRE9677 4.100PGRE9677

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#70 . #71 . #72 . #73 . #74 . #75 . #76 . #77 . #78 . #79 . #80 . #81 . #82 . #83 . #84 . #85 . #86 . #87 . #88 . #89 . #90 . #91 . #92 . #93 . #94 . #95 . #96 . #97 . #98 . #99 . #100

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387 388 390 391 392 393 394 395 397 398 400 402 403 405 406 407 409 411 412 414 416 417 418 419 420 421 423 424 426 427 428

5 PGRE0177 Solutions 5.1 PGRE0177 #1 . . 5.2 PGRE0177 #2 . . 5.3 PGRE0177 #3 . . 5.4 PGRE0177 #4 . . 5.5 PGRE0177 #5 . . 5.6 PGRE0177 #6 . . 5.7 PGRE0177 #7 . . 5.8 PGRE0177 #8 . . 5.9 PGRE0177 #9 . . 5.10 PGRE0177 #10 . . 5.11 PGRE0177 #11 . . 5.12 PGRE0177 #12 . . 5.13 PGRE0177 #13 . . 5.14 PGRE0177 #14 . . 5.15 PGRE0177 #15 . .

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429 430 432 434 435 436 437 438 439 440 441 442 443 445 446 447

9

CONTENTS

5.16 5.17 5.18 5.19 5.20 5.21 5.22 5.23 5.24 5.25 5.26 5.27 5.28 5.29 5.30 5.31 5.32 5.33 5.34 5.35 5.36 5.37 5.38 5.39 5.40 5.41 5.42 5.43 5.44 5.45 5.46 5.47 5.48 5.49 5.50 5.51 5.52 5.53 5.54 5.55 5.56 5.57 5.58 5.59 5.60 5.61 5.62 5.63

PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177 PGRE0177

CONTENTS

#16 . #17 . #18 . #19 . #20 . #21 . #22 . #23 . #24 . #25 . #26 . #27 . #28 . #29 . #30 . #31 . #32 . #33 . #34 . #35 . #36 . #37 . #38 . #39 . #40 . #41 . #42 . #43 . #44 . #45 . #46 . #47 . #48 . #49 . #50 . #51 . #52 . #53 . #54 . #55 . #56 . #57 . #58 . #59 . #60 . #61 . #62 . #63 .

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449 450 451 452 453 454 456 457 458 459 461 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 488 490 492 493 494 495 496 497 498 499 500 501 502

CONTENTS

5.64 PGRE0177 5.65 PGRE0177 5.66 PGRE0177 5.67 PGRE0177 5.68 PGRE0177 5.69 PGRE0177 5.70 PGRE0177 5.71 PGRE0177 5.72 PGRE0177 5.73 PGRE0177 5.74 PGRE0177 5.75 PGRE0177 5.76 PGRE0177 5.77 PGRE0177 5.78 PGRE0177 5.79 PGRE0177 5.80 PGRE0177 5.81 PGRE0177 5.82 PGRE0177 5.83 PGRE0177 5.84 PGRE0177 5.85 PGRE0177 5.86 PGRE0177 5.87 PGRE0177 5.88 PGRE0177 5.89 PGRE0177 5.90 PGRE0177 5.91 PGRE0177 5.92 PGRE0177 5.93 PGRE0177 5.94 PGRE0177 5.95 PGRE0177 5.96 PGRE0177 5.97 PGRE0177 5.98 PGRE0177 5.99 PGRE0177 5.100PGRE0177

CONTENTS

#64 . #65 . #66 . #67 . #68 . #69 . #70 . #71 . #72 . #73 . #74 . #75 . #76 . #77 . #78 . #79 . #80 . #81 . #82 . #83 . #84 . #85 . #86 . #87 . #88 . #89 . #90 . #91 . #92 . #93 . #94 . #95 . #96 . #97 . #98 . #99 . #100

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503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 524 525 526 528 529 530 532 533 535 537 539 540 542 543 545 546 547

6 Appendix 6.1 Equations . . . . . . . . . . . . . . . . 6.1.1 Classical Mechanics . . . . . . 6.1.2 Electricity & Magnetism . . . . 6.1.3 Optics . . . . . . . . . . . . . . 6.1.4 Thermodynamics & Stat Mech 6.1.5 Quantum Mechanics . . . . . . 6.1.6 Special Relativity . . . . . . . . 6.1.7 Electronics . . . . . . . . . . . 6.1.8 Special Topics . . . . . . . . .

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548 548 548 550 552 553 555 556 557 558

11

CONTENTS

6.2

Units 6.2.1 6.2.2 6.2.3

CONTENTS

and Conversions Units . . . . . Conversions . . SI Prefixes . .

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559 559 560 560

Chapter 1

Introduction

13

1.1. ABOUT THE OMNIBUS

1.1

CHAPTER 1. INTRODUCTION

About the Omnibus

The “Physics GRE Solutions Omnibus” is a compilation of 400 solutions to the 4 Physics GRE practice exams. In addition to the solutions, a section of useful equations, units and conversions worth memorizing has been added. A website has also been created, http://PhysicsGrad.com, as a companion to the book. PhysicsGrad.com is a community based site that differs from “The Omnibus” in that all 400 problems feature both my solutions and solution contributions from registered members. Additionally, the website offers updates on University and departmental events and admissions, help with selecting universities and advice on test taking. I gladly welcome any and all corrections and/or advice that any Omnibus reader happens to find, whether it be typos, poor explanations or even outright falsehoods. The easiest way to contact me is via email through [email protected].

14

Chapter 2

PGRE8677 Solutions

15

2.1. PGRE8677 #1

2.1

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #1

Recommended Solution (A) In a frictionless environment, this would be true. However, note that the problem tells us that friction is not negligible (in fact it is proportional to the rocks velocity −v) so it isn’t possible for the acceleration to always be g. (B) Since the frictional force is always fighting the rocks motion proportional to its speed, the only point at which frictional forces disappear will be when the rock stops moving (v = 0) at the top of its flight. At this point, the only acceleration on the rock will then be the acceleration due to gravity. (C) As we just demonstrated in (B), the acceleration can be g at the top of the rocks flight (D) In a frictionless environment, kinetic energy at the bottom of the rocks path will be converted to potential at the peak of its flight. The rock will then head back to its initial point converting all potential back to kinetic energy. However, in an environment with non-negligible friction, some of that initial kinetic energy is lost to the surroundings through heat, sound, etc and so the initial energy can never be recovered at the end of the rocks flight. (E) Even in an ideal (i.e. frictionless) environment, this can’t be true. This could only be true if there was some force applied to the rock on the way down which is not the case. Correct Answer (B)

Alternate Solution Sum the forces in the vertical direction for the rocks motion,

16

2.1. PGRE8677 #1

CHAPTER 2. PGRE8677 SOLUTIONS

Ftot = −Ff ric − FG

(2.1)

Ftot = −kv − mg

(2.2)

(A) From the equation, we can see that it is not generally true that the total acceleration on the rock is simply g (B) Plug in v = 0 and g becomes the only acceleration and so this is correct (C) As demonstrated in (B), acceleration is not less than g when v = 0 (D) This can’t be true because energy is dissipated from friction so the final velocity can’t be equal to its initial velocity (E) Conservation of energy won’t allow you to have more velocity in the end than you started out with and we would expect to have a lower final velocity than initial velocity based on dissipation of energy from friction Correct Answer (B)

17

2.2. PGRE8677 #2

2.2

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #2

Recommended Solution We can first eliminate (D) because we would need a continuous force pushing the satellite away from the Earth to get an orbit spiraling outward. We can then eliminate (E) because oscillations arent going to occur unless the potential energy of the satellite remains the same after the force is applied, which isnt the case. Next, eliminate (C) because, as a general rule, when we perturb an orbit with a brief bit of thrust, the satellite will return to the same point on the next pass but the rest of its orbit will be altered, so it cant be circular again. Finally, when choosing between (A) and (B), a small energy change from a circular orbit is more likely to become an elliptical orbit and, in general, elliptical orbits are more common for lower energy orbits, like a satellite orbiting earth. Correct Answer (A)

18

2.3. PGRE8677 #3

2.3

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #3

Recommended Solution Recall from quantum mechanics (specifically from the relativistic de Broglie relations) that wave speed is E p γmc2 γmv c2 v

vp = = =

(2.3) (2.4) (2.5)

The velocity of light through a medium is related to its relative permittivity and relative permeability by 1 µ

(2.6)

c2 c =√ µ 2.1

(2.7)

v= Plug this in and get vp =

Correct Answer (D)

19

2.3. PGRE8677 #3

CHAPTER 2. PGRE8677 SOLUTIONS

Alternate Solution It is generally true that only particles with mass (i.e. not a photon) will have phase velocities faster than the speed of light. Checking the possibilities √ (A) 3.1c ≈ 1.7c √ (B) 2.1c ≈ 1.4c √ (C) c/ 1.1 ≈ 0.95c √ (D) c/ 2.1 ≈ 0.7c √ (E) c/ 3.1 ≈ 0.56c From this you can comfortably eliminate (A) and (B) and cautiously eliminate (C) because it is nearly the speed of light. With (D) and (E), ask yourself whether it makes more sense for the permittivity and permeability to be additive (i.e. 1.0 + 2.1) or multiplicative (i.e. (1.0)(2.1)). Even if you can’t recall that the velocity of an electromagnetic wave is the inverse product of the two values, you could at least make an educated guess by recalling that every time you have ever seen the permittivity of free space (0 ), it has always acted as a scaling value. For example in the Gravitational Potential equation UG =

1 mM 4π0 r

Correct Answer (D)

20

(2.8)

2.4. PGRE8677 #4

2.4

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #4

Recommended Solution The general equation of a traveling wave is 2π y(x, t) = Asin (x ± vt) λ 



(2.9)

Where A is the amplitude, T is the period, λ is wavelength, t is time and x is position. (A) The amplitude in this instance is A. In order to get an amplitude of 2A, we would need our equation to have the amplitdue, A, multiplied by 2 (B) From the general traveling wave equation, if the ± ends up being a negative then the wave propagates in the positive direction and the converse is also true. (C) The period is T = 1/f and propagation velocity is v = f λ so T = λ/v not T = T /λ. Also, T /λ would give the wrong units for the period. (D) Propagation speed is v = f λ and the period T = 1/f so v = λ/T , not x/t. (E) Propagation speed is v = f λ and the period T = 1/f so v = λ/T Correct Answer (E)

21

2.5. PGRE8677 #5

2.5

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #5

Recommended Solution This problem asks us about an inelastic collision so we can’t assume that energy is conserved between the initial and final stages of the swinging spheres. Break the problem up into 3 separate time frames with the first phase involving just the falling of sphere A. In this phase, the initial potential energy is equal to the kinetic energy immediately before the two spheres collide 1 M gh0 = M v02 2

(2.10)

p

(2.11)

so the velocity before impact is v0 =

2gh0

In the second phase, consider the collision of the ball, which should conserve momentum M vo = (M + 3M )vf = 4M vf

(2.12)

combine Equation 2.11 and Equation 2.12 to get M

p



2gh0 = 4M vf

so the final velocity of the spheres after they’ve combined is 22

(2.13)

2.5. PGRE8677 #5

CHAPTER 2. PGRE8677 SOLUTIONS

s

vf =

2gh0 = 16

s

gh0 8

(2.14)

finally, in the third phase, we know that the kinetic energy of the combined spheres immediately after collision will be equal to the potential energy at their peak 1 (4M )vf2 = (4M )ghf 2 Combine Equation 2.14 and 2.15 to get s

1 2

(2.15)

2

gh0  8

hf

= ghf =

1 h0 16

Correct Answer (A)

23

(2.16) (2.17)

2.6. PGRE8677 #6

2.6

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #6

Recommended Solution Note that as x → ∞, the track becomes infinitely steep and the particle should free fall at an acceleration of g. Only (D) approaches g in this limit. Correct Answer (D)

Alternate Solution Consider the particle at some point on the curve and break it up into its components (Figure 2.1) The tangential acceleration is going to be the component of the acceleration due to gravity multiplied by a unit vector in the same direction as the tangent at = ~g · tˆ 24

(2.18)

2.6. PGRE8677 #6

CHAPTER 2. PGRE8677 SOLUTIONS

x

O

θ dy

at dx

y Figure 2.1: Velocity components on a particle as it moves through an arc since we are talking about a simple unit vector, take the dot product of the two and get at = gcos(θ)

(2.19)

and with the angle θ, we get at = gcos(θ) = g

dy dy = gp 2 at dx + dy 2

(2.20)

recall that dy d = dx dx

x2 4

!

=

x 2

(2.21)

so Equation 2.20 becomes gx gy 0 at = p =√ 0 2 1 + (y ) 4 + x2 Correct Answer (D)

25

(2.22)

2.7. PGRE8677 #7

2.7

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #7

Recommended Solution Draw out your force diagram with the applied force shown pulling with 10N to the right and a tension force with an x component in the opposite direction of the applied force and a y component along the dotted line (pointing up). When the box is in equilibrium, acceleration will be 0 so we are concerned with the point at which the applied force of 10N is equal to the opposing tension force FT −x = 10N

(2.23)

Take tan(θ) to get tan(θ) =

FT −x FT −y

(2.24)

and re-arrange it to get FT −x = FT −y tan(θ)

(2.25)

The force in the vertical direction will only be due to gravity so we can use FT −y = 20N as an approximation. Plug this into Equation 2.25 and then combine FT −x with Equation 2.23 to get 10N

= (20N ) tan(θ)   −1 1 θ = tan 2

26

(2.26) (2.27)

2.7. PGRE8677 #7

CHAPTER 2. PGRE8677 SOLUTIONS

Correct Answer (A)

27

2.8. PGRE8677 #8

2.8

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #8

Recommended Solution Using our kinematic equation v 2 = v02 + 2a∆x

(2.28)

and assuming the initial velocity is zero, solve for the acceleration (10 m/s)2 = 2a(0.025 m) 2

a = 2000 m/s

(2.29) (2.30)

With a mass of 5 kg, use Newton’s second law to get F = (5 kg)(2000 m/s2 ) = 10, 000 N

(2.31)

Correct Answer (D)

Alternate Solution The kinetic energy of the stone right at impact will be 1 1 EK = mv 2 = (5 kg)(10 m/s)2 = 250 J (2.32) 2 2 This calculation doesn’t account for the small potential energy the stone will have once it has hit the nail and it also makes the assumption (which the problem doesn’t necessarily give us) that energy is being conserved. However, as an approximation, let’s assume that all energy is accounted for and conserved so we can let all the kinetic energy translate into the work done when moving the nail W F

= F · ∆x = 250 J 250 J = 10, 000 N = 0.025 m 28

(2.33) (2.34)

2.8. PGRE8677 #8

CHAPTER 2. PGRE8677 SOLUTIONS

Correct Answer (D)

29

2.9. PGRE8677 #9

2.9

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #9

Recommended Solution The drift velocity of a charged particle due to an electric field is vd =

i nqA

(2.35)

where i is the current, n is the number of particles, q is the charge of the particle and A is the cross sectional area. We could calculate the value exactly but it is generally quicker and sufficient, especially with the wide range of choices, to do a quick approximation. vd =

(1 ×

1028

e− /m3 )(2

100 A ≈ 2 × 10−5 × 10−19 C)(3 × 10−3 m2 )

which is closest to (D) Correct Answer (D)

30

(2.36)

2.10. PGRE8677 #10

2.10

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #10

Recommended Solution From Gauss’s law, we have I

E · dA =

S

Q 0

(2.37)

For a sphere, the surface area dA will be 4πr2 and the charge volume Q will be the volume of a sphere, Q = 4/3 πr3 . Plug things in to get 

E 4πr

2



4 πr3 30 r 30



=

E = so the curve must be linear, i.e. (C)

Correct Answer (C)

31



(2.38) (2.39)

2.11. PGRE8677 #11

2.11

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #11

Recommended Solution You should recall from quantum mechanics (especially if you were forced to do the rigorous proof from Griffith’s canonical quantum mechanics textbook) that the divergence of the curl of any vector field is always 0. Take the divergence of each side to get 



= ∇ · (∇ × H)

(2.40)





= 0

(2.41)

∇ · D˙ + J ∇ · D˙ + J

Correct Answer (A)

32

2.12. PGRE8677 #12

2.12

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #12

Recommended Solution Without knowing much about the physics involved, we can solve this problem with a bit of musical knowledge. From the Doppler effect, we know that pitch (frequency) goes up as a source of sound moves toward you, eliminating (A) and (B). Next, recall that the next octave of any tone is twice that of the original, meaning that (C) and (D) would both be suggesting that even at near the speed of light the source hasn’t even exceeded an octave in sound. (E) is the most reasonable solution. Correct Answer (E)

Alternate Solution If you recall the equation for the frequency of a moving sound wave, you can calculate the exact solution "

f=

1 1±

vsource vwave

#

f0

(2.42)

Plug everything into Equation 2.42 to get 

f=

1 f0 = 10f0 1 − 0.9 

Correct Answer (E)

33

(2.43)

2.13. PGRE8677 #13

2.13

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #13

Recommended Solution We can eliminate (A) and (B) because we know that the interference pattern should be affected, at the very least we know this is the case for the interference extrema as the relative phase changes. Eliminate (C) because the interference pattern certainly wont be destroyed for all but two points. Finally, eliminate (D) because the monochromacity is not destroyed and, even if it stopped being monochromatic, that wouldnt stop an interference pattern. Alternatively, instead of doing process of elimination you can directly figure that (E) is correct because the Flicker-Fusion Threshold (The frequency at which an objects movement starts to become imperceptible to the human eye) occurs at roughly 60 Hz, which is a far cry from the 500 Hz given in the problem. Correct Answer (E)

34

2.14. PGRE8677 #14

2.14

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #14

Recommended Solution When you hear ideal gas law, always go straight to the ideal gas law equation (the ‘Piv-nert’ equation) P V = nRT

(2.44)

(A) Recall from the first law of thermodynamics that work done on the system is, W = P dV . Combining this with the ideal gas law shows that (A) is true.















dQ dQ dQ (B) Recall that Cv = dQ dT and Cp = dT . If (B) was correct, then dT = dT = Cp = Cv v

p

v

p

which the initial question clearly tells us is not true. (C) This isn’t necessarily true, especially with changes in volume.



(D) Recall that Cv = dU dT which tells us that the energy of an ideal gas is only dependent on temperature. (E) Presumably the correct answer will never be an ill-formed question (or it won’t be graded if it is) so exclude (E) Correct Answer (E)

35

2.15. PGRE8677 #15

2.15

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #15

Recommended Solution If we had no atoms in the box (N = 0) then the probability of not finding an atom would be 1. From this, eliminate (A) and (D). When we have only 1 atom in the box (N = 1) it would be very unlikely that the single atom would occupy the specific volume of 1.0 × 10−6 cm3 so the probability of not finding it there should be close to 1 but not exactly 1. From this, eliminate (B) and (E). Correct Answer (C)

Alternate Solution If we were to break the box up into small cubes, each with a volume of 1.0 × 10−6 cm3 , we would see that the total numbers of cubes in the container is cubes 1.0 × 10−6 cm3 = = 106 cubes (2.45) container 1 cm3 Assuming that the probability of being in any given cube in the container is equally likely we would expect the probability of finding 1 atom in a single cube to be 1 P1 atom = 6 (2.46) 10 Keep in mind, however, that this is the probability of finding the atom inside a single small cube. What we want is the probability that we DON’T find the atom inside that cube. From this, we know that the probability of the atom being outside the cube will be Pout = 1 − Pin

(2.47)



Pout = 1 − 10−6



(2.48)

Extrapolating this to multiple atoms, we use the multiplication rule to get 

Pout = 1 − 10−6

n

Correct Answer (C)

36

(2.49)

2.16. PGRE8677 #16

2.16

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #16

Recommended Solution The muon is a lepton/fermion (spin-1/2) and has charge of -1 (A) electron Electrons are leptons/fermions (spin-1/2) and have a charge of -1. They differ from the muon in that their mass is about (1/200th ) that of a muon. Muons are are frequently referred to as heavy electrons because of their similar properties. (B) graviton You could probably immediately throw this option out because we aren’t even particularly sure that these exist. However, if it exists, the graviton should be massless, have no charge, have spin-2 and is a force particle (boson). (C) photon Massless, no charge (at least for our purposes), spin-1 and a force particle (boson). (D) pion Not an elementary particle, depending on which π − meson you get spin (+1, -1 or 0), depending on which π − meson you get charge (0, +e, -e). (E) proton Not an elementary particle (hadron) and has a charge of +1. Correct Answer (A)

37

2.17. PGRE8677 #17

2.17

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #17

Recommended Solution The problem tells us that we need to get from A Z X to A ZX

A−4 Z−1 Y

in two steps,

→?? X 0 →?? X 00 →A−4 Z−1 Y

(2.50)

0 − →A ¯e Z+1 X + e + v

(2.51)

A ZX

0 →A−4 Z−2 X + α

(2.52)

A ZX

0 →A Z+1 X + γ

(2.53)

in β − decay, we have A ZX

for α decay,

and for γ decay,

From these, it should be clear that we can get our solution by the combination A ZX

→ β − decay → α decay →A−4 Z−1 Y

A ZX

A−4 A−4 0 00 →A Z+1 X →Z−2 X →Z−1 Y

From this, it should be clear that the only two step process which will give us Correct Answer (A)

38

2.18. PGRE8677 #18

2.18

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #18

Recommended Solution Recall the time-independent Schr¨ odinger equation (Equation 2.54) ¯2 2 h ∇ ψ(x) + V (x)ψ(x) 2m take the second derivative of ψ(x) to get Eψ(x) = −

ψ 0 (x) = −b2 xψ(x) 00

4 2

ψ (x) = b x ψ(x)

(2.54)

(2.55) (2.56)

plug everything in to get Eψ(x) = −

i ¯2 h 4 2 h b x ψ(x) + V (x)ψ(x) 2M

(2.57)

and simplify Equation 2.57 to get ¯ 2 b4 x2 h + V (x) 2M Plug x = 0 into Equation 2.58 to use the condition, V (0) = 0, E=−

(2.58)

¯ 2 b4 (0)2 h + V (0) = 0 (2.59) 2M Which tells us that all of the energy we found previously is accounted for entirely by V (x), E=−

39

2.18. PGRE8677 #18

CHAPTER 2. PGRE8677 SOLUTIONS

¯ 2 b4 x2 h (2.60) 2M Additionally, you can also get to this conclusion by realizing that E shouldn’t have any dependence on x but that V (x) should, so the only term we have must be accounted for by V (x). V (x) =

Correct Answer (B)

Alternate Solution We can immediately eliminate (A) because it lacks the necessary dependence on x. Next, we can eliminate (C) because the wavefunction (ψ) given makes our potential a quantum harmonic oscillator which must have a dependence on x2 rather than x4 . Next, eliminate (D) because the energy, E, must have some dependence on the mass of the particle. This will get you to the point at which you can either guess between the two solutions or do enough of the time independent Schrdinger equation to see some dependence on x emerge. Correct Answer (B)

40

2.19. PGRE8677 #19

2.19

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #19

Recommended Solution Recall from the Bohr Model of the Hydrogen atom that the total energy of the atom in relation to n is Etot =

−13.6 eV n2

which agrees with (E) when A = −13.6 eV. Correct Answer (E)

41

(2.61)

2.20. PGRE8677 #20

2.20

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #20

Recommended Solution Recall Einstein’s famous equation for energy with a resting mass E = mc2

(2.62)

and his lesser known equation of relativistic momentum E = P0 c4 + m0 c2

(2.63)

In the problem, the proton will need to use the resting mass energy equation and the kaon uses the relativistic momentum equation, giving mp c2 = Pk c4 + mk c2

(2.64)

2

mp = Pk c + mk

(2.65)

We could plug in the exact values given, however we can simplify the calculations by using masses K + = 500M eV /c2 and P + = 1000M eV /c2 , giving (1000 M eV /c2 )2 = Pk c2 + (500 M eV /c2 )2 Pk = 750, 000 M eV /c

2

(2.66) (2.67)

Recall that the relativistic momentum is Pk = p

m0 v 1 − v 2 /c2

(2.68)

substituting that in to our previous equation gives p

m0 v = 750, 000 M eV /c2 1 − v 2 /c2

(2.69)

you can get the velocity by itself by moving the denominator on the LHS to the RHS, squaring both sides and then grouping terms to get 42

2.20. PGRE8677 #20

CHAPTER 2. PGRE8677 SOLUTIONS

3 2 c 4 r 3 = c 4 ≈ 0.86c

Vk2 =

(2.70)

Vk

(2.71)

Vk

Correct Answer (E)

43

(2.72)

2.21. PGRE8677 #21

2.21

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #21

Recommended Solution The equation for the space time interval is ∆s2 = ∆r2 − c2 ∆t2

(2.73)

but since the problem specifies that c = 1, Equation 2.73 becomes ∆s2 = ∆r2 − ∆t2

(2.74)

Recall the fantastically cool fact that the Pythagorean theorem works the same in any number of dimensions, so the length ∆r is ∆r2 = ∆x2 + ∆y 2 + ∆z 2 2

2

= 2 +0 +2

2

(2.75) (2.76)

= 8

(2.77)

∆s2 = 8 − 4

(2.78)

The time difference is ∆t = 2 so

∆s = 2 Correct Answer (C)

44

(2.79)

2.22. PGRE8677 #22

2.22

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #22

Recommended Solution Recall the electromagnetic field tensors    

F µν = 

0 −Ex /c −Ey /c −Ez /c Ex /c 0 −Bz By Ey /c Bz 0 −Bx Ez /c −By Bx 0

    

Which clearly demonstrates that both an electric field and/or a magnetic field may exist. This also happens to contradict (A), (C), and (D). Correct Answer (B)

45

2.23. PGRE8677 #23

2.23

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #23

Recommended Solution Most of these options can be analyzed by knowing that copper is a good conductor and silicon is generally used as a semi-conductor (A) It is generally true that the conductivity is higher in magnitude for a conductor than a semiconductor. (B) Conductivity and temperature are inversely related for a conductor. In this specific option as the temperature goes up we would expect the conductivity to go down. (C) For semi-conductors, as temperature increases, conductivity increases and vice versa. (D) It is generally true that impurities in a conductor will decrease conductivity. (E) The addition of impurities to a semi-conductor, commonly called doping, is done to increase conductivity rather than decrease conductivity. This is the only false statement of the bunch. Correct Answer (E)

46

2.24. PGRE8677 #24

2.24

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #24

Recommended Solution Recall Kirchhoff’s voltage loop law which tells us that the sum of all voltages about a complete circuit must sum to 0 X

Vk = 0

(2.80)

Recalling that V = IR, sum the voltage of the battery, generator and resistor G − V − VR = 0

(2.81)

However, since there is an internal resistance of 1 Ω, VR = (R + 1 Ω)I G − V − (R + 1 Ω)I = 0

(2.82)

120 volts − 100 volts − R(10 amps) − 10 volts = 0

(2.83)

10 volts − R(10 amps) = 0

(2.84)

R = 1Ω Correct Answer (C)

47

(2.85)

2.25. PGRE8677 #25

2.25

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #25

Recommended Solution From the lorentz force, we know that the magnetic field vector maximizes force when it is orthogonal to the electric field

F





(2.86)

~ + qv Bsin(θ) ~ = qE

(2.87)

~ + qv × B ~ = qE

Initially, the particle is at rest so the initial velocity is v = 0 and only the electric field will provide any force. Once the particle is in motion, however, the velocity is non-zero and will be moving in the direction of the electric field. Since this problem explicitly states that the electric field and magnetic field are parallel, we know that v and B are also parallel and thus, our angle is θ=0 ~ + qv Bsin(0) ~ ~ F = qE = qE

(2.88)

so we would only expect the electric field to influence the particle and for it to provide force solely in the direction of the field. Correct Answer (E)

48

2.26. PGRE8677 #26

2.26

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #26

Recommended Solution Henry Moseley discovered, from the Bohr model, that when an external electron strikes and knocks loose one of the innermost electrons, a gap is created which allows an electron from the next Bohr level (n = 2) to fall into its place. This transition is known as a K series transition. In general, all transitions from some Bohr level n ≥ 2 down to n = 1 is one of the K series transitions (Figure 2.2)

Lβ

Kγ

Kβ

Mβ

Lγ

Kδ

Mα

Lα

n=5 n=4

N - series

n=3

M - series

n=2

L - series

n=1

K - series

Kα

Figure 2.2: K, L and M series electron transitions The specific transition we are concerned with (i.e. the minimum energy transition) is that of Kα so we can use Moseley’s law E = Re (Z − β)

2

1 1 − 2 n2f ni

!

(2.89)

where Re is the Rydberg energy at Re = 13.6 eV and β corresponds to the transition type (K, L, M , N , . . .). To make things quicker, round all of your numbers off to get 49

2.26. PGRE8677 #26

CHAPTER 2. PGRE8677 SOLUTIONS

2

E = (15 eV ) (30)



1 1 − 1 4



(2.90)

E = 10, 125

(2.91)

E ≈ 10, 000

(2.92)

Correct Answer (D)

50

2.27. PGRE8677 #27

2.27

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #27

Recommended Solution The solution to this answer should stick out like a sore thumb. Without knowing any information about any of the other possible answers, you should immediately know that any particle with a charge (like an electron) will experience deflection under a uniform magnetic field regardless of whether it has spin or not. The Lorentz force exhibits precisely this fact, 

~ + q ~v × B ~ F = qE



(2.93)

where q is the particles charge. In fact, it wasn’t until we realized that neutral particles could experience deflection (I’m speaking of the Stern-Gerlach experiment)that we realized we had to incorporate particle spin into the mix. Correct Answer (D)

Alternate Solution (A) Recall that electron spin, along with the Pauli exclusion principle, determined the arrangement of electrons in energy level diagrams and, from that, influences our understanding of the structure of the periodic table. (B) Einstein and Debye models of specific heat both eventually incorporated electron contribution into their descriptions which necessitates a discussion on spin. (C) The Zeeman effect is a method of distinguishing between electrons with the same energy level by instituting a magnetic field to influence them to different energies. The “anomalous Zeeman effect” appeared in all instances when the net spin of a grouping of electrons wasn’t 0. This effect wasn’t understood until electron spin was understood, hence why it was called “anomalous”. (D) Electron deflection could be explained using only the Lorentz force and its dependence on the electrons charge. (E) The “Gross” atomic structure describes the splitting of line spectra without factoring in the effects of electron spin. “Fine” atomic structure was a correction to this model which factored in spin and relativistic effects. 51

2.27. PGRE8677 #27

CHAPTER 2. PGRE8677 SOLUTIONS

Correct Answer (D)

52

2.28. PGRE8677 #28

2.28

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #28

Recommended Solution The condition for normalizing a wavefunction is Z 2 ψ (x) = 1

(2.94)

for this specific scenario, our wavefunction is for a rigid rotator with dependence on φ in which the dumbbell rotates from φ = −π to φ = π Z π Z π 2 A2 e2imφ = 1 ψ (φ) = −π

(2.95)

−π

Recall from doing Fourier series approximations that the RHS of Equation 2.95 is equivalent to 2πA2 , giving us 2πA2 = 1 A =

1 √ 2π

Correct Answer (D)

53

(2.96) (2.97)

2.29. PGRE8677 #29

2.29

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #29

Recommended Solution Use Fleming’s left hand rule (Figure 2.3) to figure out the direction of force relative to a moving charge

Figure 2.3: Fleming’s left hand rule Using the wire in the image as the wire in the problem, you should see that we want to rotate the disembodied left hand such that the force vector (thumb) is pointing in the same direction as the wire. Doing so results in the current vector (middle finger) pointing towards the wire. Correct Answer (A)

54

2.30. PGRE8677 #30

2.30

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #30

Recommended Solution (A) This could only be true if all subshells (S, P , D) were completely full, which isn’t the case. (B) Since the valence shell is 4s1, it should be clear that this single electron doesn’t fill out the subshell. (C) We are only concerned with angular quantum number l = 2 which corresponds to s, not l = 4 which corresponds to g. Why does ETS keep asking about the wrong quantum numbers? (D) You can figure that this isn’t correct by having the periodic table memorized, in which case you would either know that Potassium has atomic number 19 or that the element with atomic number 17 is chlorine. Alternatively, because the electron configuration is described for its ground state, just add up all of the electrons (superscripts) to get a total of 19 electrons and, therefore, 19 protons. (E) Atoms with a single electron in its outer shell can be thought of as analogous to a hydrogen atom. From this (i.e. without doing any painful calculations) you can convince yourself that the charge distribution of Potassium will be similar to that of the hydrogen atom, in fact spherical. Correct Answer (E)

55

2.31. PGRE8677 #31

2.31

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #31

Recommended Solution This problem is testing your knowledge of physics history. Phillip Lennard experimented with essentially the same setup (Figure 2.4) in the early stages of the development of the photoelectric effect. Lennard fired different frequencies of light at a photocell and collected the photoelectrons emitted. attached to his photocell was a variable power supply, a voltmeter and a micro-ammeter As the light knocked electrons from the photocell, it would gain a positive charge. The second plate also gained a positive charge because it was connected in the same circuit as the photocell. The recently freed electrons would then be attracted to the now positively charged second plate and would flow to it. Finally, the electrons would move through the circuit back to the photocell and start the process over, generating a current. The variable power supply was placed in this circuit such that it would fight the flow of electrons from moving back to the photocell (which occurs because the negative end of the potential is attached to the end of the circuit not receiving light). At low potentials, low energy electrons would be caught and pushed back while higher electrons

Figure 2.4: Phillip Lennard’s experimental photocell arrangment 56

2.31. PGRE8677 #31

CHAPTER 2. PGRE8677 SOLUTIONS

could flow through, meaning the micro-ammeter would get a decreased reading. Lennard discovered that he could keep increasing the potential of the power supply until he reached a potential that would completely cease electron movement. At this point, that is when the micro-ammeter first reads 0, we’ve found the potential V on the collector. From this, eliminate all choices but (A) and (C). Lastly, you can determine that the potential must be negative because the potential of the power supply is fighting the electron movement so it must be negative valued. Correct Answer (A)

57

2.32. PGRE8677 #32

2.32

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #32

Recommended Solution (A) The photoelectric equation was derived well before quantum mechanics came around. (B) The important aspect of the photoelectric effect was the discrete quantization of energy. Determination of an electrons wavelength won’t get you there. (C) Eliminate this choice because, aside from not being relevant, it isn’t even true. (D) This is the answer we’ve been looking for. The important distinction that the photoelectric effect gave us was that Energy is discrete and that it is dependent on the frequency of a light source, not the intensity. (E) The photoelectric effect necessitates that we think of light as particles rather than waves. Correct Answer (D)

58

2.33. PGRE8677 #33

2.33

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #33

Recommended Solution I think it’s worth noting that this question does some strong hinting, whether intentional or unintentional, that the previous 2 questions can be solved without knowing much if anything about the equation they gave you. In the photoelectric equation, W represents the work function (in my experience, φ is more commonly used) and it represents the minimum amount of work (energy) necessary to pop an electron from a material. Thus, you can read the photoelectric equation from left to right as ”The total kinetic energy of an ejected electron (|eV |) is equal to the energy of the photon that hit it (hν) minus the energy it required to pop the electron out in the first place (W )”. Reading it like this should convince you that really the photoelectric equation is nothing more than a conservation of energy argument. Correct Answer (D)

59

2.34. PGRE8677 #34

2.34

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #34

Recommended Solution Recall that Work is force integrated over the area in which the force is applied and that work has the opposite sign of the potential energy (−W = U ), −W =U =−

60

Z

F · ds

(2.98)

2.34. PGRE8677 #34

CHAPTER 2. PGRE8677 SOLUTIONS

Since we are given the energy and asked to find force, differentiate your work equation to get the force equation  d d  −Kx4 (U ) = dx dx F = −4kx3

Correct Answer (B)

61

(2.99) (2.100)

2.35. PGRE8677 #35

2.35

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #35

Recommended Solution You should remember that the Hamiltonian is the sum of kinetic energy and potential energy (as opposed to the Lagrangian which is the difference between the two). From this you should get H = T +V p2 H = + kx4 2m 62

(2.101) (2.102)

2.35. PGRE8677 #35

CHAPTER 2. PGRE8677 SOLUTIONS

In case you immediately think of T = mind/convince yourself that

1 2 2 mv

for the kinetic energy, take a second to re-

p2 1 mv 2 = 2 2m Correct Answer (A)

63

(2.103)

2.36. PGRE8677 #36

2.36

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #36

Recommended Solution You should recall from any course that was heavy into mechanics, that the equations of motion for a system come out as a result of applying a minimization on the actions of the system, which is known as the principle of least action. From this, you should recall that our equations of motion are generally always time dependent, not position dependent and you can eliminate (D) and (E). As for the other three, you will just have to remember that the principle of least action is the integral with respect to time of the Lagrangian, 64

2.36. PGRE8677 #36

CHAPTER 2. PGRE8677 SOLUTIONS

Z t2 1 t1

2

mv 2 − kx4

Correct Answer (A)

65

(2.104)

2.37. PGRE8677 #37

2.37

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #37

Recommended Solution Consider that when the radius, r, goes to 0, the tension force should be just the force as a result of an object hanging from a string, i.e. FT = mg. In this limit, only (D) and (E) are remaining. Next, compare the units in both to see that in (D) they are asking you to add a term with units of m2 /s2 with another term with units of m2 /s4 and that ain’t happening. Correct Answer (E)

Alternate Solution Draw out your force diagram (Figure 2.5), It should be clear that the tension force is FT2 = (FT −x )2 + (FT −y )2 66

(2.105)

2.37. PGRE8677 #37

CHAPTER 2. PGRE8677 SOLUTIONS

l FT

FT - y

FT - x FG Figure 2.5: Component forces on a mass rotating on a cord The vertical component of tension will be equal and opposite that of the force from gravity because it won’t be accelerating in that direction, FT −y = FG = mg

(2.106)

The tension in the horizontal will just be FT −x and it will be equal to the centripetal acceleration due to its rotation, FT −x = mac = mω 2 r

(2.107)

Plug Equations 2.106 and 2.107 into Equation 2.105 to get FT2 = FT



mω 2 r

2

+ (mg)2

q

= m ω4 r2 + g2 Correct Answer (E)

67

(2.108) (2.109)

2.38. PGRE8677 #38

2.38

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #38

Recommended Solution Construct a truth table based on the conditions of the problem Input 1 0 1 0 1

Input 2 0 0 1 1

Output 0 1 1 1

which is the truth table of an OR gate. Recall that the truth table for an AND gate is Input 1 0 1 0 1

Input 2 0 0 1 1

Output 0 0 0 1

To recall these tables, simply remember that for an OR gate, it must be true that either Input 1 OR Input 2 must be true for the output to be true. For an AND gate, however, it must be true that input 1 AND input 2 are true for the output to be true. Correct Answer (A)

68

2.39. PGRE8677 #39

2.39

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #39

Recommended Solution In the area of ω = 3 × 105 second−1 , the line is roughly linear and decreasing. From this, we can eliminate any solution which shows an increase in gain with an increase in angular frequency, i.e. eliminate (B) and (C). Next, consider that we are talking about a log-log graph and (A) won’t give us a linearly decreasing function with this type of scaling. Finally, between (D) and (E), we need to look at the slope near ω = 3 × 105 second−1 . The horizontal and vertical axes are scaled with the same step size so we can approximate the slope as moving down 2 steps and to the right roughly 4 steps, giving a slope of roughly 2/4 or 1/2. From this, we see that the gain is dependent on the squared value of the angular frequency Correct Answer (E)

69

2.40. PGRE8677 #40

2.40

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #40

Recommended Solution The standard deviation for radioactive emission is described by the Poisson noise, √ σ=

λ

(2.110)

where λ is the average number of radioactive samples. To simplify things, let λ ≈ 10, 000 so σ=

p

10, 000 = 100

Correct Answer (A)

70

(2.111)

2.41. PGRE8677 #41

2.41

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #41

Recommended Solution The binding energy for an atom is the amount of energy required to strip an electron from the atom. We can immediately eliminate (D) and (E) because the binding energy of these atoms is so low that it decomposes on its own (i.e. they are both radioactive). Next, eliminate (B) because it doesn’t have a full valence shell and will generally be more willing to give up an electron than atoms with completed shells. Finally, choose (C) because Iron has a much stronger nucleus binding electrons in their shells and it is generally true that as you move down the periodic table, binding energy increases. Correct Answer (B)

71

2.42. PGRE8677 #42

2.42

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #42

Recommended Solution If you happen to recall the mean free path equation for the probability of stopping a particle moving through a medium, then this problem is a relatively straightforward plug-n-chug problem P (x) = nσdx

(2.112)

Where P (x) is the probability of stopping the particle in the distance dx, n is the number of nuclei, σ is the scattering cross section and dx is the thickness of the scatterer. Plug everything in to get P ndx
1 × 10−6 nuclei   = 1020 nuclei cm3 (0.1cm) = 1 × 10−25 cm2

σ =

(2.113) (2.114) (2.115)

Correct Answer (C)

Alternate Solution In case you don’t recall the relevant equation, examine the units of all the information given to you to see if you can derive it on the spot. You are given, Scatterer Thickness 0.1 cm Nuclei 1020 nuclei/cm3 Quantity making it a given distance

1

nuclei

1×106

72

2.42. PGRE8677 #42

CHAPTER 2. PGRE8677 SOLUTIONS

Since we ultimately want to get a result with units of cm2 , we will have to combine our 3 known values as Quantity making it a given distance nuclei  = nuclei (Nuclei)(Scatterer Thickness) cm3 (cm) Plug in your values to get (C). Correct Answer (C)

73

(2.116)

2.43. PGRE8677 #43

2.43

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #43

Recommended Solution Ideally, you should recognize that the frequency given in the problem as being that of a Simple Harmonic Oscilattor (SHO) 1 ωSHO = 2π

s

k m

(2.117)

The simple harmonic oscillator involves a mass oscillating about a fixed point. Based on this fact, we should expect the frequency for this linearly oscillating system of springs and masses to exhibit the SHO frequency when they are also moving about a fixed point. Of the potential solutions, only (B) describes masses A and C as moving about a fixed point, specifically mass B. Correct Answer (B)

Alternate Solution In case you don’t recall the frequency of a SHO, you can calculate it by first considering all of the forces on one of the masses, Fnet = m¨ x = −kx

(2.118)

x(t) = Asin(2πωt + φ)

(2.119)

for a SHO, the position equation is

74

2.43. PGRE8677 #43

CHAPTER 2. PGRE8677 SOLUTIONS

plug this into the force equation to get the differential equation d (Asin(2πωt + φ)) = −k (Asin(2πωt + φ)) dt and when you find the general solution to the diffeq, you should get m

1 ω= 2π

s

k m

Correct Answer (B)

75

(2.120)

(2.121)

2.44. PGRE8677 #44

2.44

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #44

Recommended Solution The most difficult part of this problem isn’t performing the calculation but convincing yourself of the correct solution. When the particle strikes the stick, the particle stops which tells us that all of the momentum that the particle originally had has been transferred to the stick. Because the particle doesn’t strike the stick at its center of mass, the stick will start to spin in addition to moving to the right. This may convince you that if the momentum is conserved, that a fraction of that momentum results in a linear motion for the stick and the other fraction of that momentum goes into the rotation. However, if we consider all of the momentum on the stick, we get the following diagram. Note that at any instant of time, the momentum of the ends of the stick (and every point between the end and the COM) has a horizontal and vertical component of momentum exactly opposite that of the opposite end of the stick (Figure 2.6). Since momentum must be conserved, we sum the momentum components to get

Pnet - x = PCOM + Px + (−Px ) = PCOM Pnet - y = Py + (−Py ) = 0

(2.122) (2.123)

From this, we get that the net momentum transferred from the particle to the stick is all contained within the center of mass. Finally, compare the initial and final momentum to get 76

2.44. PGRE8677 #44

CHAPTER 2. PGRE8677 SOLUTIONS

Figure 2.6: Conservation of momentum in a rotating system

Pi = Pf

(2.124)

mv = M V mv V = M

(2.125)

Correct Answer (A)

77

(2.126)

2.45. PGRE8677 #45

2.45

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #45

Recommended Solution This problem is making reference to Compton scattering. Compton scattering is the phenomena by which photons collide with a particle, impart some of their kinetic energy to the particle and then scatters off at a lower energy (Figure 2.7).

Figure 2.7: Photon wavelength change due to compton scattering The Compton equation derived for these interactions is h (1 − cos(θ)) mp c If we plug in all the values provided in the problem, we get λ0 − λ =

h π 1 − cos mp c 2 h +λ mp c 

0

λ −λ = λ0 =

(2.127)

 

which tells us that the new wavelength, λ0 has an extra Correct Answer (D)

78

h mp c

(2.128) (2.129) added to it.

2.46. PGRE8677 #46

2.46

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #46

Recommended Solution According to Stefan’s Law (Stefan-Boltzmann’s Law), power radiation of a blackbody is only dependent on temperature according to j ∗ = σT 4

(2.130)

where σ is a constant equal to 2π 5 k 4 J = 5.67 × 10−8 2 4 2 3 15c h sm K For our problem, we are doubling the temperature so our power is σ=

j ∗ = σ(2T )4 = 16σT

4

= 160 mW Correct Answer (E)

79

(2.131)

(2.132) (2.133) (2.134)

2.47. PGRE8677 #47

2.47

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #47

Recommended Solution Without knowing anything about the Frank-Hertz experiment, you can immediately eliminate options (A) and (B) from the list because it is neither true that electron collisions are always elastic, nor is it true that they are always inelastic. By a similar line of reasoning, since we’ve already argued that some electron scattering can be elastic, it wouldn’t make since to then say that electrons always lose some specific amount of energy, so we eliminate (D). Finally, when choosing between (C) and (E), you will need to know that the Nobel Prize winning Frank-Hertz experiment demonstrated that at a specific energy range (4.9 volts to be specific) electrons begin to experience inelastic collisions and that the energies lost in the collision came in discrete amounts. Correct Answer (C)

80

2.48. PGRE8677 #48

2.48

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #48

Recommended Solution From our quantum mechanical selection rules, we know that any change of state must be accompanied by a change in the quantum angular momentum number ∆l = ±1

(2.135)

This tells us that if the wave function is spherically symmetric, i.e. l = 0, then after a change to a new wave function, l 6= 0 and so they can’t both be spherically symmetric. Correct Answer (D)

81

2.49. PGRE8677 #49

2.49

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #49

Recommended Solution The classical Hamiltonian equation is

H = T +V p2 +V = 2m

(2.136) (2.137)

The quantum mechanical Hamiltonian is ¯2 2 h ∇ +V (2.138) 2m examine both equations to see that we can move from the classical Hamiltonian to the quantum mechanical one if we allow p = i¯h∇. H=−

Correct Answer (B)

82

2.50. PGRE8677 #50

2.50

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #50

Recommended Solution If you place a flat conducting bar with a current passing through it into a magnetic field with field vectors pointing perpendicularly to the direction of current, the Lorentz force will cause the stream of electrons to curve to one face of the bar while the remaining positive charges are left on the opposite face of the bar (Figure 2.8). This results in a potential difference in the conductor known as the ”Hall Voltage” which can be calculated using IB (2.139) dne where I is the current, B is the magnetic flux density, d is the plate depth, e is the electron charge and n is the charge carrier density. One of the benefits the Hall effect has to experimental physics relates to the direction in which a current will curve depending on the direction of the magnetic field, current and charge. Effectively, the direction in which the electrons curve determines the sign of the Hall Voltage and provided the direction of the current and magnetic field are known, it is trivial to determine the sign of the charge carriers. VH = −

Correct Answer (C)

83

2.51. PGRE8677 #51

CHAPTER 2. PGRE8677 SOLUTIONS

Figure 2.8: Moving charges tend to one side of a conducting bar from the Hall effect

2.51

PGRE8677 #51

Recommended Solution The Debye and Einstein models of specific heat were essentially identical except for the way in which they were scaled. Debye scaled his model around a value he called the Debye temperature (TD ) that was dependent on a number of properties of the material. Einstein, on the other hand, scaled his model based on a single frequency value (hν) and constant k. Although they were identical in terms of their predictions that vibrational energies were dependent on 3N independent harmonic oscillators, Debye’s scaling proved to be more accurate in the low temperature range and so it prevailed. Correct Answer (B)

84

2.52. PGRE8677 #52

2.52

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #52

Recommended Solution From what you learned about electrostatics, you should recall that in a region where there are no charges and with a uniform potential on the surface, electric potential is described by Laplace’s equation ∇2 V = 0

(2.140)

It is also true that within the boundaries of any region which is satisfied by Laplace’s equations, there can be no local minima or maxima for V (courtesy of the Divergence Theorem). From this condition, we know that whatever value of potential we have on the surface, the value inside must be exactly the same, else we’ve found a maximum or minimum. Therefore, the potential at the center of the cube must be V , just as it is on the surface of the cube. Correct Answer (E)

85

2.53. PGRE8677 #53

2.53

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #53

Recommended Solution When a charged particle oscillates in 1-dimension it will radiate electromagnetic radiation in a direction perpendicular to to the motion of the particle. For an imperfect but useful analogy, think along the lines of the Lorentz Force and the orthogonality of its electric and magnetic fields. Expanding on this Lorentz Force analogy a bit more, recall that the orthogonality condition comes as a sine function from a cross product F





(2.141)

= qE + vB sin(θ)

(2.142)

~ + ~v × B ~ = q E

The dipole oscillation, much like the Lorentz Force, has this same sine dependence. Knowing only these two facts we can conclude that the point P must be in the xy-plane, because the problem has already specified that it lies in this plane, and that the amplitude is maximized when θ = 90◦ Correct Answer (C)

86

2.54. PGRE8677 #54

2.54

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #54

Recommended Solution This problem is extremely quick and easy if we can simply recall the definition of the dielectric constant, K. The dielectric constant is a scaling factor which describes the concentration of electrostatic flux in any given material. It is calculated by taking the ratio between the amount of electrical energy stored by the material when some voltage is applied to it (absolute permittivity of the material, s ) relative to the permittivity of a vacuum, 0 , s (2.143) 0 from the previous equation, it should be clear that K = 1 when the absolute permittivity of the material is equal to that of a vacuum. Since the polarization charge density, σp , is K=

σp = P · n ˆ ~ = χe E n ˆ

(2.144) (2.145)

~ is the electric field. where P is the electric polarization, χe is the electric susceptibility and E Since the electric susceptibility is χe =

P s = −1 ~ 0 0 E

(2.146)

We see that when s = 0 (i.e. when K = 1), the bound charge density should equal 0. This limit is only satisfied by (E) Correct Answer (E)

87

2.55. PGRE8677 #55

2.55

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #55

Recommended Solution You should recall the thermal energy referenced in the problem from Boltzmann’s constant, which effectively tied together the microscopic affects of fermions to the macroscopic view of thermal energy, P V = N kT

(2.147)

When dealing with individual fermions, in our case electrons, we are primarily concerned with the Fermi Energy which describes the highest energy level occupied by an electron. The highest occupied energy level can be found using energy level diagrams and continuously appending electrons to the lowest unoccupied levels. At the point that we’ve run out of electrons to assign to energy levels, the highest level electron determines the Fermi Energy and, additionally, the Fermi Velocity. If the Pauli exclusion principle weren’t true, then electrons would have no reason to build into higher energy levels (in fact they would likely all bunch into the ground state energy level) and so, without the Pauli exclusion principle, the mean kinetic energy of electrons wouldn’t be dependent on the thermal energy of the system. Correct Answer (B)

88

2.56. PGRE8677 #56

2.56

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #56

Recommended Solution From quantum mechanics, we know that the expectation value or the mean value of an operator is found by hψ ∗ |Q|ψi =

Z ∞

ψ ∗ Qψ dx

(2.148)

−∞

where Q is some operator. In this problem they tell us that Q is the operator corresponding to observable x and traditionally, the operator Q is used such that (Qψ) (x) = (xψ) (x)

(2.149)

so finding the expectation value of Q is effectively the same thing as finding the mean value of x and so the correct answer is (C). Be aware, however, that an operator which “corresponds to a physical observable x” doesn’t necessarily imply that you are finding the mean value of x, even though it does mean you are finding the mean value of the observable itself. Correct Answer (C)

89

2.57. PGRE8677 #57

2.57

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #57

Recommended Solution The eigenfunction of a linear operator is any non-zero function, f , that satisfies the condition Af = λf

(2.150)

where λ is the eigenvalue. From this, we will want to plug in all of the given functions, f , into ∂f (2.151) ∂x without doing any work, it should be obvious that both (A) and (B) won’t work because only the RHS of Equation 2.151 contains a derivative and so the trig functions will be different on each side. However, for the purposes of rigor, we can compute each of the 5 possible solutions ¯hkf = −i¯h

(A) ¯hk cos(kx) 6= i¯ hk sin(kx) (B) ¯hk sin(kx) 6= −i¯ hk cos(kx) (C) ¯hke−ikx = i2 ¯ hke−ikx 6= −¯ hkeikx (D) ¯hkeikx = −i2 ¯ hkeikx = h ¯ keikx (E) ¯hke−kx 6= i¯ hke−kx Correct Answer (D)

90

2.58. PGRE8677 #58

2.58

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #58

Recommended Solution Looking through the three options provided, we see that III, Wave-front angular frequency, i.e. the color of the object, is one of the properties which the problem claims a hologram might “record”. If you’ve ever seen a real hologram (i.e. not one from star wars) then you know that color is something that is not conserved (Figure 2.9). In fact, the question gives this away when it tells you that the resulting hologram is monochromatic.

Figure 2.9: Comparison of realistic holograms to cinematic holograms From this, we can eliminate any solutions which claim to record Wave-front angular frequency, i.e. (C), (D) and (E). When choosing between (A) and (B), we simply need to determine whether any phase information is saved by the hologram. Since a hologram is generated as the result of interference between two light beams and wave interference is dependent on wave phase, choose (B). Correct Answer (B)

91

2.59. PGRE8677 #59

2.59

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #59

Recommended Solution The group velocity, vg , is related to the angular frequency, ω, by ∂ω ∂k ω is given in the problem so we take the derivative with respect to k to get vg =

∂ω ∂k

= =

∂ p 2 2 c k + m2 ∂k c2 k √ c2 k 2 + m2

(2.152)

(2.153) (2.154)

and plugging in the limit k → 0, you should get that vg = 0 which eliminates all possible solutions but (D) and (E). Next, letting k → ∞, both terms with c and k will blow up and m will effectively be 0, giving us c2 k √ c2 k 2

c2 k ck = c =

Correct Answer (E)

92

(2.155) (2.156)

2.60. PGRE8677 #60

2.60

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #60

Recommended Solution For a simple harmonic oscillator the frequency is typically given as 1 f= 2π

s

k m

(2.157)

this gives you a mass dependence and allows you to eliminate (A) and (B). Additionally, it also demonstrates that there is no initial velocity dependence and we can eliminate (E). Finally, when choosing between (C) and (D), ask yourself where we get the potential energy equation in the first place. Recall that a mass which exhibits simple harmonic motion due to a spring has a spring force of Fs = −kx

(2.158)

Additionally, recall that Z

V =

Fs dx

(2.159)

which means we can solve for the potential energy of the SHO by integrating the spring force equation, Z

V

=

Fs dx

(2.160)

= kx2 + C

(2.161)

where C is just some constant. This expression is identical to the potential energy described in the problem and so we conclude, by Hooke’s law, that the force and, by extension, the potential energy of the system is dependent on the constant of proportionality for position, b, but not dependent on the arbitrary constant, a. Correct Answer (C) 93

2.60. PGRE8677 #60

CHAPTER 2. PGRE8677 SOLUTIONS

Alternate Solution Recall that the potential energy of a system is related to the force on that system by dV = −F (x) dx

(2.162)

so our potential energy, V , becomes  d  a + bx2 dx = −2bx

Fs = −

(2.163) (2.164)

By Newton’s second law,

Fs = m¨ x

(2.165)

−2bx = m¨ x

(2.166)

2b = mω 2

(2.167)

but since x ¨ = ω 2 x, we get

s

ω =

2b m

(2.168)

finally, since the angular frequency, ω, is related to frequency, f , by ω = 2πf 1 f= 2π

s

2b m

Correct Answer (C)

94

(2.169)

2.61. PGRE8677 #61

2.61

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #61

Recommended Solution The problem specifies that the rocket has velocity v and that the motion of the rocket is determined by the equation dv dm +u =0 (2.170) dt dt we will consider the rocket moving to the right and let this direction of motion be positive. In order for the net momentum of the system to be 0, there must be another velocity, in our case u, moving in a direction opposite to that of the rocket. For options (A), (B) and (C), each of the velocities mentioned should be positive as they are all instances of the rocket in its motion in the positive direction (to the right). Now, when choosing between (D) and (E), realize that our original equation of motion describes a stationary frame viewing a moving frame. From this, we know that u couldn’t be the exhaust speed in a stationary frame, else we wouldn’t be taking into account the varying reference frames. m

Correct Answer (E)

95

2.62. PGRE8677 #62

2.62

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #62

Recommended Solution The question gives you the initial conditions that when m = m0 we get v = 0. Substitute m = m0 into each to see which, if any, goes to 0. (A) u

m m




= u 6= 0

m

(B) ue m = ue 6= 0 (C) u sin (D) u tan

m m




m m




= u sin(1) 6= 0 = u tan(1) 6= 0

(E) (A), (B), (C) and (D) were all wrong, therefore we choose (E) Correct Answer (E)

Alternate Solution The quick method to solve this problem involves using some very irresponsible math that we physicists (and to a larger extent, chemists) like to use to simplify differentials. multiply the dt out of our equation of motion to get

96

2.62. PGRE8677 #62

CHAPTER 2. PGRE8677 SOLUTIONS

mdv + udm = 0

(2.171)

mdv = −udm

(2.172)

get dv on its own and integrate both sides Z

dv = −u

Z

dm m

(2.173)

and we will just generalize the results as ∆v = −u [ln(m)]m m0   m vf − v0 = −u ln m0

(2.174) (2.175)

since the question tells us that v0 = 0, we make the substitutions to get, 

vf = −u ln

m m0



which doesn’t match any of the options given as possible solutions. Correct Answer (E)

97

(2.176)

2.63. PGRE8677 #63

2.63

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #63

Recommended Solution Anytime you see different arrangements of variables in which all or most of the options will have different units than one another, check the units. Recall that we are looking for units of charger per area (A)

q 4πD

C ≡m

(B)

qD2 2π

≡ C · m2

(C)

qd 2πD2

C ≡m

(D)

qd 2πD3

C ≡m 2

(E)

qd 4π0 D2

C ≡m Correct Answer (D)

98

2.64. PGRE8677 #64

2.64

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #64

Recommended Solution In electronics, when we want to maximize power transfer we need to do some impedance matching in which the load impedance and complex source impedance is ZS = ZL∗

(2.177)

ZS = ZL∗

(2.178)

Rg + jXg = Rg + jXl

(2.179)

plugging in given impedance values,

In order for the RHS to be equal to the LHS, Xl = −Xg Correct Answer (C)

99

(2.180)

2.65. PGRE8677 #65

2.65

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #65

Recommended Solution A magnetic dipole occurs in any instance that you have a closed circulation of electrical current. In the case of our problem, we have an enclosed loop of wire with a current i. Any given dipole can be described by its dipole moment and in the case of a magnetic dipole, we get the magnetic dipole moment Z

µ=

i dA

(2.181)

where dA is a differential piece of the area about which the current circulates. In our case, our area is simply that of a circle with radius b and so we get a magnetic moment of µ = iπb2 2

∝ ib

Correct Answer (B)

100

(2.182) (2.183)

2.66. PGRE8677 #66

2.66

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #66

Recommended Solution Without knowing much about thermodynamics, we can figure that our final solution must contain some units of temperature, specifically we expect to get something like inverse Kelvins. Looking at the units of the 4 different variables, we get P = atm V = m3 S = J/K U =J Note that only entropy, S, has any temperature units and the only way to isolate the temperature unit is to do 

∂S ∂U



(2.184) V

Correct Answer (E)

Alternate Solution Recall, from thermodynamics, the equation dU = T dS − P dV

(2.185)

you should never forget this equation because chemistry, more than any other subject, is T dS (T dS = tedious..... GET IT!). Temperature is only well-defined for a system in equilibrium, so we let dV = 0 and we get, 101

2.66. PGRE8677 #66

CHAPTER 2. PGRE8677 SOLUTIONS

dU 1 T

= T dS   dS = dU V

Correct Answer (E)

102

(2.186) (2.187)

2.67. PGRE8677 #67

2.67

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #67

Recommended Solution Recall from thermodynamics that with excessive amounts of energy at the disposal of a system, particles will tend to equally populate all energy levels. Since the question tells us that temperature (i.e. thermal energy) is significantly larger than the energy levels, we know that every energy level must have an equal likelihood of being populated. From this, calculate the average of the 3 energy levels as 0 +  + 3 4 =  3 3 Correct Answer (C)

103

(2.188)

2.68. PGRE8677 #68

2.68

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #68

Recommended Solution This problem and the solution should be one of those ideas you know by heart. If you don’t, stop what ever you are doing and commit this to your memory. As an object with mass moves faster and approaches extremely high velocities, the mass of the object begins to increase. As you approach the speed of light, the mass approaches infinity. The only way a particle can achieve a velocity at the speed of light is if the particle has a rest mass of 0. In case the rule of thumb isn’t enough to convince you, recall E=

q

p2 c2 + m20 c4

(2.189)

let m → 0, giving

E =

q

p2 c2

(2.190)

= pc

(2.191)

pc = hν hν p = c h = λ

(2.192)

then, by the Planck relationship,

which we know, experimentally, is the momentum of the photon. Correct Answer (A)

104

(2.193) (2.194)

2.69. PGRE8677 #69

2.69

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #69

Recommended Solution In this problem, you should immediately eliminate (D) because we know the car can’t move faster than the speed of light. Next, eliminate (E) because there is sufficient information in this problem to answer the question explicitly. Lastly, look for the speed which justifies our use of relativistic equations to account for the contraction of a 2 meters, which would be 0.8c.

105

2.69. PGRE8677 #69

CHAPTER 2. PGRE8677 SOLUTIONS

Correct Answer (C)

Alternate Solution Length contraction from relativistic effects is related to the Lorentz factor by, s

L0 =

v2 L =L 1− 2 γ c

(2.195)

where L0 is the length of a contracted object in the rest frame and L is the rest length. Plug in the given values to solve s

3 m = 5 m 1− 9 25 16 9

= 1− =

v =

v2 c2

v2 c2

v2 c2 4 c 5

Correct Answer (C)

106

(2.196) (2.197) (2.198) (2.199)

2.70. PGRE8677 #70

2.70

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PGRE8677 #70

Recommended Solution Now, we are talking about the moving car being the rest frame and the garage being the contracted frame. We use the same contraction equation used in the previous problem, however we let L0 be the contracted garage reference frame and let L be the rest frame for the garage. In the previous equation, we calculated the speed of the car as v = 0.8c, so we plug everything into our contraction equation to get

107

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s

v2 L0 = 4 m 1 − 2 c √ = 4 m 0.36 = 2.4 m Correct Answer (A)

108

(2.200) (2.201) (2.202)

2.71. PGRE8677 #71

2.71

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PGRE8677 #71

Recommended Solution I’m generally not a big fan of these multi-part GRE questions, however I really like this one because it illustrates one of the classic paradoxes, or at least apparent paradoxes, in relativity. In the vehicles reference frame, the garage is too short to contain the vehicle. However, in the garage reference frame the car should have no problem fitting in the garage. The primary qualitative lesson we learned from Einstein is that we can’t prefer one reference frame to another. Based on this, we aren’t justified in saying that one solution is more “true” than the other. The only solution which 109

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alludes to this inherent duality of solutions is (E), which describes another classical qualitative result from Einstein. Specifically, I’m referring to our inability to achieve truly instantaneous events between different reference frames. Correct Answer (E)

110

2.72. PGRE8677 #72

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PGRE8677 #72

Recommended Solution This problem confuses me because it has 1 correct choice and 4 absurd choices. This means you don’t need to know why the correct answer is correct, only that the other 4 are ridiculous. (A) What? Since when? No it doesn’t! (B) What? Of course they can transmit signals! How else would we get any useful information out of x-rays in, for example, medical applications. (C) WHAT? Imaginary mass? When? Where? NO! (D) And? Why would the order in which physical theories were discovered determine the accuracy of Relativity? (E) The phase velocity and group velocities indeed are different. This is important because the refractive index of light through a medium is the ratio of the phase speed of light through the medium to that of light through a vacuum. It is, therefore, significant that the phase speed never be faster than the speed of light, which is frequently true of the group velocity. If they weren’t different, we might get n > 1. However, you don’t really need to know any of that to solve this problem, do you! Correct Answer (E)

111

2.73. PGRE8677 #73

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PGRE8677 #73

Recommended Solution If you’ve ever seen the rainbow patterns that form on water bubbles and oil slicks, then you’ve seen thin film interference. This phenomena occurs as a result of light waves reflecting off of the top surface of a medium with a higher index of refraction than that from which the light came from and, additionally, light waves reflecting off of the bottom of that surface as another transition to a higher index of refraction occurs (Figure 2.10)

Figure 2.10: Thin film interference to create a non-reflecting layer The same phenomena occurs whether the light is incident on the top surface at an angle or orthogonal to it, however I’ve intentionally drawn it at an angle to make it easier to observe the phase shift between the two reflected rays. For every reflection of a light wave off a medium of higher index of refraction, we get a shift of half the wavelength of the light,

∆air = ∆coating = 112

λ 2 

(2.203) λ + 2t 2



(2.204)

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In the previous equations, note that ∆coating has an additional 2t of phase shift because it must also pass both ways through the thickness of the coating. In order for the coating to eliminate all reflections, we need the outgoing waves to experience destructive interference, so we find the relative shift of the system 



∆coating − ∆air     λ λ − = 2t + 2 2 = 2t

∆ =




(2.205) (2.206) (2.207)

since we only get destructive interference with waves of differing phase shift at half-odd integer values , set the relative shift to 1 m+ λ 2 λm λ + 2 4



2t = t =



(2.208) (2.209)

where m = 0, 1, 2, 3, . . . . Since we are only concerned with the thinnest possible coating which will give us deconstructive intereference, we choose m = 0 and t = λ/4. Correct Answer (A)

113

2.74. PGRE8677 #74

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PGRE8677 #74

Recommended Solution Consider a beam of unpolarized light headed towards you and you have three ideal polarizers in hand

Figure 2.11: Different orientations of a polarizing filter In Figure 2.11, grey vectors indicate individual directions of oscillation, black vectors indicate the net polarization, greyed out areas indicate areas of absorbed polarization and white areas indicate unaffected polarization. If we wanted to completely eliminate all light using two of the polarizers, we could place them in series with a rotation of exactly π/2 between them, for example using the horizontal and vertical polarization. In this arrangment, anything that survived through the horizontal polarizer would be caught by the vertical polarizer and no light would be transmitted on the other side. However, imagine we were to place another polarizer between the horizontal and vertical polarizers such that this third polarizer is rotated π/4 or 45 with respect to the other two. As the light first passes through the horizontal polarizer, only half of the photons that hit the polarizer will pass through. These photons will continue to the 45 polarizer where, again, half of the remaining photons get absorbed and the other half pass through. After the photons pass through the 45 polarizer, the remaining photons will spread out from -22.5 to +67.5. This occurs because linearly polarized light will always completely fill a full 90 of angular spread, which I didn’t mention previously because the previous instances came out with a 90 spread. Finally, the remaining photons will pass through the vertical polarizer giving a final total of 1/8 the original number of photons (Figure 2.12) Correct Answer (B)

114

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Figure 2.12: Remaining light after passing through the three polarizer system

2.75

PGRE8677 #75

Recommended Solution Recall Kepler’s third law of motion ”The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit” so we have T 2 = βr3

(2.210)

The period is given as T = 80 minutes, so (80 min)2 = βRe3 6400 min2 β = Re3

(2.211) (2.212)

using the same proportionality constant, β, but for 24 hours (i.e. 1440 minutes), we get

115

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T 2 = βr3 6400 min Re3

(1440 min)2 = r3 =

(2.213) 2

!

1440 min2 Re3 6400 min2

r3

(2.214) (2.215)

For a quick and easy approximation, pretend that 1440 = 1500 and 6400 = 6000, so then we get (1440 · 1440)Re3 6400 1440 3 Re = 4 = 360

r3 =

(2.216) (2.217) (2.218)

Finally, take the cube root of both sides to get

r =

√ 3 360

≈ 7 Correct Answer (B)

116

(2.219) (2.220)

2.76. PGRE8677 #76

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PGRE8677 #76

Recommended Solution Recall the angular momentum equation L = Iω

(2.221)

in which the moment of inertia, I, of a ring is Iring = M R2

(2.222)

L = M R2 ω

(2.223)

which gives us

To solve for ω sum the total energy of the rolling ring

Utot = Uroll + Utrans     1 2 1 M gh = Iω + M v2 2 2 but since ω = v/R, we get

117

(2.224) (2.225)

2.76. PGRE8677 #76

CHAPTER 2. PGRE8677 SOLUTIONS

1 M gh = M R2 ω 2 + 2 gh = R2 ω 2 

s

ω =





1 M R2 ω 2 2



gh R2

(2.226) (2.227) (2.228)

Then combining ω into the angular momentum equation, we get the solution L = M R2 ω s

= M R2

(2.229) gh R2

p

= M R gh Correct Answer (B)

118

(2.230) (2.231)

2.77. PGRE8677 #77

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PGRE8677 #77

Recommended Solution Any object with a net force of F = −kx is a simple harmonic oscillator (via Hooke’s Law). The position equation for a SHO is x = A cos(ωt)

(2.232)

take the derivative of position to get the velocity equation dx = −A sin(ωt)ω (2.233) dt If we square both equations and get their trig functions by themselves, we can apply the identity cos2 (θ) + sin2 (θ) = 1, x2 A2 v2 A2 ω 2

= cos2 (ωt)

(2.234)

= sin2 (ωt)

(2.235)

using our trig identity and letting our position be x = A/2, cos2 (ωt) + sin2 (ωt) = 1 x2 v2 + = 1 A2 A2 ω 2 1 v2 + 2 2 = 1 4 A ω

(2.236) (2.237) (2.238)

Now solve for v v2 A2 ω 2

=

v2 =

119

3 4 3 2 2 A ω 4

(2.239) (2.240)

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Finally, since ω = 2πf , r

3 2 2 2 A 4π f √4 3πf A =

v =

Correct Answer (B)

120

(2.241) (2.242)

2.78. PGRE8677 #78

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PGRE8677 #78

Recommended Solution Total energy of the system must be the sum of the kinetic and potential energies Unet = T + V

(2.243)

We are told that the initial potential is 0 and one of the particles has an initial velocity and, therefore, an initial kinetic energy 1 U0 = mv02 (2.244) 2 Since the problem tells us to neglect radiation, we assume energy is conserved so the final energy needs to equal the initial energy 1 Uf = U0 = T + V = mv02 (2.245) 2 which tells us that the total energy of the system is constant over time and also that the energy is positive (assuming neither particle has negative mass). Correct Answer (C)

121

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PGRE8677 #79

Recommended Solution ~ = 0 tells us that divergence of the magnetic field must always Recall that Maxwell’s eqution ∇ · B be 0 and that divergence is defined as the amount of outward flux of a vector field. With this definition, we know that we need to look for any vector field that has some non-zero flux leaving the area. (A) 5 flux lines come in and 5 flux lines go out. Net flux leaving the area is 0. (B) 5 flux lines come in and 5 flux lines go out. Net flux leaving the area is 0. (C) 5 flux lines come in and 5 flux lines go out. Net flux leaving the area is 0. (D) 0 flux lines come in and 8 flux lines go out. Net flux leaving the area is NOT 0. (E) 0 flux lines come in and 0 flux lines go out. Net flux leaving the area is 0. Correct Answer (D)

122

2.80. PGRE8677 #80

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PGRE8677 #80

Recommended Solution From Gauss’s law, we get ~ = ρ ∇·E 0

(2.246)

with no charges in the region, ρ = 0 and ~ =0 ∇·E

(2.247)

take the derivative of each term with respect to its appropriate variable (i.e. x → ˆi, y → ˆj and ˆ z → k) (A) A(2y − x) 6= 0 (B) A(−x + x) = 0 (C) A(z + 0) 6= 0 (D) Ayz + Axz 6= 0 (E) Ayz Correct Answer (B)

123

2.81. PGRE8677 #81

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PGRE8677 #81

Recommended Solution Recall from Faraday’s law of induction, ε=−

dφB~ dt

(2.248)

~ ∝ I, we know we will eventually have since the magnetic field is proportional to the current, B to take the derivative of I and this means we should end up with sin(ωt) rather than cos(ωt) so we can eliminate (A) and (D). Next, since we know we want to end up with a units of volts for our EMF, check the units of (B), (C) and (E). µ0 is given in the front of the test booklet and as long as you recall that weber = volt · sec, (B)



volt·sec Amp·m



(Amp)



m2 m



1 sec



= volt

(C)



volt·sec Amp·m



(Amp)



m m2



1 sec



=

(E)



volt·sec Amp·m



(Amp)

m m

volt m2


 1  volt sec = m

124

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Correct Answer (B)

Alternate Solution From Faraday’s law of induction, ε=−

dφB~ dt

(2.249)

where the magnetic flux, φB~ , is Z

φB~ =

~ · dA = Bπa ~ 2 B

(2.250)

From the Biot-Savart law, we get the magnetic field from a current passing through a loop of wire as µ0 Idl × rˆ (2.251) 4πR2 Z µ0 Idl sin(θ) (2.252) = 4πR2 since all of the potential solutions have no θ dependence like they should, we’ll assume that the approximate solution the problem refers to is simply finding the resulting EMF at the optimized angle, θ = π/2, with radius b, Z

~ = B

µ0 Idl sin(π/2) 4πb2 Z µ0 I dl = 4πb2 µ0 I = (2πb) 4πb2 µ0 I = 2b Plug this all back into the magnetic flux to get ~ = B

Z

µ0 Iπa2 2b µ0 π a2 = (I0 cos(ωt)) 2 b with respect to time and get

φB~ =

Finally, take the derivative of φB~ V

= =

µ0 π a2 d I0 (cos(ωt)) 2 b dt µ0 π a2 I0 ω sin(ωt) 2 b

Correct Answer (B)

125

(2.253) (2.254) (2.255) (2.256)

(2.257) (2.258)

(2.259) (2.260)

2.82. PGRE8677 #82

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PGRE8677 #82

Recommended Solution This problem is describing a phenomena called the Zeeman effect, in which an atom with electrons in a degenerate state (i.e. electrons with equivalent energies but differing electron configurations) is passed through a magnetic field. Since the affect of a magnetic field is related to the electron configuration, the degenerate electrons will have their energies altered by the magnetic field in different ways and the single emission line which was initially composed of multiple degenerate electrons becomes visible as different emission lines (Figure 2.13),

Figure 2.13: The Zeeman effect for degenerate energy states This tells us that the possible solutions is either (D) or (E) but since there is no specific restriction which says the emission lines must always be doubled, we choose (E). Correct Answer (E) 126

2.83. PGRE8677 #83

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PGRE8677 #83

Recommended Solution The phrasing of this problem tells us everything we need to know in order to figure out the difference between a low-density and high-density state of the same gas. For a high density gas the mean life of an atomic state is significantly longer than the amount of time it takes for atomic collisions to occur, so by the time the atomic life is up, multiple collisions have occurred and energy has been swapped around a bunch of times. In a less dense state, not as many of these energy exchanges have occurred so there are fewer distinct energies and the ones that are there are more well-defined than the high-density gas. This is all you need to know to tell you that the high-density gas will have a broader spectral line than the low-density gas. If you want something a bit more rigorous, recall Heisenberg’s Energy-Time uncertainty principle ¯ h (2.261) 2 For the higher-density gas, as compared to a low-density gas, the decrease in time uncertainty requires an increase in energy uncertainty so we get a broader (i.e. less precise) spectral line. Alternatively, the increase in time uncertainty for the less-dense gas will give require that the energy uncertainty be less than the high-density gas and so we expect a more well-defined (i.e. thinner) spectral line. ∆E ∆t ≥

Correct Answer (C)

127

2.84. PGRE8677 #84

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PGRE8677 #84

Recommended Solution Sodium has 11 electrons, just as the problem so kindly gives us, so we construct our energy level diagram (Figure 2.14)

Figure 2.14: Energy level diagram of Sodium in its ground state With our one lone electron, we expect to get a spin of S = 1/2. This tells us that our term symbol component of 2S + 1 will become 2(1/2) + 1 = 2, so our term symbol will be of the form 2

LJ

(2.262)

which allows us to eliminate all but (B) and (D). Since the final electron is in the 3s shell, L = 0 and since J = L + S, J = 1/2. From this, our term symbol should be 2

S1 2

Correct Answer (B)

128

(2.263)

2.85. PGRE8677 #85

2.85

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #85

Recommended Solution As a general rule, Photoelectric effect: Low Energy (eV Range) Compton Scattering: Mid Energy (KeV Range) Pair Production: High Energy (> 1 MeV) Based on this general rule, we know that line #2 should correspond to pair production because it only exists at higher energy ranges, which eliminates (A), (D) and (E). Between (B) and (C), it is more likely that line #1 corresponds to the Photoelectric effect because the interaction peaks at a lower energy range. Correct Answer (B)

129

2.86. PGRE8677 #86

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PGRE8677 #86

Recommended Solution In Faraday’s Ice Pail experiment, Michael Faraday took an insulated pail which he connected to an electrometer. A brass ball with a charge on its surface was lowered down into the pail. Because the brass ball surface has a net negative charge, the pails positive charges were attracted to the inner walls of the pail and negative charges were pushed to the outside of the pail and also pushed down the wire to the electrometer. As the charged brass ball was lowered enough to touch the inner floor of the pail, the charge was completely discharged to the pail and the ball could be removed with the electrometer reading a complete transfer in charge. From this experiment, Faraday was able to determine that the inverse square law was accurate to at least one part in a billion. Correct Answer (E)

130

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PGRE8677 #87

Recommended Solution Recall that the specific heat for any atom/molecule is proportional to the degrees of freedom of the molecule. In other words, the more degrees of freedom a molecule has the higher the specific heat. (A) a specific heat of 3/2 N k corresponds to a monotomic atom with only translational degrees of freedom, so we know that these diatomic molecules must be larger than that. (B) Model I has translational and rotational degrees of freedom while Model II has tranlational, rotational and vibrational degrees of freedom. We would expect the specific heat to be larger in Model II than Model I, not the other way around. (C) At high energies (i.e. high temperatures) vibrational effects must be considered, so we can’t always treat a molecule as a rigid rotor. (D) At low energies (i.e. low temperatures) vibrational effects are minimal, so we don’t always have to treat a molecule as a springy dumbbell. (E) At low energies (i.e. low temperatures) there is minimal vibrational effect so we can treat the molecule as if it is a rigid rotor. However, at higher energies (i.e. higher temperatures) vibrational effects are involved so we want to treat it as a springy dumbbell. In other words, the model we use on a molecule is dependent on its temperature. Correct Answer (E)

131

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PGRE8677 #88

Recommended Solution Recall that fermions must obey the Pauli Exclusion principle which restricts the particles from occupying the same energy levels. Bosons, on the other hand have no such restriction. This tells us that in its lowest energy state, N bosons will all be able to occupy the minimal energy state while N fermions will be forced to occupy higher energy levels after the ground level has been filled. This tells us that the pressure, which is proportional to temperature (energy), should be greater for the fermions than the bosons, which eliminates all but (B) and (C). Since the boson pressure is the low pressure extreme and fermion pressure is the high pressure extreme, it is only logical to conclude that the classical result (think P V = nRT ) will be somewhere in between. Correct Answer (B)

132

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PGRE8677 #89

Recommended Solution For two identical particles, the wave function for those particles can only be either symmetric or anti-symmetric under interchange of x1 and x2 so our two options are ψ(x1 , x2 , t) = ψ(x2 , x1 , t)

(2.264)

ψ(x1 , x2 , t) = −ψ(x2 , x1 , t)

(2.265)

or

As it turn out, whether or not the wave function is symmetric or anti-symmetric is determined by the type of particle. Specifically, symmetric wave functions obey Bose-Einstein statistics and correspond to bosons. Anti-symmetric wave functions obey Fermi-Dirac statistics and correspond to fermions. From this, our two potential wave functions are

ψboson = ψf ermion =

1 √ [ψα (x1 )ψβ (x2 ) + ψβ (x1 )ψα (x2 )] 2 1 √ [ψα (x1 )ψβ (x2 ) − ψβ (x1 )ψα (x2 )] 2

(2.266) (2.267)

Clearly, the wave function referred to in the problem corresponds to a boson and not a fermion. Electrons, positrons, protons and neutrons are all fermions while the deuteron is a boson, so we choose (E). Correct Answer (E)

133

2.90. PGRE8677 #90

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PGRE8677 #90

Recommended Solution Without solving for anything explicitly, you can be sure that the ground state energy of any system will never be 0 eV and that the ground state energy, which is the lowest energy state, will be less than the total energy of the particle at 2 eV, so eliminate (A) and (E). Next, recall that the energy of the infinite square well is proportional to the nodes of its wave function by n2 ¯h2 π 2 2M L2 where the zero point energy, or equivalently the minimum energy value E1 , is En =

¯ 2π2 h 2M L2 so the zero point energy is related to the rest of the energy levels by E1 =

En = n2 E1 The problem tells us that the energy, En = 2 eV so with n = 2, we get

134

(2.268)

(2.269)

(2.270)

2.90. PGRE8677 #90

CHAPTER 2. PGRE8677 SOLUTIONS

E1 = E1 = =

En n2 2 eV 22 1 eV 2

Correct Answer (C)

135

(2.271) (2.272) (2.273)

2.91. PGRE8677 #91

2.91

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #91

Recommended Solution Without doing much work, you can eliminate (A) and (B) because they are much too slow to be the velocity of an electron. We can also eliminate (E) because the electrons shouldn’t be moving faster than light. As a general rule, expect electron velocities around 105 m/s which, if you are feeling lucky and a bit adventurous, then you could risk it and guess (D). To be more rigorous, you will have to recall the Bragg condition from Bragg diffraction optics 2d sin(θ) = nλ

(2.274)

and recall the de Broglie relation λ=

h p

(2.275)

combining the two gives you h (2.276) mv Since we are told that first-order reflection occurs, n = 1. Using approximate values for the remaining constants, we can solve for v, 2d sin(θ) = n

v =

h md

(2.277)

6 × 10−34 m2 kg/s (10 × 10−31 kg)(3 × 10−10 m) ≈ 5 × 107 m/s

≈

which is closest to (D). Correct Answer (D)

136

(2.278) (2.279)

2.92. PGRE8677 #92

2.92

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #92

Recommended Solution Keep in mind that we are looking for a selection rule which is NOT compatible with electric dipole emission. You don’t need to recall what any of the specific selection rules are as long as you recall what type of selection rules we have. There most definitely are selection rules for ∆j, ∆l, ∆ml and ∆n but there are no selection rules for spin, ∆s. Thus, without knowing if any of the other selection rules are correct, we can be sure that the proposed spin rule is NOT correct. Correct Answer (D)

137

2.93. PGRE8677 #93

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PGRE8677 #93

Recommended Solution Recall that the frictional force f , is equal to the product of the coefficient of friction, µ and the normal force Fn = 100 newtons

f

= µ(100 newtons) f µ = (100 newtons)

(2.280) (2.281)

Next, recall that power, P , is P = IV = F v

(2.282)

where I is current, V is voltage, F is force and v is velocity. Solve for F with the problems given values for velocity, v = 10 m/s and current and voltage I = 9 Amps and V = 120 volts

F

IV v (9 Amps)(120 volts) = (10 m/s) = 108 N =

(2.283) (2.284) (2.285)

This force is the applied force of the sander which we need for the coefficient of friction equation, so make the substitution to get

138

2.93. PGRE8677 #93

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108 N 100 N = 1.1

µ =

Correct Answer (D)

139

(2.286) (2.287)

2.94. PGRE8677 #94

2.94

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #94

140

2.94. PGRE8677 #94

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Recommended Solution Before switch S is closed, the voltage through the circuit should be 0. As soon as we throw the switch at time t = 0, the maximum voltage is reached and it continually decreases as it reaches equilibrium while passing through resistors R2 and R1 . From this, we know that our solution should start at a maximum VA and approach 0 which is not true of (C), (D) and (E). Once the switch is opened again at time t1 , the current is able to flow again and since the potential collected by the inductor only has to pass current through one of the resistors, R2 to be specific, we expect the initial potential at t1 to be greater than at t0 so (B) is correct. Correct Answer (B)

141

2.95. PGRE8677 #95

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PGRE8677 #95

Recommended Solution The Carnot cycle is, theoretically, perfectly efficient and so we should expect in a Carnot engine, the entropy of the system should be perfectly conserved, dS = 0. (C) is in disagreement with this aspect of the Carnot cycle so it is incorrect. Correct Answer (C)

142

2.96. PGRE8677 #96

2.96

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #96

Recommended Solution The perturbation described in this problem should look a little something like Figure 2.15 From this, we see that the maximum impact of the perturbation will be at the center, x = 0 and so wave functions with a node at that location will be relatively unaffected (i.e. wave functions with odd-values for n). However, because all wave functions with odd-valued n are also odd functions, 143

2.96. PGRE8677 #96

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Figure 2.15: Infinite square well subject to a perturbation we know that the total area under the curve will sum to 0 (which is something a perturbation could have corrupted but, in our case, didn’t). So, for all odd values of n, the perturbed wavefunction will still have its coefficient a0,n = 0 and we choose (B). Correct Answer (B)

144

2.97. PGRE8677 #97

2.97

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #97

Recommended Solution First off, eliminate (E) because it should have no dependence on time. Next, eliminate (B) and (D) because they are the same answer and so if they are both correct, we have a problem. Now, when choosing between (A) and (D), consider that the total angular momentum will be the sum of the translational component and the rotational component L = Ltrans + Lrot Recall that angular momentum is 145

(2.288)

2.97. PGRE8677 #97

CHAPTER 2. PGRE8677 SOLUTIONS

~ × p~ = R ~ × M~v L = Iω0 = R

(2.289)

~ × M~v with For the translational component of angular momentum, we use the fact that L = R our initial velocity ~v0 ~ × M~v0 Ltrans = R   1 = RM ω0 R 2 1 = M R2 ω0 2

(2.290) (2.291) (2.292)

For the rotational component, we use the moment of inertia given in the problem

Lrot = Iω0 1 M R2 ω0 = 2

(2.293) (2.294)

However, because the rotational component is rotating in the opposite direction to that of the translational component, we must take the difference of the components to get

L = Ltrans − Lrot 1 1 = M R2 ω0 − M R2 ω0 2 2 = 0 Correct Answer (A)

146

(2.295) (2.296) (2.297)

2.98. PGRE8677 #98

2.98

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #98

Recommended Solution Start with Coulomb’s Law for a point charge kQ (2.298) l from the perspective of point P , each differential piece of the glass rod will apply a force and it will do so proportional to the charge density ρ = Q/l. Thus, we integrate the differential pieces of glass rod, V =

V

=

Z 2l kρ

l

l

= kρ

dl

Z 2l dl

l = kρ [ln(2l) − ln(l)]

(2.299) (2.300)

l

= kρ ln(2) kQ = ln(2) l Correct Answer (D)

147

(2.301) (2.302) (2.303)

2.99. PGRE8677 #99

2.99

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #99

Recommended Solution Recall that the ground state energy of Hydrogen is equal to 1 Rydberg, −13.6 eV . Positronium involves an electron-positron pair while hydrogen involves a proton-electron pair. There is no difference between the two in terms of charge but there is a significant difference in mass. Since Rydberg’s constant is mass dependent, we have to alter the original Rydberg constant Rhydrogen =

me mp e4 me + mp 8c20 h3

(2.304)

Which becomes, Rpositronium =

e4 me me me + me 8c20 h3

me e4 2 8c20 h3

(2.305)

(2.306)

To convince yourself that this makes the Rydberg constant half as large, consider that the ratio of the proton mass to electron is approximately 1836:1. Calculating the original effective mass with this, you get mp me 1 × 1836 =⇒ ≈1 mp + me 1 + 1836 Using the same values for our new effective mass 148

(2.307)

2.99. PGRE8677 #99

CHAPTER 2. PGRE8677 SOLUTIONS

me 1 =⇒ (2.308) 2 2 Since the energy is proportional to the Rydberg constant, the ground state energy of positronium must be half of the hydrogen ground state energy Ehydrogen = −6.8eV 2 again, if you aren’t convinced, consider the Rydberg equation for hydrogen Epositronium =

1 1 1 =R − 2 2 λ n1 n2 

and since E = hν =

(2.309)



(2.310)

hc λ

E=

hc R = hc λ 2



1 1 − n21 n22

Correct Answer (B)

149



(2.311)

2.100. PGRE8677 #100

2.100

CHAPTER 2. PGRE8677 SOLUTIONS

PGRE8677 #100

Recommended Solution The quickest solution to this problem (which you probably want because it is the last problem and you are likely running out of time) is to consider which of these solutions would actually make for a reasonable “pinhole” camera. For some sample values, let’s pick λ = 400 nm and D = 50 cm = 5.0 × 108 nm to satisfy our condition that λ 0

(3.22)

and since the frictional force is f = µFN , we find the normal force of the wedge by summing the vertical forces

FN

= −FG M = − m+ 2 

(3.23) 

(3.24)

where the mass on the wedge is the wedges mass plus half of the blocks mass (i.e. M/2). combine our equations and solve to get

FG−x + f > 0  1 M Mg − µ m + > 0 2 2 M − 2µm − µM > 0

(3.25)



M (1 − µ) > 2µm 2µm M > (1 − µ) Correct Answer (D)

159

(3.26) (3.27) (3.28) (3.29)

3.7. PGRE9277 #7

3.7

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #7

Recommended Solution For the given apparatus, there are 3 possible modes. The first one they give to us is a normal mode of 0 (i.e. no frequency). The next mode given represents the 2 masses swaying in the same direction. Finally, we need to consider the last mode which occurs when masses sway in opposite directions, in which case it doesn’t matter what the masses are and we can choose (A). Correct Answer (A)

160

3.8. PGRE9277 #8

3.8

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #8

Recommended Solution The description in this problem is a little bit ridiculous but once you figure out what is going on, the problem is relatively easy. Torque is positive or negative based on the right hand rule and from this, we know that we want the cone to rotate in a clockwise direction about the z-direction ˆ Any force in the kˆ isn’t going to get our cone spinning so when viewing the cone from above (+k). eliminate (A), (B) and (E). Next, looking at (C) and (D) it should be apparent that (C) will give us a negative torque (which is what we want) while (D) gives us a positive torque. Correct Answer (C)

161

3.9. PGRE9277 #9

3.9

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #9

Recommended Solution The intent behind coaxial cable shielding is to eliminate (at least in theory) the presence of an E&M field outside of the cable to reduce interference with other electronic equipment. This leads us to choice (A). Correct Answer (A)

Alternate Solution As our distance from the cable blows up to infinity (r → ∞) we would expect the magnetic field to go to 0, which eliminates (B), (D) and (E). Next, recall that the magnetic field of a single, infinitely long cable can be found from Amperes law 162

3.9. PGRE9277 #9

CHAPTER 3. PGRE9277 SOLUTIONS

~ = µ0 i B (3.30) 2πR which is identical to (C). It is unreasonable to assume that adding a shielding element won’t alter this equation with some dependence on a & b, so we are left with (A). Correct Answer (A)

163

3.10. PGRE9277 #10

3.10

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #10

Recommended Solution First, let’s recall the inverse square law for the 2 charges q1 and q2 1 q1 q2 rˆ (3.31) 4π0 r2 because of the infinite, grounded conducting plane at x = 0, we will get image charges of −q and −2q at x = −0.5 a and x = −1.5 a respectively. This tells us that we will get three charges pushing on charge q at x = 0.5 a. The first charge will be 2q at x = 1.5 a which will oppose q to the left. The other two charges, −q and −2q will attract q to the left as well. Sum all of the forces on q to get F =

2q q −q −2q (−ˆ x) + 2 (ˆ x) + (ˆ x) 2 4π0 a a (2a)2 

F

=



164

(3.32)

3.10. PGRE9277 #10

CHAPTER 3. PGRE9277 SOLUTIONS

Figure 3.1: Mirror (image) charges induced as a result of an infinite grounding plate q −4q 2q q − 2− 2 2 4π0 2a 2a 2a 2 1 7q 4π0 2 a2 

= =

Correct Answer (E)

165



(3.33) (3.34)

3.11. PGRE9277 #11

3.11

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #11

Recommended Solution Recall that the energy of the capacitor is 1 U = CV 2 (3.35) 2 Next, use Kirchhoff’s second law which tells us that the sum of all voltages about a closed circuit is zero, to get VC + VR = 0 Q + IR = 0 C Q ˙ + QR C

(3.36) (3.37) (3.38)

where I = Q˙ because current is defined as a moving charge, Q. Rearrange the previous equation and integrate to get Q C

= − 166

dQ R dt

(3.39)

3.11. PGRE9277 #11

CHAPTER 3. PGRE9277 SOLUTIONS

−

dt RC t − RC

Z

dQ Q   Q = ln Q0 Z

=

Q = Q0 e−t/RC

(3.40) (3.41) (3.42)

From this, we also conclude that I = I0 e−t/RC

(3.43)

−t/RC

(3.44)

V

= V0 e

From our initial energy equation, U = 12 CV 2 , we get a voltage equation s

2U C

(3.45)

2U0 −t/RC e C

(3.46)

V = and so our voltage equation is s

s

2U C

=

= U0 e−2t/RC

U

(3.47)

since we are concerned with the point at which half of the energy has dissipated, substitute in U = U0 /2 U0 = U0 e−2t/RC 2 1 = 2e−2t/RC −2t/RC

e

2t − RC

(3.48) (3.49)

= 2

(3.50)

= ln(2)

(3.51)

t =

RCln(2) 2

Correct Answer (E)

167

(3.52)

3.12. PGRE9277 #12

3.12

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #12

Recommended Solution Recall LaPlace’s equation ∇2 V = 0

(3.53)

since the problem only tells us to concern ourselves with the φ component, we can integrate LaPlace’s equation to get d2 V (φ) dφ2

= 0

168

(3.54)

3.12. PGRE9277 #12

CHAPTER 3. PGRE9277 SOLUTIONS

dV (φ) = A dφ V (φ) = Aφ + B

(3.55) (3.56)

since we have the initial condition V (0) = 0, we know that B = 0 V (φ) = Aφ

(3.57)

V0 = Aa V0 A = a

(3.58)

since V (a) = V0 ,

(3.59)

compare this to the equations general form, V = Aφ, to get V0 a

=

V

=

V φ Vφ a

Correct Answer (B)

169

(3.60) (3.61)

3.13. PGRE9277 #13

3.13

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #13

Recommended Solution ~ = 0 as the If you are like me, you probably learned and memorized Maxwell’s equation, ∇ · B “There ain’t no magnetic monopoles” law. For this reason, you know you can immediately get IV as one of the laws that becomes INCORRECT. Next, consider that if we don’t require the magnetic field to curl back on itself in order to force the divergence of the magnetic field to zero, then it is possible to get the electric field to not curl which tells us II could also be INCORRECT and we choose (D). Correct Answer (D)

170

3.14. PGRE9277 #14

3.14

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #14

Recommended Solution From Stefan-Boltzmann’s Law, we get the power radiation of a black body as j ∗ = uT 4

(3.62)

which tells us that doubling the temperature of the black body will alter the power proportional to the fourth power j ∗ = u(2T )4

(3.63)

4

(3.64)

= 16(uT ) since power is energy over time and heat energy is Q = mc∆T

(3.65)

we get that a unit increase in energy will increase temperature by 0.5◦ C and, therefore, 16 units of energy increase will get a change in temperature of 8◦ C. Correct Answer (C)

171

3.15. PGRE9277 #15

3.15

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #15

Recommended Solution Heat capacity of a molecule is determined by the number of degrees of freedom of the molecule. For example, in a monatomic gas, the heat capacity is 3 CV = R (3.66) 2 where the 3 comes from our 3 degrees of translational freedom (ˆ x, yˆ, zˆ). For a springy, diatomic molecule, we have to then include additional degrees of freedom for its rotation and its vibration 7 3 CV = R + Rrot + Rvib = R 2 2 Correct Answer (C)

172

(3.67)

3.16. PGRE9277 #16

3.16

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #16

Recommended Solution The maximum efficiency of a Carnot engine (a theoretically, perfectly efficient heat engine) is η=

W TC =1− QH TH

(3.68)

where η is the maximum efficiency, W is the work done by the system, QH is the heat input, TC is the absolute temperature of the cold reservoir and TH is the absolute temperature of the hot reservoir. Convert temperatures to units of Kelvin to get TC TH 800 K = 1− 1000 K = 1 − 0.8

(3.71)

= 0.2

(3.72)

η = 1−

(3.69) (3.70)

equate η to the work over heat equation with a heat of QH = 2000 J to get W QH W

= 0.2

(3.73)

= (0.2)(2000 J)

(3.74)

= 400 J

(3.75)

Correct Answer (A)

173

3.17. PGRE9277 #17

3.17

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #17

Recommended Solution The problem tells us that the frequency in the x-direction is twice that of the y-direction, so we know that the oscilloscope will be plotting f (t) = sin(ωt) + sin(2ωt)

(3.76)

Using your copy of Mathematica provided in the front of your test booklet, plot the function to get the figure below

However, in the case that your test booklet doesn’t have Mathematica, we can eliminate options (E) and (D) because the superposition of two sine waves shouldn’t give us either of the two curves. 174

3.17. PGRE9277 #17

CHAPTER 3. PGRE9277 SOLUTIONS

Next, eliminate (B) and (C) because both represent just a function of sin(ωt) or cos(ωt), not a superposition of trig functions. Correct Answer (A)

175

3.18. PGRE9277 #18

3.18

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #18

Recommended Solution With coaxial cables, impedance matching is necessary because differences in characteristic impedance can result in signal reflection, particularly in the case of a damaged/kinked line or an incorrectly/damaged termination to the cable. Correct Answer (C)

176

3.19. PGRE9277 #19

3.19

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #19

Recommended Solution Use Newton’s second law and our gravitational force law, F F

= ma mM = G 2 r

(3.77) (3.78)

combine the two and cancel m to get mM r2 M a = G 2 r gr2 M = G

ma = G

(3.79) (3.80) (3.81)

to simplify the mental math, assume that G = 6×10−11 m3 /kg·s2 , g = 10 m/s2 and r = 6×106 m

M

=

gr2 G 
2 10 m/s2 6 × 106 m

=

(6 × 10−11 m3 /kg · s2 ) 6 × 1013 = 6 × 10−11 = 1 × 1024 kg

which is closest to (A). Correct Answer (A)

177

(3.82) (3.83) (3.84) (3.85)

3.20. PGRE9277 #20

3.20

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #20

Recommended Solution based on the given diagram, we know it can’t be true that d < ω so (C) and (E) can be eliminated. Next, recall that the equation for constructive interference in double slit diffraction is d sin(θ) = m1 λ

(3.86)

additionally, we know that we will get a “missing” interference maximum when the constructive double slit equation coincides with the single slit diffraction, so we use ωsin(θ) = m2 λ

(3.87)

get both equations equal to sin(θ) and set them equal to one another and solve for d, ω m2

=

d =

178

d m1 m1 ω m2

(3.88) (3.89)

3.20. PGRE9277 #20

CHAPTER 3. PGRE9277 SOLUTIONS

comparing this to (A) and (B), we aren’t going to get an irrational number (i.e. with some fraction of integers so we can confidently choose (D). Correct Answer (D)

179

√

2 or

√

3)

3.21. PGRE9277 #21

3.21

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #21

Recommended Solution In thin film optics, and most optics in general, I is unequivocally silly so eliminate all choices which include it, i.e. (A) and (D). Next, consider III and IV and recognize that both are correct. More specifically, it is true that we will get a phase change as the light transitions from a lower index of refraction to a higher one (as it enters the bubble) and no phase change as it transitions from a higher index of refraction to a lower one (as it exits the bubble). Eliminate any options that don’t include both of these choices, specifically (C). Finally, when considering option II, recall that equations for thin film optics Constructive Interference 2t = λ/2 Destructive Interference 2t = λ which tells us that in either case, the thickness of the bubble is generally less than the wavelength (i.e. half or a quarter) Correct Answer (B)

180

3.22. PGRE9277 #22

3.22

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #22

Recommended Solution The quick and easy solution, dare I say the “Plug-n-chug” method, is to use the magnification equation for a convex lens M=

f0bj feye

(3.90)

We are given the objective focal length, fobj = 1.0 meter and the magnification M = 10 so we solve for feye , fobj M 1.0 m = 10 = 0.1 m

feye =

(3.91) (3.92) (3.93)

Lastly, we get the total distance from the sum of the two focal lengths,

d = fobj + feye

(3.94)

= 1.0 m + 0.1 m

(3.95)

= 1.1 m

(3.96)

Correct Answer (D)

181

3.23. PGRE9277 #23

3.23

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #23

Recommended Solution The average speed of a conduction electron is described by the Fermi velocity, with equation s

vf =

2Ef m

(3.97)

where the Fermi energy, Ef , is related to the Fermi temperature, Tf , by Ef = kTf

(3.98)

where k is Boltzmann’s constant. Plug everything into our equation to get s

vf

= s

=

2Ef m

(3.99)

2kTf m

(3.100)

simplify the values for the electron mass, fermi temperature and Boltzmann’s constant s

vf

= s

= s

≈

2kTf m

(3.101)

2(1 × 10−23 J/K)(80, 000 K) 10 × 10−31 kg

(3.102)

10 × 1031 16 × 1019 m2 /s2

(3.103)

≈ 1 × 106 m/s Correct Answer (E)

182

(3.104)

3.24. PGRE9277 #24

3.24

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #24

Recommended Solution Argon, like the other noble gases, has a full valence shell. Ionic bonding is bonding between a metal and a non-metal and also requires that one atom lack electrons from its valence shell and the other have an excess charge (think along the lines of the salt molecule with N a+ + Cl− ) so we can eliminate (A). Covalent bonding, on the other hand, involves the sharing of electrons to fill out the valence shell when an atom is lacking electrons in its valence shell, but again Argon isn’t missing any electrons so we eliminate (B) and then eliminate (C). Finally, since Argon isn’t a metal, eliminate (D). As it turns out, argon bonds to other argon atoms by induced dipoles via the Van der Waals force. Correct Answer (E)

183

3.25. PGRE9277 #25

3.25

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #25

Recommended Solution The ability of a particle to pass through solid material is increased as the size of the particle decreases and as the speed of the particle increases. From this, we would expect larger particles like protons and neutrons to struggle to pass “deep underground” and we can eliminate (A), (B) and (C). Between (D) and (E), all of the listed particles are sufficiently fast and small but neutrinos have no charge, unlike positrons, electrons and muons so we should expect them to not be caught or repelled away from electrons in the matter they are trying to pass through and we should expect them to be in the final answer. Correct Answer (D)

184

3.26. PGRE9277 #26

3.26

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #26

Recommended Solution At time t = 0, the number of counts is at 103.5 counts. We can approximate this value by √ √ 103.5 = 107/2 = (107 )1/2 = 100 · 100 · 100 · 10 = 10 · 10 · 10 · 10 √ if we approximate 10 ≈ 3 then we get 103.5 = 3000

(3.105)

(3.106)

This means that half the counts will be 1500 and so at about 17 min, where the counts is = 1000 counts, we’ve passed our half way point and we can eliminate (D) and (E). For (A), (B) and (C), the counts are approximately 103

185

3.26. PGRE9277 #26

CHAPTER 3. PGRE9277 SOLUTIONS

(A) 103.4 ≈ 2500 counts (B) 103.2 ≈ 1600 counts (C) 103.1 ≈ 1300 counts so we choose (B) which is the closest Correct Answer (B)

186

3.27. PGRE9277 #27

3.27

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #27

Recommended Solution The width of the wave-function is determined by the relative size of ∆p and ∆x in the Heisenberg uncertainty principle ∆x ∆p ≥

¯ h 2

(3.107)

Recalling our momentum equation p=h ¯k

(3.108)

where ¯h is a constant, we get ∆p = ¯h∆k. Combine this with the Heisenberg uncertainty principle and solve for ∆k

∆x ∆p ≥ (∆x) (¯h∆k) ≥ ∆k ≥

¯ h 2 ¯h 2 1 2(∆x)

which gives us the inverse relationship between ∆k and ∆x like in (B). Correct Answer (B)

187

(3.109) (3.110) (3.111)

3.28. PGRE9277 #28

3.28

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #28

Recommended Solution In quantum mechanics, the probability of finding the system in a certain state is given by the integral over the squared wave function, or in our case Ylm (θ, φ). The problem asks us about the state with azimuthal orbital quantum number, m=3, so we take the squared values of the form Yl3 (θ, φ). Since our wave function is already normalized, we just need to square the values of the first two terms of ψ ψm=3 = 52 + 12 = 26

(3.112)

ψtotal = 52 + 12 + 22 = 30

(3.113)

therefore, the probability of ψm=3 out of the total ψtotal is ψm=3 /ψtotal = 26/30

(3.114)

= 13/15

(3.115)

Correct Answer (E)

188

3.29. PGRE9277 #29

3.29

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #29

Recommended Solution Questions regarding the infinite square well (particle in a box) and its related plot show up on nearly every test. For this reason, and simply because it is something you oughta know, you should memorize some of the fundamental aspects of the infinite square well graph. First, the solution to the infinite square well is sinusoidal, which allows us to eliminate (A), (C) and (D). Next, when the function impinges on the infinite barrier at x1 and x2 , the amplitude continually decreases toward 0 and stops oscillating, which then allows us to eliminate (E). Correct Answer (B)

189

3.30. PGRE9277 #30

3.30

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #30

Recommended Solution Considering how often the ground state energy of positronium comes up in the GRE, and because it’s trivial to memorize, recall that its value is half that of hydrogen’s ground state E0,hyd 13.6 eV = = 6.8 eV (3.116) 2 2 Keep in mind that this is for the ground state but we need the binding energy in state n = 2. Using the Bohr equation, we see that the energy is inversely proportional to the squared value of the fundamental quantum number, n E0,pos =

En =

Z 2 E0,pos n2

(3.117)

E0 8

(3.118)

since Z = 1, the final answer is E2 =

Correct Answer (E)

190

3.31. PGRE9277 #31

3.31

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #31

Recommended Solution Recall our generic form for the Russel-Saunders term symbol 2s+1

LJ

(3.119)

the problem specifies that the helium atom has term symbol 3

S

(3.120)

so we know that 2s + 1 = 3, J = 1 and L = S. Solve for s to get s = 1 and then recalling the angular momentum quantum number can be found by j =l+s

(3.121)

since S = L, and S corresponds to 0, (i.e. (S, P, D, F, . . .) → (0, 1, 2, 3, . . .)), we finally get j =0+1=1 Correct Answer (B)

191

(3.122)

3.32. PGRE9277 #32

3.32

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #32

Recommended Solution First, recall that the equivalent resistance of resistors in parallel can be found by 1 1 1 = + + ··· Req R1 R2

(3.123)

From Equation 3.123, we can see that the equivalent resistance of the R3 -R4 system and R2 -R5 system will have less resistance than R1 on its own. Additionally, we know that power is related to voltage and current by V2 (3.124) R Which tells us that in addition to R1 having the highest resistance, it will also be the case that the highest current will interact with this resistor and so we should expect R1 to have the biggest amount of current to dissipate. P = I2 · R =

Correct Answer (A)

192

3.33. PGRE9277 #33

3.33

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #33

Recommended Solution Using the rules for resistors and in parallel, find the equivalent resistance of all resistors in the circuit as Req = 75 Ω. This tells us that the current of the entire circuit should be V 3.0 V = (3.125) R 75 Ω Next, finding the equivalent resistance of resistors R3 , R4 and R5 , you’ll find that you have the same resistance in the RHS of the circuit as you do in the LHS (i.e .just resistor R2 ). This tells us that half the current (3/150 A) will go through resistor R2 and the other half will pass through R3,4,5 . Finally, find your equivalent resistance between R3 and R4 to get 2o Ω and then solve for the voltage I=



V = IR =

3 A (20 Ω) = 0.4 V 150 

Correct Answer (A)

193

(3.126)

3.34. PGRE9277 #34

3.34

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #34

Recommended Solution Because we are talking about electromagnetic waves in a waveguide, we can throw out all conditions which don’t give the results as being orthogonal for the electric and magnetic fields separately, i.e. (A) and (C). Then, we can eliminate (B) and choose (C) because the electric field doesn’t propagate in the same direction as the direction of current so the transverse electric field, Et , should be 0. Correct Answer (D)

194

3.35. PGRE9277 #35

3.35

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #35

Recommended Solution An optical diffraction grating works in effectively the same way that a double slit, triple slit, etc interference except with more slits. Based on this similarity, we should try to use our generic equation for double slit diffraction

d sin(θ) = mλ

(3.127)

mλ θ = arcsin d 



(3.128)

where d is the distance between slits, θ is the angle of incidence, λ is the wavelength and m = 0, 1, 2, . . .. The wavelength is given as 5200 angstroms and we can find the distance between slits by assuming that the 2000 slits are evenly spaced across each centimeter of the diffraction grating. Plug this all into Equation 3.128 to get mλ θ = arcsin d   (1)(5200angstroms) = arcsin (0.0005 cm) = arcsin(0.1) 



◦

≈ 6

(3.129) (3.130) (3.131) (3.132)

Correct Answer (B)

195

3.36. PGRE9277 #36

3.36

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #36

Recommended Solution The problem tells us, quite explicitly might I add, that the surface which the E&M wave interacts with is a perfect conductor. This tells us that the net electric field must go to 0 as a result of the interaction. Since the electric field impinges on the surface with some value E0 , we know that the value afterwards must be equal and opposite, E1 = −E0

(3.133)

or, in words, its direction must reverse while maintaining the same magnitude. The magnetic field, however, won’t change direction because of the conductor so (C) becomes the obvious choice. Correct Answer (C)

196

3.37. PGRE9277 #37

3.37

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #37

Recommended Solution If you get this problem wrong, you probably should give up any aspirations you’ve ever held which involve you being a physicist. The problem tells us that the π 0 meson decays into 2 photons which head off in opposite directions. Arguably, the most fundamental and important aspect of relativity is that photons travel at the speed of light, 1 C, in all reference frames. From this, we know the only solutions can be (A) or (D). Since the problem tells us that the photon, γ2 , proceeds in the “backwards” direction, the sign should be negative and we choose (A). Correct Answer (A)

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PGRE9277 #38

Recommended Solution Before starting, let’s take a look at the general time dilation equation ∆t0 = γ∆t = p

∆t 1 − v 2 /c2

(3.134)

from this, we can see that ETS has, quite rudely, written things in terms of inverse Lorentz factors. At this point, I highly recommend that you quietly curse ETS under your breath and then re-write the equations in a more standard form, (A) ∆t1 = γ12 ∆t2 (B) ∆t3 = γ13 ∆t1 (C) ∆t3 = γ23 ∆t2 (D) ∆t2 = γ23 ∆t3 (E) ∆t2 = γ23 ∆t1 Right off the bat, eliminate (E) because it incorporates a Lorentz factor with frame 3 in it when the time for frame 3 isn’t even represented. Next, we know from the time dilation effect that the 198

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time of a moving frame in relation to a stationary frame will appear to be longer in the stationary frame. If we let the stationary frame be S1 , then we see that (A) incorrectly concludes that time in the stationary frame would be longer in the stationary frame than the moving frame. (C) and (D) don’t involve the stationary frame at all and we aren’t given enough information to conclude anything about the relation between the two inertial frames, so those are both likely to be wrong. Only (B) correctly predicts the stationary vs inertial relationship Correct Answer (B)

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PGRE9277 #39

Recommended Solution Recall that the sine function is an odd function, just like the plot given in the problem, while the cosine function is an even function. Based on this, we should eliminate all solutions which utilize cosine functions, i.e. (C), (D), and (E). I should point out that there is always a possibility that the cosine function could be shifted to produce and sine-esque plot, however in our case none of the solutions feature the necessary shift. Next, to choose between (A) and (B), plug in t = π/ω which should give us an amplitude of V (t) = −1. in (A), we get

V (t) = =

  ∞ 4X 1 nπω sin π 1 n ω

(3.135)

∞ 4X 1 sin (nπ) π 1 n

(3.136)

From Equation 3.136, it should be clear that V (π/ω) will be 0 for all values of n, so our solution must be (B). Correct Answer (B)

200

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PGRE9277 #40

Recommended Solution The acceleration at any point on the cylinder will be equal to the sum of all its accelerations. Since the problem explicitly specifies that the cylinder doesn’t slide, we know that there are no lateral forces to contribute. The only acceleration we have is the centripetal acceleration from its rotation which will be pointing toward the center of the cylinder. When the point under consideration is touching the surface of the plane, the acceleration must point up to point towards the cylinder center. Correct Answer (C)

201

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PGRE9277 #41

Recommended Solution The kinetic energy of a rotating object is related to angular frequency, ω, and moment of inertia, I, by 1 EK = Iω 2 (3.137) 2 The moment of inertia is given as I = 4 kg·m2 and we are told that the initial angular frequency of 80 rad changes down to 40 rad. Using Equation 3.137 and accounting for the change in ω, we calculate the kinetic energy as EK

 1  2 I ωf − ωi2 2   1 = 4 kg · m2 (80 rad/s)2 − (40 rad/s)2 2  1 4 kg · m2 (4800 rad/s) = 2 = 9600 J

=

Correct Answer (D) 202

(3.138) (3.139) (3.140) (3.141)

3.42. PGRE9277 #42

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PGRE9277 #42

Recommended Solution Recall our equation for torque τ = Iα

(3.142)

where α is the angular acceleration. We are given the change in angular velocity as ω = 40 rad/sec. Since the rate of change of angular velocity is given, we can find the average angular acceleration as ∆ω 40 rad/s = = 4 rad/s2 ∆t 10 s Plug our angular acceleration value from Equation 3.143 into Equation 3.142 to get α=

τ

= Iα =



(3.143)

(3.144) 2

4 kg · m



16 rad/s

= 16 N · m Correct Answer (D)

203

2



(3.145) (3.146)

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PGRE9277 #43

Recommended Solution This problem is one of those infuriating exam questions that you either know, or you don’t. In this instance, you can only be sure you’ve got the right answer if you recall that Noether’s theorem tells us that pn is a constant under the condition pn =

∂L ∂qn

(3.147)

Even if you don’t know this, we can try to eliminate some of the options based on some common sense (A) An ignorable coordinate is a coordinate that doesn’t show up in the Lagrangian which is not the case (B) I can’t think of a compelling reason to eliminate this one (C) There is no reason to assume that differentiating the Lagrangian with respect to qn will be undefined except when ∂qn = 0 which would be a poor assumption (D) There is no mention made of a time dependence so it is unlikely that

∂L ∂qn

=

d dt



∂L ∂qn



.

(E) Keep in mind that the Lagrangian and Hamiltonian are both measures of Energy and it is not likely that you can differentiate only one but have them each keep the same units. Correct Answer (B)

204

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PGRE9277 #44

Recommended Solution Recall that the Langrangian is the difference between the kinetic and gravitational potential energy, L=T −V

(3.148)

We can first eliminate (E) because the potential term most oppose the kinetic term and we need at least one minus sign. Next, we know the solution must have some potential energy term mgy so that all of the energy is kinetic at bottom of the parabola, so we eliminate (C) and (D). Finally, we know that the kinetic energy pieces should be adding together, not fighting one another, so we choose (A). Correct Answer (A)

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PGRE9277 #45

Recommended Solution Before the ball is dropped, the net energy of the system is all potential equal to mgh. Once the ball is released, the potential is converted to kinetic until the ball hits the ground and all mgh of the energy is now kinetic equal to 1/2mv02 . The problem tells us that the velocity after collision is only 4/5 of its initial velocity, v0 , so the kinetic energy on its way back up is 1 4 2 m v 2 5 0   16 2 1 m v 2 25 0 

Tf

= =



(3.149) (3.150)

If we compare this to the original kinetic energy, it is clear that the final kinetic energy and, therefore, potential energy is 16/25 = 0.64 times as big as its initial energy. Correct Answer (D)

206

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PGRE9277 #46

Recommended Solution The critical isotherm refers to a curve that has the property that the derivative of the pressure with respect to the volume is 0 207

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∂P =0 (3.151) ∂V Of the curves shown, only curve 2 has a point where taking the tangent to the curve results in a horizontal line (i.e. the derivative of the curve is 0). This occurs at precisely the point where the vertical and horizontal dashed lines cross. Correct Answer (B)

Alternate Solution Without knowing anything about isotherms, we can eliminate some options through a bit of reasoning. First, eliminate curves 3, 4 and 5 because they are all effectively the same, especially when discussing everything in qualitative terms. Next, you can eliminate curve 1 because, unlike curve 2, it is the same as all other similar curves above the horizontal dashed line and we would expect the solution to be unique. Correct Answer (B)

208

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PGRE9277 #47

Recommended Solution This problem asks us which region will have vapor and liquid in equilibrium which tells us that we expect to have both states present in the region. The quickest way to determine the answer is to 209

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consider each region in its limit (A) In region A, the volume extends to extremely small sizes and this would cause vapor to get compressed to liquid. This likely won’t support vapor and liquid phases at the same time. (B) Region B represents a middle ground of pressure and volume so there are no glaring limit issues. (C) In region C, the volume can blow up to infinity which would likely force everything to a vapor phase. (D) Region D allows the volume and the pressure to blow up to infinity, meaning there will absolutely be states which either force everything to vapor or everything to liquid. (E) Region E will allow the pressure to blow up to infinity which will force any vapor present into liquid. Only (B) lacks a limiting value that could potentially ruin our equilibrium. Correct Answer (B)

210

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PGRE9277 #48

Recommended Solution At some point in your undergraduate career, you were probably forced to calculate some standard deviations by hand. If you did, then you likely used this equation v uN uX ¯)2 σ = t pi (xi − x

(3.152)

i=1

From Equation 3.152, we know to look for a square root and we can then eliminate (A), (D) and (E) based on this condition. Next, we should expect to see σ 2 values rather than simply σ so we eliminate (B) and choose (C). Correct Answer (C)

211

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PGRE9277 #49

Recommended Solution Considering the amount of variation between the possible solutions, let’s do an approximation. Muons move very quickly and the problem tells us that the motion is relativistic, let’s approximate the speed of a muon going nearly the speed of light as just the speed of light vµ = 3.0 × 108 m/s

(3.153)

Since the scintillators are 3.0 meters apart, we can find the time scale as

∆t =

∆x vµ

(3.154)

3.0 m 3.0 × 108 m/s = 1 × 10−8 s

(3.156)

= 10 nanoseconds

(3.157)

=

so we will want to choose the nanosecond range, i.e. solution (B). Correct Answer (B)

212

(3.155)

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PGRE9277 #50

Recommended Solution First of all, eliminate (E) because we would never be so lucky that we could expand a wave function into basis states under any and all circumstances. Next, eliminate (C) and (D) because A and B as well as α and β are qualitatively identical to one another so if (C) was true, (D) should also be true and we can’t choose both. Finally, eliminate (A) because whether or not the two eigenvalues are non-degenerate should have nothing to do with the basis functions of the wave function. Correct Answer (B)

213

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PGRE9277 #51

214

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Recommended Solution Starting with classical momentum, P = mx, ˙ recall that the analog for the expectation value of momentum is dhxi dt in the infinite square well, the expectation value for position is hpi = m

(3.158)

a (3.159) 2 since hxi is nothing but constants, if we take the derivative of it then it goes to 0 and so does hpi. hxi =

Correct Answer (B)

Alternate Solution More rigorously, we can use the general equation for expectation value hpi =

Z ∞

¯h ψ ∗ ψ dx i −∞

(3.160)

we are given the wave function so if we plug everything in, we get

hpi = =

Z ∞

¯h ψ ∗ ψdx i −∞     Z nπx nπx 2nπ¯h ∞ sin cos a2 i −∞ a a

(3.161) (3.162)

and since cosine and sine are orthogonal with respect to each other, integrating over all of x will result in each one canceling out the other and the total area is 0. Correct Answer (B)

215

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PGRE9277 #52

216

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Recommended Solution The condition for orthonormality, hn|mi = δnm

(3.163)

is a function of the Kronecker delta type, i.e.

δnn = 1

(3.164)

δnm = 0

(3.165)

This is precisely the description in the problem so we choose the orthonormality condition. Correct Answer (B)

217

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PGRE9277 #53

218

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Recommended Solution The energy of the infinite square well isn’t constant so we can immediately eliminate (C) and (E). Next, we can eliminate (A) because, in theory, there shouldn’t be an upper bound to the energy and (A) suggests that there is such a maximum energy. Lastly, you can quickly check the coefficients for (B) and (D) by recalling the energy equation derived form the Schrodinger equation, E=

p2 +V 2m

(3.166)

since V = 0 inside the potential, we get E = p2 /2m which tells us the coefficient should be 1/2 rather than 1/8 and so we can choose (B). Correct Answer (B)

219

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PGRE9277 #54

Recommended Solution This problem is a left/right hand rule paradise (or nightmare depending on your familiarity with the rules). First of all, start by applying the right hand rule to the vertical wire with your thumb in the up direction. This tells us that the magnetic field is pointing into the plane on the side of the loop of wire. Now, apply the right hand rule of a magnetic field into the plane and into the loop such that your thumb is pointing into the plane and your hands are looping clockwise (Figure 220

3.54. PGRE9277 #54

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3.2), allowing us to eliminate (A), (B) and (C). Lastly, use the left hand rule (Figure 3.3) with your thumb in the direction of the current, pointer finger into the plane and middle finger in the direction of the resulting force to find that the left side of the loop goes to the left and the right side of the loop goes right, leaving us with (E). Correct Answer (B)

B

I

I B

Figure 3.2: Right hand rule for a magnetic field passing through a loop of wire

Figure 3.3: Left hand rule for a current through a magnetic field

221

3.55. PGRE9277 #55

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PGRE9277 #55

Recommended Solution The quickest way to solve this problem is to consider the limits of the lengths a and b. If either of these go to 0 then the flux goes to 0 and so too does the force. For this reason, we can eliminate any solution that doesn’t have some dependence on both a and b, i.e. (A) and (B). Next, note that (C) and (E) blows up to infinity when a → 0 so eliminate both of these.

222

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Correct Answer (D)

223

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PGRE9277 #56

Recommended Solution The general equation for energy of a quantum harmonic oscillator in state n is 

E=

1 + n hν 2 

(3.167)

In its ground state, n = 0 so the solution should be 

E = =

1 + 0 hν 2 

1 hν 2

Correct Answer (D)

224

(3.168) (3.169)

3.57. PGRE9277 #57

3.57

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PGRE9277 #57

225

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Recommended Solution Recall Faraday’s law which states that a current will be induced in a conductor due to a change in magnetic flux dφB |ε| = dt

(3.170)

from the description, we know that the half circle is rotating “uniformly” so the induced current should be constant and we can eliminate (C), (D) and (E). As the half circle begins to enter the rectangle, it will have a constantly increasing induced current and once it begins to exit the rectangle, it should have a constantly decreasing induced current. Option (A) gives us this feature but (B) suggest a constantly increasing increase in induced current, which is not what we want. Correct Answer (A)

226

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PGRE9277 #58

Recommended Solution In ground state, the number of electrons on the atom should be the same as Z = 11. The quickest way to figure out the number of electrons proposed in each of the 5 options is to sum up all of the superscripts in each configuration. Doing so will eliminate option (E). Next, we can eliminate (A) because we should completely fill 2p to 2p6 before moving to the next energy level. Next, eliminate (B) because the s level can’t have 3 electrons in it. Finally, recall your energy level diagrams (Figure 3.4) to see that we should progress to 3s after 2p as opposed to going from 2p to 3p

Figure 3.4: Energy level diagram of Sodium

Correct Answer (C)

227

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PGRE9277 #59

Recommended Solution In its ground state, the helium atom has 2 electrons in the first shell which, by the Pauli Exclusion Principle, has one spin up and one spin down. The spin multiplicity, which is 2S + 1, is what determines whether an atom is a singlet, doublet, triplet, etc. Singlet 2(0) + 1 = 1 Double 2(1/2) + 1 = 2 Triplet 2(1) + 1 = 3 In the case of helium, since we have two electrons with opposite directions of spin, they cancel to give us S = 0 which is a singlet Correct Answer (A)

228

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PGRE9277 #60

Recommended Solution If you recall the equation for cyclotron resonance frequency, this problem is a quick plug-n-chug problem eB (3.171) m from the problem description, we are given values for B and m and we can get the elementary charge value e from the front of the test booklet. Plug these values into Equation 3.171 and solve ωc =

1.6 × 10−19 coulombs (1 tesla) = (0.1) (9 × 10−31 kg) 12 ≈ 2 × 10 rad/s


ωc

Correct Answer (D)

229

(3.172) (3.173)

3.61. PGRE9277 #61

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PGRE9277 #61

Recommended Solution Start by recalling the frequency equation for a pendulum, r

ω=

mgrcom I

(3.174)

where the moment of inertia will always be that of a point mass, I = mr2 . for the first pendulum, all of the mass is located at the bottom of the pendulum which makes our center of mass at a distance r, making our frequency equation r

r

2mgr g ωI = = (3.175) 2 2mr r for the second pendulum, however, the masses are separated at a distance of r/2 and r which forces the center of mass to be 3/4 r, so the frequency equation becomes s

ωII

=

2mg(3/4)r (m(1/2 r)2 ) + (mr2 ) 230

(3.176)

3.61. PGRE9277 #61

CHAPTER 3. PGRE9277 SOLUTIONS s

= r

=

(6/4)g (5/4)r

(3.177)

6g 5r

(3.178)

at which point it should be clear that pendulum II has a frequency of Correct Answer (A)

231

p

6/5 that of pendulum I

3.62. PGRE9277 #62

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PGRE9277 #62

Recommended Solution Anytime you get an increase in volume, you will be doing work so we should first eliminate (A). By the same type of reasoning, if V1 = V0 then the work should be 0 so we can eliminate (B). Next, because the problem tells us that we can treat the gas as an ideal gas, which should mean that the type of gas is irrelevant, we should be able to ignore specific heats because they are dependent on the type of gas. From this, eliminate (D). Lastly, we need to use the thermodynamic work equation and ideal gas law to see that the solution should have a natural log, W =−

Z V1

P dV

(3.179)

V0

nRT V Combine Equation 3.179 and Equation 3.180, and integrate to get P =

W

= −

 Z V1  nRT

dV V = nRT [ln(V1 ) − ln(V2 )]   V1 = nRT ln V0

(3.180)

(3.181)

V0

so we should choose (E) Correct Answer (E)

232

(3.182) (3.183)

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PGRE9277 #63

Recommended Solution If you’re clever, you’ll notice that (A) and (B) are exactly opposite so they can’t both be wrong and we know it must be one or the other. In order to choose between the two, recall that a system of maximal probability is in its most stable state so we would expect no spontaneous changes and we choose (D). Correct Answer (D)

233

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PGRE9277 #64

Recommended Solution Immediately eliminate (D) because the presence of Ez = kz guarantees we will have some sort of electric field. Next, eliminate (C) because there is nothing about (Ex , Ey , Ez ) that forces the electric field to vary, especially considering that the only non-zero component is scaled by a constant k and z can be constant as well. Next, eliminate (A) because nothing about what is given demonstrates any time dependence and, for that matter, it says essentially the same thing as (C) which we already decided wasn’t correct. If (B) is also untrue, then we choose (E). However, recall Gauss’ law ~ = ρ ∇·E 0

(3.184)

which clearly states that we should get a charge density in the region of the electric field. Correct Answer (B)

234

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PGRE9277 #65

Recommended Solution We are looking for angular frequency so we should expect to get units of inverse time, recall the SI units for each variable/constant used in the solutions

Q/q = C

(3.185) 2

0 = C /N · m

(3.186)

m = kg

(3.187)

R = m

(3.188)

check each of the potential solutions 235

3.65. PGRE9277 #65

(A)

Qq 2π0 mR2

m = sec 2

(B)

Qq 4π0 mR2

m = sec 2

(C)

Qq 2π0 mR3

1 = sec 2 √

Qq 4π0 mR2

m = sec

Qq 2π0 mR3

1 = sec = sec−1

(D)

q

(E)

q

CHAPTER 3. PGRE9277 SOLUTIONS

Correct Answer (E)

Alternate Solution If you insist on doing this problem in the rigorous fashion, start with Coulomb’s law 1 qQ 4π0 R2

(3.189)

since there are two charges pushing on the central charge q, we change this to account for both with

F

= =

1 2qQ 4π0 R2 1 qQ 2π0 R2

(3.190) (3.191)

Next, recall that the force for an oscillating spring is m¨ x = −kx

(3.192)

and has angular frequency k m re-arrange Equation 3.192 to get everything in terms of k/m to get ω2 =

ω2 =

k x ¨ =− m x

(3.193)

(3.194)

or equivalently, ω2x = x ¨

(3.195)

Finally, since F = m¨ x, use Equation 3.195 with Equation 3.191 (and let x = R) to solve for angular frequency

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m¨ x = mω 2 x = ω2 =

qQ 2π0 R2 qQ 2π0 R2 qQ 2π0 mR3 s

ω =

qQ 2π0 mR3

Correct Answer (E)

237

(3.196) (3.197) (3.198) (3.199)

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PGRE9277 #66

Recommended Solution Recall from the Work-Energy theorem W = ∆EK

(3.200)

Since energy is conserved, the kinetic energy used to move the chain up will be equal to the total potential energy at the top of the axle. The potential energy can be found by W = EG = mgh = (10 kg)(10 m/s2 )(10 m) = 1000 J

(3.201)

Correct Answer (C)

Alternate Solution You could, if you aren’t as industrious as I, set up a differential to account for the changing mass of the chain as it is lifted up. However, to do this quickly let’s first consider our simple work equation W = F ∆x

(3.202) 2

To make a quick approximation, assume that g = 10 m/s and make measurements for every 1 m of change in the chain which will account for a decrease in 20 N. The initial change of of 1 m with a 10 meter long chain with 2 kg per meter is 200 N · m. The next bit of work will be 180 N · m and then 140 N · m and so on to get Wnet = 200 + 180 + 160 + 140 + 120 + 100 + 80 + 60 + 40 + 20 = 1060 N · m which is closest to (C). Correct Answer (C) 238

(3.203)

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CHAPTER 3. PGRE9277 SOLUTIONS

Alternate Solution If you insist on doing things the hard way, start out with the integral form of work W =−

Z

F · dx

(3.204)

In our problem the force changes as the length changes and it changes proportional to F = mg = (20 kg − 2l)g

(3.205)

that is to say that the mass is initially 20 kg and then decreases by 2 times the length of the chain. Plug Equation 3.204 into Equation 3.205 to get

W

= −

Z 0

(20 kg − 2l)gdl

(3.206)

10

Z 10

=

20g − 2gl dl

(3.207)

0

=

h

20gl − gl2

i10 0

(3.208)

= (2000 N · m) − (1000 N · m)

(3.209)

= 1000 N · m

(3.210)

Correct Answer (C)

239

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PGRE9277 #67

Recommended Solution The intensity of the light that gets transmitted through the polaroid is given as I = A + B cos(2θ)

(3.211)

Which tells us that one term can go to 0 when cos(2θ) goes to 0 while the other term, the A term, can’t. This tells us that the light is composed of two different types of polarizations and we eliminate (A), (B) and (E). Lastly, to distinguish between (C) and (D), recall Malus’ law which states that plane polarized light has intensity proportional to I = I0 cos2 (θ)

(3.212)

which we can re-write in a similar form as Equation 3.211 by the double angle identity 2 cos2 (a) = 1 + cos(2a) which gives us (C). Correct Answer (C)

240

(3.213)

3.68. PGRE9277 #68

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PGRE9277 #68

Recommended Solution To calculate optical resolution, we need to use the Rayleigh Criterion, λ (3.214) d the angle and wavelength are given so we can re-arrange Equation 3.214 to solve for d, sin(θ) = 1.22

d=

1.22λ sin(θ)

(3.215)

By a small angle approximation, which we can make because the angle is in microradians, let sin(θ) = θ and then convert all of the values into the same units to get 1.22λ θ 1.22(5.5 × 10−7 m) = 8.0 × 10−6 m ≈ 1 × 10−1 m

(3.218)

≈ 10 cm

(3.219)

d =

(3.216) (3.217)

which is closest to (C). Correct Answer (C)

Alternate Solution Even without remembering the necessary equation, you eliminate some choices by a bit of common sense. Because we are talking about a telescope reflecting mirror, we can probably eliminate (A) and (B) as being ridiculously small to be a reflecting mirror on a telescope. On the other end of the spectrum, a 100 m reflecting mirror would be ridiculously too big. In fact, the largest telescopes on earth peak at or just sightly above 10 m so 100 m is very unlikely and we can eliminate (E). At this point, you can now guess between (C) and (D). 241

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Correct Answer (C)

242

3.69. PGRE9277 #69

3.69

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #69

Recommended Solution A photon travels through a medium with an index of refraction, n, according to the equation c n The index of refraction of the glass is, n = 1.5, so we plug that in and solve v=

v = =

c 3/2 2 c 3

Correct Answer (D)

243

(3.220)

(3.221) (3.222)

3.70. PGRE9277 #70

3.70

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #70

Recommended Solution Start off with the relativistic equation E 2 = p2 c2 + m2 c4

(3.223)

The problem tells us that the energy is E = 100mc2 so we plug that into Equation 3.223, (100mc2 )2 = p2 c2 + m2 c4 2 4

10000m c

2 2

2 4

= p c +m c

(3.224) (3.225)

We could then combine terms with m2 c4 but doing so will make almost no change to the 10000m2 c4 so let’s just ignore it. Finally, solve for the p in Equation 3.225 to get p2 c2 = 10000m2 c4 2

p

2 2

= 10000m c

p = 100mc which is (D). Correct Answer (D)

244

(3.226) (3.227) (3.228)

3.71. PGRE9277 #71

3.71

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #71

245

3.71. PGRE9277 #71

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Recommended Solution Start off by recalling that as temperature and, therefore, net energy of a system blows up to infinity, energy levels will start to become equally populated. Based on this, get rid of any solution that doesn’t account for a temperature dependence, specifically (A). Also from this fact, we can eliminate any solution that doesn’t give the average number as N0 /2 when T → ∞, which would be (C). Finally, when the temperature is minimized (i.e. let T = 0), we would expect all N0 of the particles to be at energy level E1 so plug this into the remaining options to find (B)

N0 1+e−k/(0)

(D)

N0 1+ek/(0)

(E)

N0 e/kT 2

= N0

=∞

=∞

So we choose (B). Correct Answer (B)

246

3.72. PGRE9277 #72

3.72

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PGRE9277 #72

247

3.72. PGRE9277 #72

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Recommended Solution Recall that heat capacity is the derivative of energy with respect to temperature, ∂U ∂T the problem gives us the energy so we take the derivative of it CV =

CV

dU dT   N0  d d = (E1 N0 ) + dT dT 1 + e/kT  2  e/kT = N0 k kT (1 + e/kT )2 =

(3.229)

(3.230) (3.231) (3.232)

which is (A). The worst part of this problem is doing the quotient rule under pressure but you can recognize certain pieces that should be there and only do part of the derivative to get the right answer. Correct Answer (A)

248

3.73. PGRE9277 #73

3.73

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PGRE9277 #73

249

3.73. PGRE9277 #73

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Recommended Solution We can immediately eliminate (B) because it is generally true that entropy increases as temperature increases. We can also eliminate (D) because entropy should approach 0 as temperature approaches 0, not approach some non-zero value. Eliminate (E) because we have more than enough information to pick a choice. Finally, we need to decide whether or not entropy has an upper limit (i.e. option (C)) or goes off to infinity (i.e. option (A)). Because there is some temperature at which all energy levels get equally populated, we also have a temperature at which any further increases in temp won’t result in a wider dispersion of the particles so we should choose (C). Correct Answer (C)

250

3.74. PGRE9277 #74

3.74

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #74

Recommended Solution We start off with the angular frequency equation, s

ω=

mgR I

(3.233)

The moment of inertia can be found by using the parallel axis theorem I = ICOM + mR2

(3.234)

where ICOM is the moment of inertia of the object about an axis passing through its center of mass. In the case of a loop, that moment of inertia about the center of mass is the same as a point particle at distance R so we get I = ICOM + mR2 2

= mR + mR = 2mR

2

2

(3.235) (3.236) (3.237)

so the moment of inertia for rings X and Y is IX IY

= 2(4m)(16R2 ) = 2mR

2

(3.238) (3.239)

plugging these into the angular frequency, ω, gives s

ωX

= r

=

(4m)g(4R) 2(4m)(16R2 )

(3.240)

4g 32R

(3.241)

and 251

3.74. PGRE9277 #74

CHAPTER 3. PGRE9277 SOLUTIONS

s

ωY

mgR 2mR2 r g = 2R

=

(3.242) (3.243)

comparing the two, we get ωX ωY

p

g/8R g/2R r 1 = 4 1 = 2 =

p

which is (B). Correct Answer (B)

252

(3.244) (3.245) (3.246)

3.75. PGRE9277 #75

3.75

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #75

Recommended Solution By conservation of momentum, we should have the original momentum equal to the sum of the individual momentums of the two atoms, Pnet = PT + PH 0 = mT vT + mH vH mT vH = − vT mH

(3.247) (3.248) (3.249)

If we combine Equation 3.249 with the kinetic energy equation for helium, we get

KH

= = = =

1 2 m H vH 2 1 m2 mH 2T vT2 2 mH 2 1 mT 2 v 2 mH T   mT 1 2 mT vT mH 2

(3.250) (3.251) (3.252) (3.253)

and since mT > mH , the kinetic energy of the Helium atom must be larger than the kinetic energy of the Thorium atom. Correct Answer (E)

253

3.76. PGRE9277 #76

3.76

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #76

Recommended Solution The total angular momentum quantum number, j, is the sum of the spin angular momentum, s, and the orbital angular momentum number, l, j =l+s

(3.254)

Since we have three electrons and all electrons have a spin of 1/2, the total spin angular momentum must be

s = s =

1 1 1 + + 2 2 2 3 2

(3.255) (3.256)

Then, recalling the orbital angular momentum rules (S, P, D, F, . . .) → (0, 1, 2, 3, . . .), we get l = 0+1+1

(3.257)

l = 2

(3.258)

Sum the values from Equation 3.256 and 3.258 to get

s = 2+ s =

3 2

7 2

Correct Answer (A)

254

(3.259) (3.260)

3.77. PGRE9277 #77

3.77

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #77

Recommended Solution If we keep in mind that the magnetic moment is a measure of the tendency of an object to align itself with a magnetic field. Although the magnetic moment of the nucleus and electrons are both non-zero (which let’s us eliminate (A)), we can determine just from common sense that a very small and light weight particle will more easily change alignment to conform to the magnetic field than will a “heavy nucleus”. This means that the ratio of magnetic moment between nucleus and electron should be less than 1, which eliminates all but (D) and (E). Between the two, we can comfortably choose (E) because, as I said previously, the mass of the nucleus is what resists the change in alignment. Correct Answer (E)

255

3.78. PGRE9277 #78

3.78

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #78

Recommended Solution Note that at time t = 0, the velocity should be equal to the initial velocity 2v0 that it had right before grabbing the pole. To check this limit, take the derivative of all of the possible solutions at time t = 0 to see which of these correctly predicts the condition. (A) x0 = v = 2v0 (B) x0 = v +



0.5b(3v0 ) b

(C) x0 = 0.5v0 + (D) x0 = v0 +







0.5b(3v0 ) b

0.5b(6v0 ) b

(E) x0 = 0.5v0 +



cos(0) = 2.5v0





cos(0) = 2v0

cos(0) = 4v0

0.5b(6v0 ) b



cos(0) = 3.5v0

Only (A) and (C) meet this criteria and since (A) doesn’t properly account for the rotation with a sine function, choose (C). Correct Answer (C)

256

3.79. PGRE9277 #79

3.79

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #79

Recommended Solution The group velocity is the velocity with which the overall wave travels while the phase velocity is the rate at which the phase propagates. The relevant equations for group velocity and phase velocity are

vg = vp =

∂ω ∂k ω k

(3.261) (3.262)

take the derivative (tangent on the curve) between k1 and k2 to get a roughly constant negative value for the group velocity. The phase velocity, however, has a positive value because it’s a negative slope with an inverse relationship. Since one is positive and the other is negative, they should be in opposite directions and we choose (A). Correct Answer (A)

257

3.80. PGRE9277 #80

3.80

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #80

Recommended Solution Start with the equation for the energy of an electromagnetic wave, E=

hc λ

(3.263)

and now adjust it to solve for the wavelenth hc (3.264) λ we know the energy is 25 kilovolts and can utilize Planck’s constant and the speed of light constant from the front of our test booklet. Plug everything in to get λ=

(4 × 10−15 eV · s)(3.0 × 108 m/s) 2.5 × 104 eV −7 12 × 10 eV · m = 2.5 × 104 eV = 4 × 10−11 m

(3.267)

= 0.4 Angstroms

(3.268)

λ =

which is nearly (B). Correct Answer (B)

258

(3.265) (3.266)

3.81. PGRE9277 #81

3.81

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #81

Recommended Solution In electronics, we will reach the max steady-state amplitude at the point when impedence is matched between ZL = ZC

(3.269)

ZL = jωL

(3.270)

The inductor impedance is

and the capacitor impedance is 259

3.81. PGRE9277 #81

CHAPTER 3. PGRE9277 SOLUTIONS

ZC =

1 jωC

(3.271)

so applying Equations 3.270 and 3.271 to Equation 3.269 gives us

ZL = ZC 1 jωL = jωC = ω2j 2 1 = LC r 1 ωj = LC which gives the same inverse LC dependence that (C) suggests. Correct Answer (C)

260

(3.272) (3.273) (3.274) (3.275) (3.276)

3.82. PGRE9277 #82

3.82

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #82

Recommended Solution First, recall the angular impulse H is proportional to the moment of inertia by H = Iω

(3.277)

the moment of inertia of a plate through its center is (1/12)mL2 , so with our length of 2d, we get

261

3.82. PGRE9277 #82

CHAPTER 3. PGRE9277 SOLUTIONS

I = = =

1 mL2 12 1 m(2d)2 12 1 md2 3

(3.278) (3.279) (3.280)

now solve for ω in Equation 3.277 with Equation 3.277 plugged into it to get our final answer

ω = = =

H I H (1/3)md2 3H md2

Correct Answer (D)

262

(3.281) (3.282) (3.283)

3.83. PGRE9277 #83

3.83

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #83

Recommended Solution We can get the right proportionality to figure out the solution by making a few small angle approximations. Start by summing the forces in both dimensions FA = −FT −x FT −y = −FG

(3.284) (3.285)

by geometry, we can figure the length of d as d/2 L d = 2L sin(θ)

sin(θ) =

263

(3.286) (3.287)

3.83. PGRE9277 #83

CHAPTER 3. PGRE9277 SOLUTIONS

we can also use some trigonometry to a relationship between tensions sin(θ) =

FT −x FT

(3.288)

plug Equation 3.288 into Equation 3.287 to get d = 2L

FT −x FT

(3.289)

Finally, get a relationship between FT −y and FT and apply the small angle approximation cos(θ) ≈ 1 to get FT −y FT FT −y 1 = FT FT = FT −y

cos(θ) =

(3.290) (3.291) (3.292)

but since FT −y = −FG = −mg and FT −x is the force from Coulomb’s law, substitute these values into Equation 3.289, FT −x FT 2Lkq 2 d2 mg

d = 2L = which matches (A).

Correct Answer (A)

264

(3.293) (3.294)

3.84. PGRE9277 #84

3.84

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #84

Recommended Solution For choices (A) and (B), we can see that these are both true by the Larmor formula P =

e2 a2 6π0 c3

(3.295)

Next, we know that (C) must be true by the LinardWiechert potential which states, in a big hot mess, ~ x, t) = q E(~

~n − β~ γ 2 (1 − β~ · ~n)3 R2

!





˙ ~ × ~β] q ~n × [(~n − β)  +  ~˙ c 3 (1 − β · ~n) R

(3.296)

~ ∝ 1/R2 . Finally, we know that (E) is true because as we go off to of which, we only car that E infinity, both fields tend to 0. We are left with (D) so that must be our correct answer. Correct Answer (D)

265

3.85. PGRE9277 #85

3.85

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #85

Recommended Solution Start of with the relativistic energy equation E = γm0 c2

(3.297)

the energy is given to us as E = 1.5 MeV and the mass is me = 0.5 MeV/c2 , 1.5 MeV = γ(0.5 MeV/c2 )c2 γ = 3

(3.298) (3.299)

now, we want to use the relativistic momentum equation but to do so, we need the velocity of the electron. Using our result from Equation 3.299, solve for v,

γ = 3 =

1 q

v2 c2

q

v2 c2

1− 1 1−

v2 9 1− 2 c

(3.300) (3.301)

!

= 1

v2 c2

=

v =

(3.302)

8 9√

8 c 3

(3.303) (3.304)

Plug results from Equation 3.299 and 3.304 into the relativistic momentum equation to get the final answer

P

= γm0 v

(3.305) 266

3.85. PGRE9277 #85

CHAPTER 3. PGRE9277 SOLUTIONS √ = 3(0.5 MeV/c2 )( 8/3 c √ = (0.5) 8 MeV/c

(3.306)

= 1.4 MeV/c

(3.308)

Correct Answer (C)

267

(3.307)

3.86. PGRE9277 #86

3.86

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #86

Recommended Solution First, eliminate any option that includes V0 (i.e. (A) and (C)) as the oscilloscope provides this data and therefore wouldn’t be dependent on any of the other pieces of data being known. Next, recall that a capacitor discharges according to V = V0 e−t/RC

(3.309)

which tells us that we will need R and we can eliminate (C) and (E). Additionally, we will need the time which we can derive from the sweep rate, s, so we get our final solution as (B). Correct Answer (B)

268

3.87. PGRE9277 #87

3.87

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #87

Recommended Solution To find the net energy, we need to sum the kinetic energy and potential energy of the particle H =T +V

(3.310)

since the orbit is circular, we know that the centripetal force must be equivalent to the attractive force

Fc = FK K mv 2 = r r3

(3.311) (3.312)

In Equation 3.312, if we just multiply a 1/2 to the LHS and cancel out the r, this becomes our kinetic energy equaiton 1 K mv 2 = 2 2 2r Now for the potential energy, we use 269

(3.313)

3.87. PGRE9277 #87

CHAPTER 3. PGRE9277 SOLUTIONS

V

= − = − =

Z Z

F · dr

(3.314)

K dr r3

(3.315)

K 2r2

(3.316)

but because the potential is attractive, it becomes negative. We then sum the 2 potentials from Equation 3.313 and 3.316 to get

H = T +V K K = − 2 2 2r 2r = 0 which is (C). Correct Answer (C)

270

(3.317) (3.318) (3.319)

3.88. PGRE9277 #88

3.88

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #88

Recommended Solution According to the problem, this parallel plate capacitor is connected to a batter. As long as it is not removed, the voltage and electric field should not be altered, even if a dielectric is put in place. This tells us that (A), (B) and (D) must all be wrong. Next, we can find the before and after charge as

Q0 = C0 V0

(3.320)

Qf

(3.321)

= κC0 V0

so (C) must be wrong. This only leaves (E). Correct Answer (E)

271

3.89. PGRE9277 #89

3.89

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #89

Recommended Solution Recall the solution to the Infinite square well, which gives us a set of sinusoidal waves (Figure 3.5) The first plot given represents n = 0, the second is n = 1, the third is n = 2 and so on. It should become clear that all even values for n peak at x = 0 and so these will be disrupted by the infinite potential at this point. The odd valued quantum numbers won’t so they will remain. This description only matches (E). Correct Answer (E)

272

3.90. PGRE9277 #90

CHAPTER 3. PGRE9277 SOLUTIONS

Figure 3.5: Plots of the solution to the infinite square well

3.90

PGRE9277 #90

Recommended Solution It’s worth memorizing the scale of energy spacing for the different energy levels E = Etrans + Erot + Evib + Eelec

(3.322)

Erot ≈ 0.001 eV

(3.323)

Evib ≈ 0.1 eV

(3.324)

Eelec ≈ 1 eV

(3.325)

those being

from which we see that the rotational energy level should be around 10−3 which is (B). Correct Answer (B)

273

3.91. PGRE9277 #91

3.91

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #91

Recommended Solution Eliminate (A) because the pion isn’t a lepton. Next, eliminate (B) because the Λ particle is a baryon so it must have spin 1/2. We can also quickly eliminate (D) because angular momentum is conserved. Lastly, eliminate (C) because it isn’t true that all interactions that don’t produce a neutrino are weak. Correct Answer (E)

274

3.92. PGRE9277 #92

3.92

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #92

Recommended Solution For a single coil of wire, it is relatively clear that a rotating magnet with frequency of 10 Hz will give us an alternating voltage of 10 Hz. However, for three coils, for every third of a rotation 10 Hz will have been generated for a single coil and a full rotation will have done 3 of these, making the net frequency 40 Hz Correct Answer (D)

275

3.93. PGRE9277 #93

3.93

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #93

Recommended Solution When the weight is released at time t = 0 and angle θ = 0, the weight is essentially in free fall so it should have an acceleration of a = g. Plug in the θ = 0 condition to check the limit (A) g sin(0) = 0 (B) 2g cos(0) = 2g (C) 2g sin(0) = 0 p

(D) g 3 cos2 (0) + 1 = 2g q

(E) g 3 sin2 (0) + 1 = g and only (E) meets our criteria. Correct Answer (E)

276

3.94. PGRE9277 #94

3.94

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #94

Recommended Solution The Lorentz transformation always takes the form of t0 = γ (t − vx)

(3.326)

x

0

= γ (x − vt)

(3.327)

y

0

= y

(3.328)

z

0

= z

(3.329)

which tells us that whatever the coefficients we have on x0 and t0 , they should be the same with the variables swapped. This is only true of (C) so this must be our solution. Correct Answer (C)

277

3.95. PGRE9277 #95

3.95

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #95

Recommended Solution Assuming that ETS hasn’t given us a bunch of useless information in this problem, which is probably a pretty good assumption, we can arrive at the answer quickly with a bit of dimensional analysis. We are given, and should try to use, 1012 proton/sec

(3.330)

2

(3.331)

20

10

nuclei/cm

2

10 proton/sec −4

10

steradians

(3.332) (3.333)

The only way to arrange these 4 values to get a final unit of cm2 /steradian is by

(1020

(102 protons/sec) = 10−26 cm2 /steradian nuclei/cm2 )(1012 proton/sec)(10−4 steradians) Correct Answer (C)

278

(3.334)

3.96. PGRE9277 #96

3.96

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #96

Recommended Solution If you recognize that this setup is precisely the setup used to measure the index of refraction of air and you recall that the index of refraction of air is 1.000293, then you can quickly see that (C) is the correct answer. Correct Answer (C)

Alternate Solution In a gas interferometer, a beam of light is passed to a partially silvered mirror which splits the beam into two parts. One part continues through the mirror and the other is reflected at a right angle. Ultimately, both beams arrive at the observer and create an interference pattern. We know 279

3.96. PGRE9277 #96

CHAPTER 3. PGRE9277 SOLUTIONS

that the optical path length is related to the index of refraction by nL, but since the light travels the distance L twice, we re-write it as 2nL. Now, if we remove the gas from the system, our index of refraction must change (∆n) and the interference pattern will its fringes shift according to mλ (3.335) 2L Since the index of refraction of most gases is nearly 1, we typically define the index of refraction of any gas as the index of refraction of a vacuum (i.e. n = 1) plus some additional factor kp, where k is some constant and p is the air pressure. ∆n =

ngas = 1 + kp

(3.336)

changes in index of refraction are proportional to changes in air pressure by ∆p = ∆nk

(3.337)

so we combine Equations 3.335, 3.336 and 3.337 to get n=1+

mλp 2Lδp

(3.338)

to make a quick approximation, and because this information is given, lose the dependence on p (i.e. p = ∆p) and solve to get mλ 2L (40 fringes)(500 nm) = 1+ 2(5 cm) = 1 + 0.0002

(3.341)

= 1.0002

(3.342)

n = 1+

which is (C). Correct Answer (C)

280

(3.339) (3.340)

3.97. PGRE9277 #97

3.97

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #97

Recommended Solution To start off, check the units of each potential solution to see that only (A) and (D) give some sort of mass 

(A) m∗ = 12 h ¯ 2k (B) m∗ =

dk dE



= kg

4 6 = mseckg 4 )

¯2k h

(

dk dE

(C) m∗ = h ¯ 2k (D) m∗ = 



¯2 h d2 E dk2

d2 k dE 2

1/3

= A big mess that clearly has extra units

 = kg

5 8 = kgsecm 4 At which point you can either make an educated guess or recall that you should be differentiating the energy with respect to wave number, as in (D), rather than differentiating wave number with respect to energy, like in (A).

¯ 2 m2 (E) m∗ = 12 h



d2 E dk2



Correct Answer (D)

281

3.98. PGRE9277 #98

3.98

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #98

Recommended Solution Recall that we find the eigenvalues from a matrix by finding the determinant of the characteristic equation 



0−λ 1 0   0−λ 1  det  0 1 0 0−λ You can, and for speed you should, use the quick method of finding the determinant of a 3 dimensional matrix (which I once heard called the “shoe string method”) 



−λ 1 0 −λ 1   0 −λ 1 0 −λ   1 0 −λ 1 0 multiplying the diagonals and summing them (via “shoe string method”), you get (−λ3 + 1 + 0) − (0 + 0 + 0) = 0 3

(3.343)

−λ + 1 = 0

(3.344)

3

(3.345)

λ

= 1

Then from complex analysis, since we clearly only have 1 real solution (i.e. λ = 1) the rest of the solutions must be complex and, therefore, (B) must be a false statement. Correct Answer (B)

282

3.99. PGRE9277 #99

3.99

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #99

Recommended Solution For this problem, you should be able to immediately recognize that the correct answer is (A). This is because the Hydrogen atom, unlike nearly every other problem in all of quantum mechanics, is an ideal and exactly solvable system. For this reason, there is no correction factor for the hydrogen atom in its ground state. Correct Answer (A)

283

3.100. PGRE9277 #100

3.100

CHAPTER 3. PGRE9277 SOLUTIONS

PGRE9277 #100

Recommended Solution In order to figure out the balancing point of the system, we need to find its net center of mass. m1 x1 + m2 x2 (3.346) mtotal We know the masses of each block are m1 = 20 kg and m2 = 40 kg and the positions are x1 = −5 m and x2 = 5 m, with respect to the center. We also know the total mass is the sum of the 2 blocks and the mass of the rod (i.e. 20 kg + 40 kg + 20 kg = 80 kg) so we can plug everything in and solve to get (20 kg)(−5 m) + (40 kg)(4 m) 80 kg 100 = m 80 = 1.25 m

COM =

which is (C). Correct Answer (C)

284

(3.347) (3.348) (3.349)

Chapter 4

PGRE9677 Solutions

285

4.1. PGRE9677 #1

4.1

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #1

Recommended Solution First, consider the state of this circuit before the switch, S, is changed from point a to point b. When connected to the potential, V , a constant current is passed through the capacitor. The capacitor will continue to gain a potential until the potential difference on the capacitor is equal to that of the potential. Once the switch is moved over to b, that potential is gone and what is left is the potential stored on the capacitor, which will proceed through the resistor, R. Unlike the potential, the capacitor won’t be able to maintain a constant current so we would expect the current to decline as the resistor “resists” the current and energy dissipates. At this point you can cross off any curve on the plot which isn’t decreasing over time. Now, all you have to do is decide whether the initial potential energy provided by V will be V /R or V /(R + r). Considering that resistor r is isolated from resistor R, it is reasonable to conclude that only resistor R will have an influence on the current. Correct Answer (B)

Alternate Solution Using the same reasoning as in the “Recommended Solution”, eliminate choices (A),(C) and (E) due to the fact that a current supported only by a charged capacitor must decline as it is forced to pass through a resistor. Again, we need to figure out what the initial current must be for the charged capacitor-resistor circuit. Recall Kirchhoff’s first rule (also commonly known as the Loop Rule) Kirchhoff ’s First Rule The sum of the changes in potential energy encountered in a complete traversal of any loop of a circuit must equal zero 286

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Applying the Loop Rule to the second stage of our circuit (that is to say, once the capacitor has reached an equal potential to that of V and the switch, S, has been thrown to point b) we must account for two potentials summing to 0

Vcap + Vres = 0

(4.1)

iR + q/C = 0

(4.2)

R and C are both constants but i and q are functions of time and so with 1 equation and 2 variables, we are stuck. And yet, we are too industrious to leave things at that. We can relate the charge, q, to the current, i, because current is merely the change in current. i=

dq dt

(4.3)

which gives us dq R − q/C = 0 (4.4) dt Now that the only variable is q, we can solve this ordinary homogenous differential equation. Doing so, we get q = q0 e−t/RC

(4.5)

Where e is Euler’s number, not an elemental charge number. Taking the derivative of q then gives us the current as dq q0 =i=− dt RC 



e−t/RC

(4.6)

Setting t = 0 for the initial time and utilizing our equation for capacitance (q0 = CV0 ) we can find the initial current as 

i0 = −

CV0 −0/RC e = −V0 /R RC 

(4.7)

Which states that the initial current of the final stage of the circuit is V /R. We can ignore the negative on the initial current as it is merely indicating that the capacitors charge is decreasing. Correct Answer (B)

287

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CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #2

Recommended Solution In this problem we are asked to determine the reading of an ammeter attached to a circuit with a potential (V) of 5.0 V , a resistor of 10 Ω while sitting in a uniform but changing magnetic field of 150 tesla/second. In this scenario we have to account for two different phenomena which are generating a current. The first phenomena generating a current is the 5.0 V potential which will create a current according to Ohm’s Law I=

V R

(4.8)

5.0 volts = 0.5 A (4.9) 10 Ω This current will be moving in a counterclockwise direction because the “positive flow of current” moves in the opposite direction of the electron flow. In other words, the current will be moving from the positive terminal to negative terminal (The larger bar of the potential V to the smaller bar of the same potential V). The second source of current comes from the changing magnetic field being directed “into the page”. We know that a changing magnetic field generates a current thanks to Maxwell’s/Faraday’s laws, and in particular we know from Faraday’s law of induction that a changing magnetic flux will induce a potential (emf) ε I=

dφB ε= dt

(4.10)

where φB is Z Z

φB =

~ r, t) · dA ~ B(~

S

288

(4.11)

4.2. PGRE9677 #2

CHAPTER 4. PGRE9677 SOLUTIONS

which simply states that the magnetic flux (φB ) is equal to the magnetic field (B) through some total surface area (i.e. the surface integral over differential pieces of area dA ) Taking the surface integral for this simple loop of wire is the same as just calculating the area of the square with sides of 10 cm × 10 cm, or in more standard units, .10 m × .10 m which gives a total surface area of A = .01 m2 . This makes the flux equation ~ r, t) · A ~ φB = B(~

(4.12)

~ is the vector perpendicular to the surface area and it is in the same direction Since the vector A as the magnetic field lines, ~ ·A ~=B A B

(4.13)

substituting that in for our equation calculating the emf (ε)

ε = ε =

d(φB ) dB = dt dt A   tesla 

150

sec

(4.14) 

0.01 m2 = 0.15 A

(4.15)

Finally, calculate the difference between the 2 potentials (Equations 4.9 and 4.15) we found and you have 0.5 A − 0.15 A = 0.35 A Correct Answer (B)

289

(4.16)

4.3. PGRE9677 #3

4.3

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #3

Figure 4.1: Electric potential at point P in relation to ring of radius R

Recommended Solution Point p in Figure 4.1 represents a “test charge” location and we want to know what the electric potential is at that point due to a uniformly charged ring with charge Q. Using Coulombs Law in Gaussian units, we have Z

U= s

1 dQ 4π0 r

(4.17)

where r is the distance between any point on the ring and point P, and dQ is a differential piece of the ring. because the distance from every point on the ring to our point P is the same, r is a constant. We can pull all constants out from the integral, giving

U U

= =

1 4rπ0 Q 4rπ0

Z

dQ

(4.18)

s

(4.19)

Lastly, we need to solve for r. Drawing a line between the ring and P gives us a right triangle which allows us to use the Pythagorean theorem (Figure 4.2) 290

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CHAPTER 4. PGRE9677 SOLUTIONS

Figure 4.2: Constructing a right triangle between points P and R

r=

p

R 2 + x2

(4.20)

and substituting Equation 4.20 and 4.19 gives us U=

Q √ 4π0 R2 + x2

(4.21)

Correct Answer (B)

Alternate Solution The trick to this alternate solution involves manipulating the variables involved. No specific R or x values are given, so the correct solution must work for any choice of R and x. For example, letting x go to infinity, we would expect the electric potential to go to 0. For solutions (C) and (D), letting x go to infinity would make the potential go to infinity. Similarly, if we let R go to infinity, again potential should go to zero. This isn’t the case for solution (E), so cross it off. Finally, you just have to decide whether or not the distance between the ring (or rather differential pieces of the ring, dQ) and√point P is in agreement with (A), in which r = x, or if it is in agreement with (B), in which r = R2 + x2 . By the Pythagorean theorem, the hypotenuse can’t be the same length as one of its sides so r 6= x. Correct Answer (B)

291

4.4. PGRE9677 #4

4.4

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #4

Recommended Solution To solve this problem we will need Coulomb’s law, Hooke’s Law and the general equation for angular frequency Coulomb’s Law: F =

1 q1 q2 4π0 r2

Hooke’s Law: F = −kx Angular Frequency: ω =

p

k/m

In our situation we have a charged surface, specifically a ring, so we will need to replace one of our point charges with this value, which can be found by integrating a differential piece of charge over the entire ring.

F F

= =

−q 4π0 r2 −qQ 4π0 r2 292

Z

dQ

(4.22)

S

(4.23)

4.4. PGRE9677 #4

CHAPTER 4. PGRE9677 SOLUTIONS

Figure 4.3: Component forces at point P Equation 4.23 represents the equation for the net force. However, we will need to decompose the net force into its horizontal and vertical components. Fortunately, for every force in the vertical direction from one piece of the ring, the piece on the opposite side of the ring will cause an equal and opposite force. The vertical forces cancel each other out and we are left with the horizontal forces. Hopefully it is clear from Figure 4.3 that the horizontal force component should be Fx = F cos(θ) = F (x/R)

(4.24)

substitute 4.23 with 4.24 to get F =

−qQ x 4π0 r2 R

(4.25)

Finally, we can calculate the distance between P and any point on the ring using the pythagorean theorem r 2 = R 2 + x2

(4.26)

combining 4.26 with 4.25 gives us F =

−qQx 4π0 (R2 + x2 )R

(4.27)

Now taking Hooke’s law, solve for k to get k = −F/x

(4.28)

and substitute our force equation into it. After simplifying, you should get k=

qQ 4π0 (R2 + x2 )R

(4.29)

and then substitute this in for the k in our angular frequency equation and simplify to get s

ω=

4π0

qQ + x2 )Rm

(R2

(4.30)

Finally, take into account that the problem asked you to consider when R >> x, meaning x effectively goes to 0, giving 293

4.4. PGRE9677 #4

CHAPTER 4. PGRE9677 SOLUTIONS

s

ω=

qQ 4π0 mR3

Correct Answer (A)

294

(4.31)

4.5. PGRE9677 #5

4.5

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #5

Recommended Solution To start off with, figure out every force that will be on this car as it travels through the arc. The problem identifies for us the force from air resistance, Fair . Additionally, we can identify that there will be a net centripetal force pointing towards the center of the arc. Keep in mind that the centripetal force is not one of the component forces but the sum total of all the forces. drawing out the force diagram gives us Figure 4.4

Figure 4.4: Net force diagram on a turning vehicle Now we must ask ourselves which direction the horizontal force must point such that when it is added to Fair , we will get a net centripetal force pointing down. Check each of the 5 choices in Figure 4.5 and you will quickly see that only (B) can be correct. Correct Answer (B)

295

4.5. PGRE9677 #5

CHAPTER 4. PGRE9677 SOLUTIONS

Figure 4.5: Potential force combinations

Alternate Solution A quick and qualitative way of figuring out the solution is to consider why the car is turning at all. The force from air resistance cant be responsible for the motion and the centripetal force is merely a description of the net force (i.e. the sum of the component forces) acting on the car. As such, the force that the road is applying to our car tires must have some vertical component of force pointing down. You can eliminate choices (C), (D), and (E) based on this. Next, consider that the air resistance force has a horizontal component of force (technically it only has this horizontal force) and so for our net centripetal force to have no horizontal force, the allusive force we are finding must have an equal and opposite horizontal component. (A) has no horizontal component so cross it off. Correct Answer (B)

296

4.6. PGRE9677 #6

4.6

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #6

Recommended Solution This problem can be solved quickly by utilizing conservation of energy laws. At the top of the incline, the only energy for the system is gravitational potential energy UG = mgh

(4.32)

According to the description, the block slides down the incline at a constant speed. This means that the kinetic energy Uk = 12 mv 2 , is the same at the beginning of the blocks motion as it is at the end and thus none of the potential energy we started with changed to kinetic energy. However, because energy must be conserved, all of the energy that didn’t become kinetic energy (i.e. all mgh of it) had to have dissipated from friction between the incline and the block. Correct Answer (B)

Alternate Solution Start by drawing out a force diagram for the block at the top of the slide (Figure 4.6) take the sum of all of the forces in the x and y direction. Note that the acceleration in the x direction is 0 because the speed is constant, therefore the net force is 0.

297

4.6. PGRE9677 #6

CHAPTER 4. PGRE9677 SOLUTIONS

Figure 4.6: Force diagram of a block/incline system

X

X

Fnet,x = FGX − f = 0

(4.33)

Fnet,y = FN − FGY = 0

(4.34)

Using trigonometry, you should see that FGX = FG sin(θ) and FGY = FG cos(θ). Since F = mg, FGX = mg sin(θ) and FGY = mg cos(θ). Additionally, f = µFN so we get f = FGX = µFN = mg sin(θ)

(4.35)

FN = mg cos(θ)

(4.36)

and

Combine Equations 4.35 and 4.36 and get µmg cos(θ) = mg sin(θ) µ=

sin(θ) = tan(θ) cos(θ)

(4.37) (4.38)

Now, recall that work, which is equivalent in magnitude to energy, is W = F (∆X)

(4.39)

and since we are concerned with the energy (work) generated from friction, the force in 4.39 must be f . Make the substitution to get Wf = f (∆X)

(4.40)

Wf = mg sin(θ)(∆X)

(4.41)

298

4.6. PGRE9677 #6

CHAPTER 4. PGRE9677 SOLUTIONS

where ∆X is the length of the ramp. Since sin(θ) = h/∆X, we know that ∆X = h/sin(θ)

(4.42)

mgsin(θ)h sin(θ)

(4.43)

finally, combine 4.41 and 4.42 Wf =

Wf = mgh Correct Answer (B)

299

(4.44)

4.7. PGRE9677 #7

4.7

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #7

Recommended Solution Because the collision is elastic, we know that energy and momentum are conserved. From this, we know that the initial potential at height h will equal the kinetic energy immediately before the ball strikes the brick. Additionally, we know that the momentum of the ball/brick system must be the same before collision as after. This gives us the equations 1 1 2 2 2 mVb,0 = mVb,f − mVB,f 2 2

(4.45)

mVb,0 = mVb,f + 2mVB,f

(4.46)

and

With two equations and three unknowns, we can get a relationship between any of the two variables. Combine 4.45 and 4.46 in order to get a relationship for Vb,0 with Vb,f and Vb,0 with VB,f .

VB,f

=

Vb,f

=

2 Vb,0 3 1 Vb,0 3

(4.47) (4.48)

This tells us that the final velocity of the ball b is 1/3 of its initial velocity and the block B leaves the collision at 2/3 the initial velocity of the ball. Once the ball leaves at its velocity of 300

4.7. PGRE9677 #7

CHAPTER 4. PGRE9677 SOLUTIONS

Vb,f , it will move up to its final height as the kinetic energy becomes potential energy. The kinetic energy is 1 1 2 mVb,f = m 2 2   1 1 2 m V = 2 9 b,0 1 mgh 9

2 1 Vb,0 3   1 1 2 mVb,0 9 2



Uf

=

Uf

=

Uf

=



(4.49) (4.50) (4.51)

Comparing the final energy to the intial, we get mghf

=

hf

=

1 mgh0 9 1 h0 9

Correct Answer (A)

301

(4.52) (4.53)

4.8. PGRE9677 #8

4.8

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #8

The quickest way to solve this problem is to recognize that the problem is describing a damped harmonic oscillator. The object, in this case a particle with mass m, oscillates with a damping force proportional to F = −bv applied to the particle. We know that the force applied to the particle is fighting against the oscillation because it is always in the opposite direction of the velocity and this confirms that this is a damped harmonic oscillator. If the force was positive and adding to the force of the oscillation, then we would have a driven harmonic oscillator. Now, consider the period of damped harmonic oscillator in comparison to an unhindered SHO without the opposing force. Strictly speaking, a damped oscillator doesn’t have a well-defined period and without knowing the specific values for mass, spring constant, etc we don’t know whether we are talking about a system that is underdamped, over damped or critically damped. Nevertheless, we can see that all of the damped oscillations (Figure 4.7) will experience an increase in the period.

Figure 4.7: Damped harmonic oscillators

Correct Answer (A)

302

4.9. PGRE9677 #9

4.9

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #9

Recommended Solution The Lyman and Balmer series both refer to different types of transitions of an electron in a hydrogen atom from one radial quantum level (n) to another. The Lyman series is a description of all such transitions from n=r to n=1, such that r ≥ 2 and is an integer. The first Lyman transition (commonly called Lyman-alpha) is n=2 going to n=1, the second (Lyman-beta) involves a transition of n=3 to n=1, etc. The Balmer series, on the other hand, involves transitions from some n=s to n=2, such that s ≥ 3 and is an integer. The longest wavelength for both series involves the smallest transition, i.e. n=2 going to n=1 for the Lyman Series and n=3 going to n=2 for the Balmer. The Rydberg formula can then be used to find the wavelength for each of the two transitions 1 =R λ

1 1 − 2 2 nf ni

!

(4.54)

For this problem we won’t need to compute anything, just compare λL and λB . Doing this for the shortest Lyman transition gives 1 λL 1 λL

1 1 = R 2− 2 1 2 3 = R 4 



(4.55) (4.56)

λL = 4/(3R)

(4.57)

and for the Balmer transition 1 λB 1 λB

1 1 = R 2− 2 2 3 5 = R 36 

λB = 36/(5R) 303



(4.58) (4.59)

(4.60)

4.9. PGRE9677 #9

CHAPTER 4. PGRE9677 SOLUTIONS

making the ratio λL /λB =

4/(3R) = 5/27 36/(5R)

Correct Answer (A)

304

(4.61)

4.10. PGRE9677 #10

4.10

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #10

Recommended Solution In internal conversion, one of the inner electrons of the molecule is ejected. Because of this, one of the outer electrons will drop down a level to fill the space and some form of electromagnetic radiation is released. From this, you can immediately remove (A), (D) and (E). Now you simply have to decide whether the electromagnetic radiation will correspond to the energy level of X-Rays or γ rays. As it turns out, this electronic transition will release X-rays. However, if you don’t know this, you may be able to reason through to the answer. Consider that γ rays are the highest energy form of electromagnetic radiation and these will generally be a result of an energy transition for a nucleus. An electron transition, on the other hand, involves smaller exchanges of energy and correspond to the lower energy X-rays. Correct Answer (B)

305

4.11. PGRE9677 #11

4.11

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #11

Recommended Solution In this problem, ETS is testing your knowledge of physics history. If you recall the Stern-Gerlach experiment (1922), this problem becomes quite easy. The Stern-Gerlach experiment involved firing neutral silver atoms through an inhomogenous magnetic field. The classical understanding (i.e. before Stern-Gerlach) would have suggested no deflection because the atoms are neutral in charge and have no orbital angular momentum and thus generate no magnetic dipole. However, this experiment showed that the beam split into two distinct beams, adding evidence to the ultimate conclusion that electrons have a spin property of 1/2. Keep in mind that, in general, the number of distinct beams generated after hitting the magnetic field will be equal to 2S + 1. Silver has a single unpaired electron and so S = 1/2, giving the original Stern-Gerlach experiment its two beams. In this problem, ETS went easy on us and gave us hydrogen to work with, which also has S = 1/2 and thus will split into two beams. Correct Answer (D)

306

4.12. PGRE9677 #12

4.12

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #12

Recommended Solution When positronium questions pop up on the GRE, they generally can and should be solved by its relation to the hydrogen atom. In this case, you have to recall that the ground state energy of Hydrogen is equal to 1 Rydberg = −13.6 eV. Positronium involves an electron-positron pair while hydrogen involves a proton-electron pair. There is no difference between the two in terms of charge but there is a significant difference in mass. Since Rydberg’s constant is mass dependent (Equation 4.62), we have to alter the original Rydberg constant Rhydrogen =

me mp e4 me + mp 8c20 h3

(4.62)

Which becomes,

Rpositronium = =

me me e4 me + me 8c20 h3 me e4 2 8c20 h3

(4.63) (4.64)

To convince yourself that this makes the Rydberg constant half as large, recall that the ratio of the proton mass to the electron mass is approximately 1836:11 . Calculating the effective mass with an electron and proton, we get 1 ∗ 1836 mp me = ≈1 mp + me 1 + 1836

(4.65)

Calculating the effective mass for the electron/positron pair, with their equivalent masses, gives us me me me = me + me 2

(4.66)

Since the energy is proportional to the Rydberg constant, the ground state energy of positronium must be half of the hydrogen ground state energy 1

It isn’t necessary to know this quantity in order to arrive at this simplification. All that is really important is that the difference between the two masses is significant

307

4.12. PGRE9677 #12

CHAPTER 4. PGRE9677 SOLUTIONS

Ehydrogen = −6.8eV 2 again, if you aren’t convinced, consider the Rydberg equation for hydrogen Epositronium =

1 1 1 =R − 2 2 λ n1 n2 

and since E = hν =

(4.67)



(4.68)

hc λ

E=

hc R = hc λ 2



1 1 − 2 2 n1 n 2

Correct Answer (C)

308



(4.69)

4.13. PGRE9677 #13

4.13

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #13

Recommended Solution The problem gives us the specific heat of water as 4.2 kJ/kg which should be a strong hint that you will need to use the equations for heat absorption into a solid, Q = cm∆T

(4.70)

We are also given the power of the heating element, meaning we know the amount of energy input. Additionally, because the problem states that the water never manages to boil, even if it comes close, the water must be outputting energy at an equal rate or at least very close to it. We then just need to know how long it will take for energy to dissipate from the system equivalent to a change in temperature of 1◦ C. Using the definition of power, P = W/∆t

(4.71)

W = P ∆t

(4.72)

Or equivalently

We can then combine the two equations (Q=W) to get cm∆T = P ∆t

(4.73)

When making substitutions, keep in mind that 1 L of water in mass is 1 kg, giving 

4.2

kJ kg



(1 kg)(1◦ C) = (100 watts)(∆t)

(4.74)

We can get everything into the same units by converting watts to kJ. Specifically, 1 watt = 1 J/s = 0.001 so 100 watt = 100 J/s = 0.1 kJ/s.

309

4.13. PGRE9677 #13

CHAPTER 4. PGRE9677 SOLUTIONS

4.2kJ = 0.1

kJ ∆t s

(4.75)

Then solving for ∆t gives ∆t ≈ 40 sec Correct Answer (C)

310

(4.76)

4.14. PGRE9677 #14

4.14

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #14

Recommended Solution This is one of those rare “Plug-n-Chug” problems on the PGRE. Cherish it! In this problem we have 2 copper blocks in an insulated container. This tells us that the total energy of the system is conserved. Since both blocks are of the same mass, we know the final temperature of both blocks will reach equilibrium at 50 kcal each. Since heat travels from high temperatures to low temperatures, all energy transfer will take place from the block with T2 = 100◦ C to the block T1 = 0◦ C and so we only need to consider this one direction of energy transfer. Using our equation for heat absorption for a solid/liquid body, we can plug in all known values

Q = cm∆T

(4.77)

Q = (0.1kCal/kg K)(1kg)(50 K)

(4.78)

Q = 5 kCal

(4.79)

Correct Answer (D) Additional Note In this specific problem ETS has been very nice to us by making both blocks of the same mass. In the case that they gave us blocks of different masses, we wouldn’t be able to easily conclude that the final temperature of each block would be at 50◦ C. If, for example, all other values were the same but the block at T1 = 0◦ C had a mass of m1 = 1 kg and the block at T2 = 100◦ C had a mass of m2 = 2 kg, then we would have to solve for the final temperature. To do so, consider that the

311

4.14. PGRE9677 #14

CHAPTER 4. PGRE9677 SOLUTIONS

system is insulated so no energy can leave. This means that the total energy transfer will be 0 and the sum of the energy transfers between the two blocks will sum to 0. This gives us Q1 + Q2 = cm1 (Tf − Ti,1 ) + cm2 (Tf − Ti,2 ) = 0

(4.80)

c is the same for both so cancel it out and solve for Tf , Tf = 33.3◦ C

312

(4.81)

4.15. PGRE9677 #15

4.15

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #15

Recommended Solution The thermodynamic process goes through its entire cycle, so energy is conserved, ∆U = 0. From this and the first law of thermodynamics, ∆Q = ∆U + ∆W , we know that the total heat will just be the sum of its work terms. ∆Qnet = ∆Wnet = ∆WAB + ∆WBC + ∆WCA The work equation for a thermodynamic system is W = P V = nRT , so

R

(4.82)

P dV and we have the ideal gas law,

Z V2

WAB =

P dV V1

313

(4.83)

4.15. PGRE9677 #15

CHAPTER 4. PGRE9677 SOLUTIONS

=

Z V2 nRT V1

V

= nRT ln(

dV

V2 ) V1

(4.84) (4.85)

Next, moving from B to C, we get WBC = P ∆V = nR∆T

(4.86)

Finally, taking C to A, there is no change in volume so we would expect to get a work of 0, i.e. WCA = P ∆V = P (0) = 0

(4.87)

Adding up all of the components (Equations 4.85, 4.86 and 4.87), we get

Wnet = WAB + WBC + WCA = nRT ln(V2 /V1 ) + nR∆T + 0

(4.88) (4.89)

Since the problem specifies we have one mole of gas, Equation 4.89 becomes Wnet = RT ln(V2 /V1 ) + R∆T

(4.90)

Finally, consider the work equation for path BC to realize that ∆T = (Tc − Th ). Reversing the two heats, as we see in all of the possible solutions, will result in a negative sign coming out, giving a final result of Wnet = RT ln(V2 /V1 ) − R(Th − Tc ) Correct Answer (E)

314

(4.91)

4.16. PGRE9677 #16

4.16

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #16

Recommended Solution 1 The problem gives us the equation for the mean free path as ησ . To get the density of air, use the 3 Atm m −5 ideal gas law and use the gas constant R = 8.2 × 10 K Mol

PV n V

= nRT P = RT

(4.92) (4.93)

Which should give you 6 × 1023 Mol Mol = 3 × 1025 3 (4.94) −2 3 2.4 × 10 m m Approximate the size of any given air molecule as being about 1 nm so the collision cross section is 1 nm2 which is 1 × 10−18 m2 . Put this into the mean free path equation provided and you have 1 1 = ≈ 1 × 10−7 25 ησ (3 × 10 Mol/m3 )(1 × 10−18 m2 ) Correct Answer (B)

315

(4.95)

4.17. PGRE9677 #17

4.17

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #17

Recommended Solution Recall from quantum mechanics that we can get the probability of finding a particle in any position by taking the integral of the squared wave function, Z b

Pab =

|Ψ(x)|2 dx

(4.96)

a

The integral of a curve is just the area underneath it and since a plot of the function is provided, we can quickly find the area. However, be aware that the curve represented is Ψ, and we want the area under the curve for Ψ2 . For this reason, we must square every piece of the plot and then take the area under the curve. Since we are concerned with the probability of the particle being located between x = 2 to x = 4, we need to compare that with the total area of the squared wave function. Doing so, from left to right, we have Area2→4 = (2)2 + (3)2 = 13 2

2

(4.97) 2

2

2

2

Area0→6 = (1) + (1) + (2) + (3) + (1) + (0) = 16 thus the probability is 13/16

316

(4.98)

4.17. PGRE9677 #17

CHAPTER 4. PGRE9677 SOLUTIONS

Correct Answer (E)

317

4.18. PGRE9677 #18

4.18

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #18

Recommended Solution Recall the “infinite square well” that everybody does as their first, and frequently only, exactly solvable quantum system. You should recall that the wave function with infinite potential barriers 318

4.18. PGRE9677 #18

CHAPTER 4. PGRE9677 SOLUTIONS

on each side restricted the wave function to the area between the potential walls, with the exception of a small bit of tunneling on each side of the infinite potentials. This should tell you that no matter what the wave function looks like, it better have an amplitude that is lessened by interacting with a greater potential wall than it would in the open space. From this, eliminate (A), (D) and (E). Now, you just need to decide if the the wavefunction will be able to maintain its amplitude, frequency etc as its trying to tunnel through the wall. Again, you should recall from the infinite square well that this wasn’t the case. Instead, the amplitude of your wave function continually dropped and approached Ψ(x) = 0. (C) shows this characteristic drop but (B) does not. Correct Answer (C)

319

4.19. PGRE9677 #19

4.19

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #19

Recommended Solution This problem is a simpler case of Ruterford scattering. We are told that the scattering angle is 180◦ which is the maximum possible scattering angle. This means that the alpha particle is contacting the silver atom head on and all of the kinetic energy of the electron is becoming potential energy, resulting in Equation 4.99 1 1 q1 q2 mv 2 = 5 M eV = 2 4π0 r

(4.99)

0 is given in the list of constants in the beginning of the test. q for an alpha particle is 2(1.6 × 10−19 ) C and q for silver is 50(1.6 × 10−19 ) C. Converting the known kinetic energy into more convenient units gives, 5 M eV ≈ 8 × 10−13 J and plugging everything in and solving for r, you get r=

2(1.6 × 10−19 C) 50(1.6 × 10−19 C) ≈ 2.9 × 10−14 m 4π(8.85 × 10−12 )(8 × 10−13 J) Correct Answer (B)

320

(4.100)

4.20. PGRE9677 #20

4.20

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #20

Recommended Solution In this problem we are told that an elastic collision occurs, which tells us that energy and momentum will be conserved. From this, write the equation for each. Momentum: Ptotal = mV0 = −mV1 + µV2 Energy: Etotal = 21 mV02 = 12 mV12 + 12 µV22 Substitute in 0.6V0 = V1 and simplify to get the momentum equation mV0 = −0.6 mV0 + µV2 

1.6 mV0 = µV2  1.6 mV0 = V2 µ

(4.101) (4.102) (4.103)

and the energy equation mV02 = 0.62 mV02 + µV22 0.64 mV02 = ! 0.64 mV02 µ

(4.104)

µV22

(4.105)

= V22

(4.106)

For your momentum equation, square both sides and set the resulting equation equal to your energy equation 2.56 m2 V02 0.64 mV02 = µ2 µ Now simplify and you are left with only m and µ. Solving should give you

321

(4.107)

4.20. PGRE9677 #20

CHAPTER 4. PGRE9677 SOLUTIONS

µ=4m

(4.108)

µ = 16 u

(4.109)

and since m = 4 u,

Correct Answer (D) Additional Note The method by which you solve this problem is identical to that used in problem #7 (See Section 4.7) on this same test (PGRE9677). It’s all a matter of recognizing that energy and momentum are conservered in an elastic collision.

322

4.21. PGRE9677 #21

4.21

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #21

Recommended Solution From the Parallel-Axis theorem, we know that the moment of inertia of any object is equal to the sum of the objects inertia through its center of mass and M h2 I = Icom + M h2 = M R2 + M R2 = 2M R2

(4.110)

plug the provided values into Equation 4.110 to get s

T = 2π

2M R2 = 2π M gR

s

T = 2π

s

2R g

(4.111)

2(0.2 cm) ≈ 1.2 10 m/s2

(4.112)

Correct Answer (C)

323

4.22. PGRE9677 #22

4.22

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #22

Recommended Solution The problem tells us that the golf ball is orbiting mars, which tells us that the height of the golf ball relative to the surface of the planet is constant. The easiest way to deal with the provided information is to utilize the kinetmatic equation 1 y − y0 = v0 t − at2 2 We can assume the initial velocity is 0 and the vertical change is 2 m, giving 1 2 m = − (−0.4g)t2 2 Solving for t to figure out how much time passes for the orbital motion gives t=

q

4 m/0.4g ≈ 1 sec

(4.113)

(4.114)

(4.115)

Since velocity is change in position over time, which is 1 second, we get a final velocity of v=

∆X = 3600 m/s = 3.6 km/s 1 sec Correct Answer (C)

324

(4.116)

4.23. PGRE9677 #23

4.23

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #23

Recommended Solution The problem specifies that , what ever value it might be, is small. Assume that the value of epsilon is so small as to be effectively irrelevant. Under this condition, look for statements which conform to known orbital phenomena. (A) There’s no reason to assume that energy isn’t conserved in this scenario. Mechanical energy R (Work) is W = F · ds and this won’t interfere with the conservation of energy. Additionally, we should only expect conservation of energy to fail if there is some fricitional force applied to the object and the problem never mentions such a force. (B) Angular momentum is conserved as long as there are no external torque on the system.  wouldn’t impose an external torque. (C) This is consistent with Kepler’s third law of planetary motion (i.e. P 2 ∝ a3 ) (D) Noncircular orbits are rare and from process of elimination (see (E)) we can see that this is the only false proposition (E) Circular type orbits are relatively common and so we should expect this to be true. Correct Answer (D)

325

4.24. PGRE9677 #24

4.24

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #24

Recommended Solution This problem can be solved with nothing more than Coulomb’s law, F =

1 q1 q2 4π0 r2

(4.117)

We start with two spheres, each with a charge of q. When the uncharged sphere touches sphere A, electrons are passed to the uncharged sphere until they each reach equilibrium. In this case, equilibrium involves half of the charge ending up on each sphere. Now, the initially uncharged sphere has a charge of 21 q and sphere B has a charge of 1q. When these two come in to contact, they equilibrate again. The average for these two charges is ( 12 + 1) q = 3/4q (4.118) 2 The initially uncharged sphere leaves itself and sphere B having a charge of 3/4 q. Knowing the charge for sphere A and sphere B, plug this into the equation for Coulomb’s Law to get 1 (1/2 q)(3/4 q) 3 = 2 4π0 r 8



1 qA qB 4π0 r2

i.e. the force is now 3/8 of its original charge. Correct Answer (D)

326



(4.119)

4.25. PGRE9677 #25

4.25

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #25

Recommended Solution Generally, I would recommend going through each possible solution to make sure you are finding the “best solution”. However, in this problem one of the options stood out as being clearly false. Before the switch is thrown, current flows through the path of C1 and charges it to energy U0 . After the switch has been thrown, C1 will still charge to energy U0 so we know U0 = U1 . Without knowing, U2 , we can see that (E) violates this result (unless U2 = 0 which is clearly not the case). I’d also like to point out that (D) and (E) disagree with one another (again, unless one of the energies is 0) so you can immediately determine that your answer must be one of these two. Correct Answer (E)

Alternate Solution Before the switch is thrown, current is flowing through the path of C1 and charges it to U0 which means that the initial capacitance referred to in the problem is C0 = C2 . Also, from the problem, 327

4.25. PGRE9677 #25

CHAPTER 4. PGRE9677 SOLUTIONS

C1 = C2 so all three capacitors are equivalent. Additionally, all potentials are equal in a parallel circuit (i.e. V0 = V1 = V2 ), giving Q0 /V = Q1 /V = Q2 /V (A) Q0 = Q1 = Q2 (which will be shown in part (B)) and half of the sum of two identical things will equal itself. (B) Since Q0 /V = Q1 /V = Q2 /V , multiply the V out and get Q0 = Q1 = Q2 (C) As was mentioned previously, the potential in a parallel circuit is always equivalent across all capacitors in the circuit. V0 = V1 = V2 (D) Since all capacitors have the same capacitance and the same voltage, by U = 12 CV 2 , we get U0 = U1 = U2 (E) As was demonstrated in (D), U0 = U1 = U2 so it has to be the case that U1 + U2 = 2U0 . (E) is false. Correct Answer (E)

328

4.26. PGRE9677 #26

4.26

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #26

Recommended Solution If you know the equation for frequency in an RLC circuit, then this problem is relatively straightforward. This question is primarily testing your ability to do unit conversions and multiply very large and/or very small numbers. The equation for the resonance frequency in an RLC circuit is f=

1 √ 2π LC

f2 =

1 4π 2 LC

(4.120) (4.121)

We know that the final frequency should be 103.7 MHz which we can simplify as 100 MHz. Squaring this value gives us 1.0 × 104 MHz. Converting this to hz, we get 1 × 1010 hz. The inductance is 2.0 microhenries which is equivalent to 2.0 Ω · s. Rearranging the previous equation to solve for capacitance, gives C=

1 4 π 2 (2.0 Ω · s)(1.0 × 1010 hz)

(4.122)

To make things easier, let’s set π 2 = 10, which makes 4π 2 = 40. Substituting everything into Equation 4.122, C=

1 Farads 8 × 1011

(4.123)

Finally, convert to pF by recalling that 1 × 1012 pF = 1 F, giving a final value of C = 0.8 pF which is closest to (C) 1 pF. Correct Answer (C)

329

4.27. PGRE9677 #27

4.27

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #27

Recommended Solution Logarithmic scaling is great for exponential curves and inverse curves (i.e. negative exponentials). From this, you know that (A), (C), (D) and (E) all should be either log-log or semilog. (A), (C), and (E) fulfill this requirement but (D) doesn’t. Just to be sure that (D) is the “Best Solution” check (B) to see that we would expect it to utilize a linear graph just as the solution suggests. Correct Answer (D)

330

4.28. PGRE9677 #28

4.28

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #28

Recommended Solution The wave on this oscilloscope clearly displays the sum of two separate waves with different frequencies. The wave with the larger frequency has a wavelength of about 6 cm. The speed of the wave is given as v = 0.5 cm/ms. Using the relationship of a linearly traveling wave through a homogeneous medium, we can calculate the frequency λ = f

=

v f v λ

Plug in the values from the problem into Equation 4.125 331

(4.124) (4.125)

4.28. PGRE9677 #28

CHAPTER 4. PGRE9677 SOLUTIONS

f=

(0.5 cm/ms) 1 = (6 cm) 12 ms

(4.126)

However, we want the frequency in Hz, so convert Equation 4.126 to get 1 f= 12 ms



1000 ms 1s



= 83 Hz

Which agrees with option (D). Correct Answer (D)

332

(4.127)

4.29. PGRE9677 #29

4.29

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #29

Recommended Solution The units for the Planck length should be a length (probably meters). The units for the three constants are m3 kg·s2

Gravitational Constant: G ≡

Reduced Planck’s Constant: ¯ h≡

m2 ·kg s

m s

Speed of Light: c ≡

We know that there can’t be a unit of seconds in the final result and since all instances of seconds shows up as a denominator, we know at some point we will have to do some division. This only happens in option (E) . Correct Answer (E)

Alternate Solution Using the units for the three constants, multiply out each one to figure out which of these gives us a result in meters (A) G¯hc =



(B) G¯h2 c3 = (C) G2 ¯hc =

m3 kg·s2







m3 kg·s2

m3 kg·s2

m2 ·kg s



2 



m s

 m2 ·kg 2 s m2 ·kg s






=

m6 s4


m 3 s m s




=

=

m10 ·kg s7

m9 kg·s6

333

4.29. PGRE9677 #29

(D) G1/2 ¯h2 c = (E)


1/2 G¯h/c3



m3 kg·s2

=

CHAPTER 4. PGRE9677 SOLUTIONS

1/2 

r

m3 kg·s2

 m2 ·kg 2 s



m2 ·kg s



m s

/




=

m13/2 kg 3/2 s4


m 3 s

=m

Correct Answer (E)

334

4.30. PGRE9677 #30

4.30

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #30

Recommended Solution This problem can be solved with just a bit of clever reasoning. First, consider that the initial state of the system is water at equilibrium and at 20 cm from the bottom of the curve. We can get a good approximation of the total mass of the water by assuming that all of the water is accounted for by the 40 cm of vertical tube length (i.e. ignoring the curve). There is likely not much of a difference between the two values and as long as we are consistent with this assumption, deviations won’t present themselves in the final ratio. So, we assume that with 40 cm of water at 1 g/cm3 , we have roughly 40 g of water (Technically we would have 40 g/cm2 but because the tube is the same size throughout, the cross-sectional slices with units of cm2 will be the same for all parts and we might as well treat it as 1). Now, we can add to that the more dense liquid which we know accounts for 5 cm of the tube and has a density 4 times greater than the water. The total mass of the more dense liquid is 20 g giving a grand total of 60 g of liquid. The system will be at equilibrium when pressure on both sides of the tube is equal. Keeping in mind that pressure is proportional to density (i.e. P = ρgh) and density is proportional to mass (i.e. ρ = m/V ) we can conclude, and it should seem reasonable that, the system will be at equilibrium when we have equal amounts of mass in each side of the tube. On the left side, we have 20 g of the denser liquid, leaving the 40 g of water. Leaving 10 g of water on the left side and the remaining 30 g of water on the right, we get 30 g of liquid on each side. Now that it is in equilibrium, figure the amount of height taken up by each liquid. On the left, 20 g of the dense liquid is taking up 5 cm of space (as marked on the 335

4.30. PGRE9677 #30

CHAPTER 4. PGRE9677 SOLUTIONS

diagram) and the 10 g of water is adding an extra 10 cm, giving 15 cm on the left. On the right, we just have 30 g of water which makes 30 cm of liquid. The ratio of the right side to the left side is 30 cm/15 cm or 2/1. Correct Answer (C)

Alternate Solution A more rigorous method of tackling this problem is to use the equation for pressure as P = ρgh with the condition that the system is at equilibrium when the pressure on the left side is the same as on the right. Setting P1 = P2 , we have to note that P1 is actually composed of two different liquids with different heights and different densities, i.e. P1 = Pdense + Pwater,1 = Pwater,2 . Substituting in values, we have (ρdense g (5 cm)) + (ρwater g (h1 − 5 cm)) = ρwater g h2

(4.128)

(4 g/cm3 )(5 cm) + (1 g/cm3 )(h1 − 5 cm) = (1 g/cm3 )h2

(4.129)

Then using the fact that h1 + h2 = 45 cm, you will have 2 equations and 2 unknowns. When you solve for h1 and h2 , you should get h1 = 15 cm and h2 = 30 cm, making the ratio h2 /h1 = 30 cm/15 cm = 2/1. Correct Answer (C)

336

4.31. PGRE9677 #31

4.31

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #31

Recommended Solution (A) Any object which falls through a medium which resists its motion will continually increase its velocity until it reaches some terminal velocity. If it helps, think of the mass falling through air resistance as the ”viscous medium”. In order for the retarding force to decrease kinetic energy, velocity would have to decrease at some point, which it doesn’t. (B) As in part (A), the mass will reach some terminal velocity, however after it has done that it will maintain that velocity, not slow down and stop. (C) The terminal velocity of an object, by definition, is the maximum speed which the mass can reach in a given medium. In other words, the maximum speed and the terminal speed are the same, so the object can’t decrease its speed from a maximum speed to a terminal speed. (D) For (D) and (E) we finally have an accurate description of terminal velocity, however we now need to decide whether the speed of the object is dependent on b and m or just b. You might be inclined to think that speed isn’t dependent on mass because of the classical, well known results of Galileo which demonstrates that objects fall at the same speed regardless of their mass. Keep in mind that this is only true in a vacuum and drawing out a force diagram should make it quite clear that the speed is dependent on mass (i.e. Fnet = ma = mg − bv). If you still aren’t convinced, ask yourself why a feather falls more slowly than a brick in a real world scenario. (E) This solution is the same as (D) except that it correctly identifies b and m as being variables of velocity Correct Answer (E) 337

4.32. PGRE9677 #32

4.32

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #32

Recommended Solution Rotational kinetic energy is Uk = 12 Iω 2 . The moment of inertia for a point mass about a radius R is I = mR2 . For the rotation about point A, we need to determine the length between each mass and point A. Since the line between B and A bisects a 60◦ angle, we can make a right triangle and use trigonometry to find the length as l/2 l =⇒ RA = √ R 3 Which gives us the rotational kinetic energy equation as cos(30◦ ) =

Uk−A

1 l2 1 = (3m)R2 ω 2 = (3m) 2 2 3

!

1 ω 2 = ml2 ω 2 2

(4.130)

(4.131)

The rotational kinetic energy around point B is easier to calculate because we have just two masses rotating at length l, giving us 1 Uk−B = 2ml2 ω 2 = ml2 ω 2 2 Comparing Equations 4.131 and 4.132, we can see that Uk−B is twice as big as Uk−A Correct Answer (B)

338

(4.132)

4.33. PGRE9677 #33

4.33

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #33

Recommended Solution From quantum mechanics, recall that the probability of finding the state of any given operator can be found by Equation 4.133 Z

P =

hψ|A|ψi

(4.133)

There are two terms with quantum number l = 5 which have coefficients of 2 and 3. This gives a total of 32 + 22 = 13. Thus, the probability is 13 out of a total sum of 38 (i.e. 22 + 32 + 52 = 38) so the probability is 13/38. Correct Answer (C)

339

4.34. PGRE9677 #34

4.34

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #34

Recommended Solution (A) Gauge invariance deals with an invariance of charge. This isn’t violated simply due to a preferential direction. (B) Time invariance doesn’t deal with spin or Electromagnetic interactions. Besides, time invariance is rarely violated in any context. (C) Translation invariance deals with the invariance of system equations under any translational frame. This isn’t related to this problem. (D) Reflection invariance is violated in this instance because the preferential direction in one frame changes if we were to create a reflection of that frame. Consider, for example, the way in which things will look reversed when viewed in a mirror. If particles are moving in a preferential direction, say +x, then there exists some reflected frame in which the particles move in the preferential direction of -x. (E) Rotational invariance deals with maintaining system equations under any rotation of the system frame. This isn’t violated in this example. Correct Answer (D)

340

4.35. PGRE9677 #35

4.35

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #35

Recommended Solution (A) Recall from doing atomic energy level diagrams that the Pauli exclusion principle tells us that electrons in an atom can’t have the same set of 4 quantum numbers: n, l, ml and ms . In general, it also tells us that fermions may not simultaneously have the same quantum state as another fermion. More rigorously, it tells us that for two identical fermions, the total quantum state of the two is anti-symmetric. (B) The Bohr correspondence principle says that quantum mechanical effects yield classical results under large quantum numbers. Definitely not the answer. (C) The Heisenberg uncertainty principle tells us that the information of two related aspects of a quantum state can only ever be known with inverse amounts of certainty. Said plainly, the more you know about one component of the state of a system, the less you know about another2 . This isn’t related to the quantum state of fermions. (D) A Bose-Einstein condensate is a gas of weakly interacting bosons which can be cooled sufficiently to force them to their ground state energies. This doesn’t have anything to do with fermions. (E) Fermi’s Golden Rule involves the rate of transition from one energy eigenstate into a continuum of eigenstates. This is clearly not the right answer. Correct Answer (A)

2

The typical example is momentum and position

341

4.36. PGRE9677 #36

4.36

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #36

Recommended Solution When an object with mass moves at relativistic speeds, the mass of the object increases. You can always remember this because as an object approaches the speed of light, the mass approaches infinity, hence why massive objects can’t travel at the speed of light. When the two lumps of clay hit one another the sum of the masses in non-relativistic terms would be 8 kg and thus at relativistic speeds, the mass must be higher. We can eliminate (A), (B) and (C) from this fact and, if you can’t figure out the next step, then you at least have the problem down to two solutions. Relating the rest mass of the combined lumps (M ) to the total energy of the system for the two masses separately (2m)gives Enet = 2γmc2 = M c2 2q

1 1−

v2 c2

m=M

(4.134) (4.135)

Plug all of the values provided in the problem into Equation 4.135 to get M=q

1 1 − ( 53 c)2 /c2

(4kg) = 10kg

Correct Answer (D)

342

(4.136)

4.37. PGRE9677 #37

4.37

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #37

Recommended Solution Hooray for plug-n-chug physics problems. Relativistic addition of velocities is solved with u0 =

u+v 1 + vu c2

(4.137)

Plug in the values given and, shortly thereafter, chug to get u0 =

0.3c + 0.6c 1+

(0.3c)(0.6c) c2

=

0.9c = 0.76c 1.18

(4.138)

Correct Answer (D)

Alternate Solution Velocity addition “sort of” still works in relativistic terms but it isn’t quite as simple as finding the sum of velocities. We know that the if we could just add up the velocities, then the velocity of the particle would be 0.9c. However, our ability to continually gain additional velocity drops off as we approach the speed of light, so the speed must be less than 0.9c and we eliminate (E). Since addition still “sort of works”, we would expect the speed to at least be greater than the speed of just the particle at 0.6c, so we can eliminate (A) and (B). Finally, you just have to decide whether (C) or (D) is a more reasonable speed. 0.66c is barely larger than the speed of the particle on its own so (D) is a more reasonable solution. Correct Answer (D)

343

4.38. PGRE9677 #38

4.38

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #38

Recommended Solution Consider Einstein’s equations for relativistic energy and relativistic momentum, Relativistic energy: Erel = γmc2 Relativistic momentum: Prel = γmv From these 2 equations, it should be clear that the only way we will get velocity from them is to divide Prel over Erel , which results in γmv Prel 5 M eV /c 1 = = = c Erel γmc2 10 M eV 2 Correct Answer (D)

344

(4.139)

4.39. PGRE9677 #39

4.39

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #39

Recommended Solution Ionization potential is lowest for atoms with full valence shells or nearly full valence shells because they generally don’t want to lose electrons. However, atoms with 1 or 2 additional electrons will be very likely to lose an electron and will have high ionization potential. (A) Full valence shell (B) Nearly full valence shell (C) Nearly full valence shell (D) Full valence shell (E) Cs has one additional electron, so it is most likely to lose that electron Correct Answer (E)

345

4.40. PGRE9677 #40

4.40

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #40

Recommended Solution From the Rydberg Formula,we get 

E = E0

1 1 − λ21 λ22



(4.140)

Finding the values needed for the equation gives us, λ2 = 470 nm

(4.141)

E0 = 4 × 13.6eV ≈ 55eV

(4.142)

hc ≈ 2 eV (4.143) λ Then, plug in our recently calculated values (Equations 4.141 4.142 and 4.143) to solve for λ1 E=



2eV = 55eV

1 1 − 2 λ1 16



=⇒ λ1 ≈ 3

Which is only true of solution (A) Correct Answer (A)

346

(4.144)

4.41. PGRE9677 #41

4.41

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #41

Recommended Solution You should be able to immediately eliminate (D) and (E). Option (D) clearly can’t be correct because it suggests that the electron has a spin quantum number of s = 3/2 which is never true (it must always be s = ±1/2). Option (E) can’t be true because you can’t, by definition, move to a different energy level and maintain the same energy. Next, recall that the angular quantum number for P corresponds to L = 1 and, in general (S, P, D, F, . . .) −→ (0, 1, 2, 3, . . .)

(4.145)

so (C) should have L = 1, not L = 3. Finally, between (A) and (B), you must recall your quantum number selection rules, specifically that transitions are allowed for ∆L = ±1 so (A) is allowed because L moves from P = 1 to S = 0 while (B) violates this allowed transition. Correct Answer (A)

347

4.42. PGRE9677 #42

4.42

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #42

Recommended Solution From the photoelectric effect, the equation for maximum kinetic energy is hc −φ (4.146) λ where φ is the Work Function. Plug all of your known values in and round everything to 1 significant figure to simplify things, UK = hν − φ =

UK

(4 × 10−15 eV · s)(3 × 108 m/s) −φ 5 × 10−7 m 12 × 10−7 eV · m = −φ 5 × 10−7 m 12 −φ = 5 = 2.4 − φ =

(4.147) (4.148) (4.149) (4.150)

Finally, Plug in your value for the work function into equation 4.150 and solve UK = 2.4 − 2.28 = 0.12 which is closest to (B). Correct Answer (B)

348

(4.151)

4.43. PGRE9677 #43

4.43

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #43

Recommended Solution The line integral in this problem moves about a circle in the xy-plane with its center at 0. Since we will be doing a line integral, we should probably know the general solution for a line integral about a curve, C Z b

I

f (t) f 0 (t)

(4.152)

f (x, y, z) f 0 (x, y, z)

(4.153)

f (s) =



a

C

So in our case, we need to solve the integral Z b





a

Since the circle is contained entirely in the xy-plane, z = 0. Additionally, since we are dealing with a circle, we will want to convert our values for x and y into their parametric equivalents. Specifically, x = R cos(θ) and y = R sin(θ). Replacing and x and y with our parametric equations gives us 4.155 Z b dx a

Z 2π

dy y −x dθ dθ



dθ

(4.154)

[(R sin(θ))(−R sin(θ)) − (R cos(θ))(R cos(θ))] dθ

(4.155)

0

Multiplying out and simplifying, you should get Z 2π

h

i

−R2 sin2 (θ) + cos2 (θ) dθ

(4.156)

0

Since sin2 (θ) + cos2 (θ) = 1, Equation 4.156 becomes Z 2π

−R2 dθ = −2πR2

(4.157)

0

The negative sign in 4.157 (and in fact the sign in general) is dependent on the direction in which you traverse the curve so we are primarily concerned with the magnitude of the solution. Correct Answer (C) 349

4.43. PGRE9677 #43

CHAPTER 4. PGRE9677 SOLUTIONS

Alternate Solution Kelvin-Stokes theorem tells us that the surface integral of the curl over a vector field is equivalent to the line integral around that vector field in a Euclidean 3-space. Mathematically, that is Z

(∇ × f ) · da =

I

f · dl

(4.158)

Which means that if we can take the curl of our function, u, and the integral of it over our area, then we have the line integral of the same function. Recall that the curl of a function can be calculated by taking the determinant of the matrix

i

j

∂ ∂x

∂ ∂y

Fx Fy



k

∂ ∂z

Fz

Applying this to our function, u and taking the determinant gives ˆi ∂z + ∂x ∂y ∂y 



∂z ∂y + ˆj − ∂x ∂z 



−∂x ∂y + kˆ − ∂x ∂y 



(4.159)

Since x, y and z are not functions of one another, everything in Equation 4.159 goes to 0 with the exception of the final term, which becomes −2. Substituting ∇ × u into Equation 4.158, Z 2π

−2 · dA = −2A

(4.160)

0

where the area of the circle is πR2 , so I

u · dl = −2A = −2πR2 Correct Answer (C)

350

(4.161)

4.44. PGRE9677 #44

4.44

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #44

Recommended Solution Velocity is given and since the derivative of velocity is acceleration, take the derviative. Take note, however, that v is a function of position (x) and position is a function of time (t) so use chain rule to get dv dx dx dt dv = (v) dx 
= −nβx−n−1 βx−n

(4.163)

= −nβ 2 x−2n−1

(4.165)

a =

Correct Answer (A)

351

(4.162)

(4.164)

4.45. PGRE9677 #45

4.45

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #45

Recommended Solution A simple method of generating a low-pass filter in a circuit involves placing a resistor in series with a load and a capacitor in parallel with that same load. low-frequency signals that attempt to pass through the circuit will be blocked by the capacitor and will be forced to pass through the load 352

4.45. PGRE9677 #45

CHAPTER 4. PGRE9677 SOLUTIONS

instead. Meanwhile, high-frequency signals will be able to bypass the capacitor with little to no effect. Of the possible solutions, only (D) provides us with a resistor in series with a load and a capacitor in parallel with that load. Correct Answer (D)

353

4.46. PGRE9677 #46

4.46

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #46

Recommended Solution From Faraday’s law of induction, we know that the potential (Electromotive Force) for a loop of wire is dφB |ε| = N dt

(4.166)

Since there is only one loop, it simplifies to dφB |ε| = dt

(4.167)

The problem tells us that the potential is, ε = ε0 sin(ωt), so we can set that equal to Equation 4.167 and substitute in φB = B · dA, to get ε0 sin(ωt) =

d (B · dA) dt

(4.168)

move dt over and integrate both sides Z

Z

ε0 sin(ωt) dt =

B · dA

− ε0 cos(ωt)/ω = B · A = BπR2 Finally, rearrange to solve for ω, 354

(4.169) (4.170)

4.46. PGRE9677 #46

CHAPTER 4. PGRE9677 SOLUTIONS

−ε0 cos(ωt) BπR2 and consider when the angular velocity is maximized, t = 0, to get ω=

ω=

−ε0 BπR2

Correct Answer (C)

355

(4.171)

(4.172)

4.47. PGRE9677 #47

4.47

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #47

Recommended Solution Faraday’s law gives us potential for a changing magnetic field as dφB dt

|| =

(4.173)

Where φB is the flux of the magnetic field through some area which is Z

φB =

B · dA

(4.174)

The area through which the flux passes is just the area of a circle with radius R so φB = BπR2 || =

 d  BπR2 dt

and since the flux is changing from the the rotation and the loops rotate at N rev s , 356

(4.175)

4.47. PGRE9677 #47

CHAPTER 4. PGRE9677 SOLUTIONS

|| = N BπR2

(4.176)

Correct Answer (C)

Alternate Solution Alternatively, you can go through a process of elimination by removing unlikely or impossible choices. (A) is clearly wrong because, by Faraday’s law, a changing magnetic field will generate a potential. (D) is also clearly wrong because it suggests that potential decreases as the rate of variance in the magnetic field increases. For (B), (C) and (E), check the units. rev kg m A s2

(B) 2πN BR ≡ (rev)(kg/s2 A)(m) ≡ (C) πN BR2 ≡ (rev)(kg/s2 A)(m2 ) ≡ (E) N BR3 ≡ (rev)(kg/s2 A)(m3 ) ≡

rev kg m2 A s2

=

rev A

J

rev kg m3 A s2

Of these, only (C) has the right units. Correct Answer (C)

357

4.48. PGRE9677 #48

4.48

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #48

Recommended Solution Mesons are hadrons so they have mass, which means they can’t exceed the speed of light and also can’t match the speed of light (i.e. the mass-less photon) so we eliminate (D) and (E). Next, let’s try to treat the motion as if it is non-relativistic. Only half of the π + mesons make it through the 15 meters, thus the amount of time to move the 15 meters is 1 half life or 2.5 × 10−8 seconds. From this, try ∆X 15 m = ≈ 5 × 108 m/s (4.177) ∆t 2.5 × 10−8 The value gives us a speed faster than light, meaning that what ever the actual speed is, it needs to be analyzed using relativistic equations. This tells us that the velocity must be very near c but less than it. v=

(A) (B) (C)

1 2

C: Not nearly fast enough for relativistic influence to take effect

q

2 5

C ≈ 0.6 C: Very similar to (A) and likely not fast enough to have significant relativistic effect.

√2 5

C ≈ 0.9 C: This value is the closest to justifying relativistic influence so it is the most likely to be correct. Correct Answer (C)

358

4.48. PGRE9677 #48

CHAPTER 4. PGRE9677 SOLUTIONS

Alternate Solution To calculate the value exactly, start with your equation for “proper time”, ∆τ 2 = ∆t2 − ∆X 2

(4.178)

∆t2 = ∆τ 2 + ∆X 2

(4.179)

We know, from the half-life given in the problem and the fact that only half of the sample makes it through the 15 meters, that it takes one half-life of time to move 15 meters. Substitute in 15 for the position and 15/2 for the time, 2

2

2

2

"  2

∆t = (15/2) + 15 = 15

1 2

#

+ 1 = 225



1 +1 4



(4.180)

Take the square root of both sides to get s

∆t =



225

  q 1 5/4 + 1 = 15 4

(4.181)

Now using the basic definition for velocity (i.e. v = ∆X/∆t), with distance ∆X = 15, we get ∆X 15 1 2 = p =p =√ ∆t 15 5/4 5/4 5 Correct Answer (C)

359

(4.182)

4.49. PGRE9677 #49

4.49

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #49

Recommended Solution Relativistic influences on Electricity and Magnetism occurs proportionally to the lorentz factor, γ. Since Ez = 2σ0 , we would expect the Electric field to be influenced by Ez γ =

σ 1 p 20 1 − v 2 /C 2

Correct Answer (C)

360

(4.183)

4.50. PGRE9677 #50

4.50

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #50

Recommended Solution The space time interval equation is ∆S 2 = −(c ∆t)2 + ∆X 2

(4.184)

The space time interval is S = 3C · minutes and the position interval is ∆X = 5c · mintues. Plug these into Equation 4.184 9C 2 = −(C ∆t)2 + 25C 2

(4.185)

∆t = 4 minutes

(4.186)

Then, solve for ∆t to get

Correct Answer (C)

361

4.51. PGRE9677 #51

4.51

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #51

Recommended Solution The solution to the infinite square well (also known as the particle in a box) is a sine function in which the number of nodes on the wave is n + 1. In the ground state, you will have 2 nodes at the end of the infinite walls with the peak of the wave at exactly the middle. For n = 2, we get 3 nodes with two peaks (or a peak and a trough if you insist) and the middle of the wave falls on a node. If you keep checking all of the values for n, you will find that all states with even values for n result in a node in the middle of the well.

Correct Answer (B)

Alternate Solution The solution to the infinite square well is 

ψn (x, t) = A sin 362

nπx L



(4.187)

4.51. PGRE9677 #51

CHAPTER 4. PGRE9677 SOLUTIONS

If we are only concerned with the middle position, set x = L/2, giving 

ψn (x, t) = A sin

nπ 2



(4.188)

and this equation will go to zero any time you are taking the sine of integer values of π (i.e. sin(0), sin(π), sin(2π), . . .). Correct Answer (B)

363

4.52. PGRE9677 #52

4.52

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #52

Recommended Solution For the spherical harmonics, we get a sine term for the harmonic Y11

1 =− 2

r

3 sin(θ)eiπφ 2π

(4.189)

and r

1 3 sin(θ)eiπφ (4.190) = 2 2π which means that m = ±1. The eigenvalues of a spherical harmonic can be found with Y1−1

LZ ψ = m¯hψ

(4.191)

LZ ψ = ±1¯hψ

(4.192)

Plugging in our value for m gives

which is (C). Correct Answer (C)

364

4.53. PGRE9677 #53

4.53

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #53

Recommended Solution Positronium atoms can only decay into even numbered groupings of photons (to conserve spin) so we can eliminate choices (B) and (D). Positronium atoms are very unstable and since energy must be conserved, there is no way that the atom will decay without releasing some photons so we can eliminate (A). Finally, between 2 photons and 4 photons, consider how silly a feynman diagram will look with 4 photon lines emanating from the interaction. In case I wasn’t being clear, it is wicked silly. Correct Answer (C)

365

4.54. PGRE9677 #54

4.54

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #54

Recommended Solution Electromagnetic waves are typically thought of as being composed of a magnetic and electric wave moving in the same direction and oscillating orthogonally (Figure 4.8). The summation of any two or more of these waves will occur in one of two forms. One of the forms involves waves oscillating in the same plane as one another and the other involves oscillations in different plans. In general, any waves in the same plane will result in field vectors also in the same plane, resulting in a trajectory that moves in only 2 dimensions. Field vectors in an alternate plane will sum to a 3 dimensional rotational trajectory. For the specific problem here, the second wave is rotated π or 180◦ so the field vectors lie in the same plane and the final trajectory must be linear. Based on this, you can eliminate (C), (D), and (E). In order to decide between (A) and (B), realize that two waves perfectly in phase (i.e. with no rotation) will have field vectors in the same quadrant and so the angle will be 45◦ . On the other hand, a wave rotated π from the other will be in different quadrants and so the angle will be 135◦ . 366

4.54. PGRE9677 #54

CHAPTER 4. PGRE9677 SOLUTIONS

Figure 4.8: Classical view of elctromagnetic wave behavior Correct Answer (B)

367

4.55. PGRE9677 #55

4.55

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #55

Recommended Solution By Malus’ Law, the intensity of a an electromagnetic wave after passing through a perfect polarizer is 1 I = c0 E02 cos2 (θ) (4.193) 2 The optical path difference for the second wave is z = 2π/k. Plugging this into the original equation and letting z → 0 and t → 0. E=x ˆE1 ei(kz−ωt) + yˆE2 ei(kz−wt+π) = x ˆE1 e0 + yˆE2 ei(2π+π) Recall Euler’s identity,

eiπ

(4.194)

= 1, so the Equation 4.194 becomes E = E1 + E2 368

(4.195)

4.55. PGRE9677 #55

CHAPTER 4. PGRE9677 SOLUTIONS

Since the two waves are decoupled, the magnitude of the entire wave will be the magnitude of each individual wave added separately I = (E1 )2 + (E2 )2 Correct Answer (A)

369

(4.196)

4.56. PGRE9677 #56

4.56

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #56

Recommended Solution Immediately eliminate choice (A) because there is no way you are going to get an angle of 0◦ between the water surface and the light source. Next, eliminate (E) because an angle of 90◦ will result in no bending of the light and so there will be no total internal reflection. As for the last 3, you can probably intuitively figure that it won’t be at 25◦ simply because that isn’t a very steep angle. However, if we want to be more rigorous, we must use Snell’s law. Recall that total internal inflection occurs in any instance in which using Snell’s law would require you to take the sine of an angle and get a value that isn’t possible. In our scenario, if we apply an angle of 50◦ , we get sin(θ1 ) n2 = sin(θ2 ) n1 sin(θ2 ) = 1.33 sin(50◦ )

(4.198)

= 1.02

(4.199)

(4.197)

However, there is no angle at which the sine function will give you a value larger than 1 and this means you’ve reached total internal reflection. Admittedly, it is a bit rude on the part of ETS to simultaneously give you relatively complicated decimals and lesser used angles3 with no calculator, but at the very least you could quickly get rid of the angle 25◦ by using this method. Then, knowing that ETS is looking for the minimum angle for total internal reflection, do an approximation (e.g. 45◦ as an approximation for θ = 50◦ ) to determine which potential angle is closer to pushing our value over 1. Correct Answer (C)

3

i.e. not one of the angles we’ve all memorized from the unit circle

370

4.57. PGRE9677 #57

4.57

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #57

Recommended Solution The equation relating slit width for a single slit diffraction to wavelength is

d sin(θ) = λ d =

λ sin(θ)

(4.200) (4.201)

Convert the wavelength into meters, giving d=

(4 × 10−7 m) sin(4 × 10−3 rad)

(4.202)

For small angles, we can make the approximation sin(θ) = θ, which gives us d=

(4 × 10−7 m) (4 × 10−3 )

d = 1 × 10−4 m Correct Answer (C)

371

(4.203) (4.204)

4.58. PGRE9677 #58

4.58

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #58

Recommended Solution We are trying to convert a “well-collimated” laser with diameter of 1 mm to a “well-collimated” laser of 10 mm, which represents a magnification of 10×. Lens magnification must have the same proportionality for the focal length as it does for the separation distance, so we would expect a focal length that is ten times larger than the 1.5 cm lens, i.e. 15 cm. We can eliminate (A), (B) and (C) from this. Finally, we need to decide whether the distance for the new lens will be 15 cm or 16.5 cm. Assuming everything is done properly, the first lens will have its focus at 1.5 cm and the second lens will have its focus at a distance of 15 cm. For ideal collimation, we will want the foci to be at the same location, so 15 cm + 1.5 cm = 16.5 cm Correct Answer (E)

372

(4.205)

4.59. PGRE9677 #59

4.59

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #59

Recommended Solution The energy of a single photon with wavelength, λ = 600 nm can be found by hc (4.206) λ The total number of photons emitted per second will be equal to the total energy generated per second divided by the energy of a single photon, i.e. E=

# of photons =

100 watts hc/λ

(4.207)

Plug everything in and rounding all of our numbers to simplify the mental math, you should get # of photons =

(1.0 × 104 W)(6 × 10−7 m) 1 ≈ 3 × 1023 −34 8 (6.63 × 10 J · s)(3 × 10 m/s) s

(4.208)

However, the question asks for the number of photons per femtosecond, so convert the previous solution to the right units to get # of photons ≈ 3 × 1023

1 1 ≈ 3 × 108 s fs

Which is closest to (B) Correct Answer (B)

373

(4.209)

4.60. PGRE9677 #60

4.60

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #60

Recommended Solution Keep in mind while we work this problem that we don’t necessarily want the most accurate answer, just the quickest method to the correct choice. The Lyman alpha line is the spectral line corresponding to a hydrogen atom transition from level n=2 to n=1. The question poses this problem in a way that would indicate that we should use the Doppler shift equation for light. If we wanted to be extremely accurate, and let me stress how much we DON’T want this, then we would need to use the relativistic equations for the Doppler shift. Note, however, that the largest possible speed for the particle given in the answers, is 2200 km/s. Convert that to m/s and compare it to the speed of light, for our purposes we will just call it C = 3.0 × 108 m/s. The fastest this particle could possibly be moving, according to the potential solutions, would be roughly 0.73% the speed of light 2.2 × 106 m/s = 0.0073 3.0 × 108 m/s

(4.210)

Relativistic effects will be “relatively” negligible at these speeds so let’s just use the nonrelativistic equation, ∆λ v = (4.211) λ C With, λ = 122 nm = 1.22 × 10−7 m −12

∆λ = 1.8 × 10

m

8

C ≈ 3.0 × 10 m

(4.212) (4.213) (4.214)

Since we only care about an approximation, pretend that the two wavelength values of 1.22 and 1.8 are just 1. Plug this all in to get (3.0 × 108 m)(1.0 × 10−12 m) = 3.0 × 103 m/s (1.0 × 10−7 m) 374

(4.215)

4.60. PGRE9677 #60

CHAPTER 4. PGRE9677 SOLUTIONS

CAREFUL! The problem asked for the answer in km/s, not m/s like we’ve solved for. Convert 3.0 × 103 m/s to get 3.0 km/s which is closest to answer (B). Correct Answer (B)

375

4.61. PGRE9677 #61

4.61

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #61

Recommended Solution Gauss’s law gives us ~ · dA ~ = q/0 E

(4.216)

Taking the integral of both sides and substituting the equation for surface area of a sphere for As Z

~ · dA~s = EA = E

Z R/2 ρAs

0

0



E 4πr2



=

Z R/2 1 

A(r)2

dr



0  5 4πA R 50 2

(4.217) 

4π(r)2 dr

(4.218)

0

R E4π 2 

2

=

(4.219)

Start canceling things out and you get A E= 50



R 2

3

=

AR3 400

Correct Answer (B)

376

(4.220)

4.62. PGRE9677 #62

4.62

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #62

Recommended Solution We can calculate Q1 and Q2 when the battery is in the system, using Q1 = C1 V = (1.0 mF)(5.0 V) = 5 mF V

(4.221)

Q2 = C2 V = (2.0 mF)(5.0 V) = 10 mF V

(4.222)

Once the battery is removed the problem tells us that the capacitors are connected to one another such that the “opposite charges are connected together”. Doing this, we get Figure 4.9

Figure 4.9: Circuit with opposite charges connected together The potential for both capacitors will be the same so we can calculate V for both capacitors with V =

Qeq Ceq

(4.223)

For a parallel circuit, the orientation of this circuit has its capacitors flipped so Qeq = Q2 − Q1 V =

Q2 − Q1 (10 mF V ) − (5 mF V ) 5 = = V ≈ 1.7 V C1 + C2 (1 mF ) + (2 mF ) 3 Correct Answer (C)

377

(4.224)

4.63. PGRE9677 #63

4.63

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #63

Recommended Solution (A) Muon: The Muon is one of the leptons, which are the fundamental particles. Muons are similar to electrons in that they have a negative charge and a spin of 1/2. The Muon IS NOT a composite object (B) Pi-Meson: All Pi-Mesons (also known as a Pion) are composed of some combination of the first generation quarks (Up quark and Down Quark). The Pion IS a composite particle (C) Neutron: Neutrons have a tri-quark arrangement, 1 Up quark and 2 Down quarks. The Neutron IS a composite particle (D) Deuteron: A deuteron, the nucleus of a Deuterium atom, is composed of a proton and a neutron as opposed to hydrogen which has just a proton. The Deuteron IS a composite particle (E) Alpha particle: Alpha particles are composed of 2 Neutrons and 2 Protons. The Alpha particle IS a composite particle. Correct Answer (A)

378

4.64. PGRE9677 #64

4.64

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #64

Recommended Solution In symmetric fission, the heavy nucleus is split into two equal halves that we are going to assume are both an example of a “medium-weight nucleus”, per the problem description. The kinetic energy after fission will be the difference between the initial total energy and the energy remaining in the two ”medium-weight” nuclei. ∆E = Eheavy N − E2 medium N

(4.225)

The energy of the heavy nucleus is given in the description as 8 million eV/nucleon and the energy of the 2 medium nuclei is 7 million eV/nucleon. MeV MeV ∆E = (1 nucleus) 8 (N ) − (2 nuclei) 7 nucleon nucleon 







1 N 2

∆E = (8 MeV − 7 MeV)N = N (1 MeV)



(4.226) (4.227)

Then taking some kind of heavy nucleus, say Uranium-238 (N=238), you get ∆E = 238 MeV Which is roughly (C). Correct Answer (C)

379

(4.228)

4.65. PGRE9677 #65

4.65

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #65

Recommended Solution The quickest method for solving this problem is to consider when the mans mass goes to infinity. When this occurs, the man won’t move at all, do to his infinite mass, and so the only energy will involve the movement of the boat with kinetic energy Uk = 12 M v 2 . The only choice that gives this equation at the limit is (D). Correct Answer (D)

Alternate Solution Energy and momentum are conserved, so we get the following equations

Ptot = mv − M V = 0

(4.229)

mv = M V

(4.230)

1 1 Utot = mv 2 + M V 2 2 2 All of the possible solutions involve only M , v and m, so solve for V . 380

(4.231)

4.65. PGRE9677 #65

CHAPTER 4. PGRE9677 SOLUTIONS

V =

mv M

(4.232)

Substitute this into Equation 4.231 Utot

1 1 = mv 2 + M 2 2 1 m2 m+ 2 M



mv M

(4.233)

!

v2

Correct Answer (D)

381

2

(4.234)

4.66. PGRE9677 #66

4.66

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #66

Recommended Solution The problem clearly states that the spacecraft is “on a mission to the outer planets” meaning that it must have an orbit with an escape trajectory. Only Parabolic and Hyperbolic orbits are escape trajectories so eliminate (A), (B) and (C). Choosing between (D) and (E), go with (E) because hyperbolic orbits occur at high velocities, like 1.5 times the speed of Jupiter, while parabolic orbits happen at lower speeds and are generally more rare. Correct Answer (E)

382

4.67. PGRE9677 #67

4.67

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #67

Recommended Solution Recall the equation for the event horizon (Schwarzschild radius) of an object is 2GM (4.235) c2 The mass of the earth is given and G and C can be found in the list of constants in your GRE booklet. Plug these all in to get Re =

Re =

2(6.67 × 10−11 m3 /kg · s2 )(5.98 × 1024 kg) (3.0 × 108 m/s)2

(4.236)

Rounding all of the numbers and multiplying out gives Re =

84 × 1013 m3 /s2 ≈ 1 × 10−3 m = 1 cm 9 × 1016 m2 /s2 Correct Answer (C)

383

(4.237)

4.68. PGRE9677 #68

4.68

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #68

Recommended Solution A quick trick to figuring out the solution to this is to consider which variables the Lagrangian must be dependent on. Looking over the diagram and considering that the Lagrangian is the difference between Kinetic and Potential energy, you should be able to convince yourself that it will have a dependence on (m, s, θ, ω). Look through each of the potential solutions to see that only (C) and (E) match this criteria. Now, consider when the angle between the rod and the vertical is 0 (θ = 0). In this case, the potential energy term should go to 0 because the bead won’t be able to move up the rod if the axis of rotation is parallel to the rod. Thus, we must have a term with a sine or tangent function in it so that the angle is forced to 0 in the potential energy term, which is only the case for (E). Correct Answer (E)

384

4.69. PGRE9677 #69

4.69

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #69

Recommended Solution Remember our good friend the right hand rule, which gives us, among other things, the direction of magnetic field vectors from a moving current (Note: this is the general principle behind how a solenoid works). From the diagram given in the problem, we have a current going “into the page” on the right and coming out of the page on the left. If you’ve done everything right and aren’t too embarrassed to be making hand gestures at your computer screen/test booklet, then you are giving the problem a thumbs up (I’m going to assume that this isn’t a case of you approving of the GRE). From this, only (A) and (B) could be correct. The only difference between (A) and (B) is a dependence on r. If we were talking about current I, rather than current density, J, then we would care about the radius (i.e. the size) of the conductive cables. However, since the problem talks about everything in terms of a constant current density, J, we know there should be no dependence on r. In case you aren’t convinced, compare Maxwell’s equations for magnetic fields with dependence on I vs dependence on J. B=

µ0 I 4π

Z

385

dl × rˆ r2

(4.238)

4.69. PGRE9677 #69

CHAPTER 4. PGRE9677 SOLUTIONS

as compared to ∇ × B = µ0 J + µ0 0

∂E ∂t

(4.239)

Correct Answer (A)

B

I

I B

Figure 4.10: Magnetic fields generated as the result of a moving electric field

386

4.70. PGRE9677 #70

4.70

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #70

Recommended Solution The Larmor Formula can be used to calculate the power radiated in non-relativistic motion of a charged particle P =

q 2 a2 6π0 c3

(4.240)

The problem states that particle B has half the mass ( 12 m), twice the charge (2q), three times the velocity (3v) and four times the acceleration (3a). The Larmor Formula isn’t dependent on mass or velocity so we are only concerned with charge and acceleration. Since the denominator of the Lamor Formula won’t be altered for either particle, we only care about the numerator. 2 16a2 ) PB (q 2 a2B ) (4qA A = B = = 64 2 a2 ) 2 a2 ) PA (qA (q A A A

Correct Answer (D)

387

(4.241)

4.71. PGRE9677 #71

4.71

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #71

Recommended Solution The angle of deflection for this particle can be calculated as y˙ x˙ We already know the velocity in the x direction as v, so tan(θ) =

(4.242)

y˙ (4.243) v Force due to gravity will be minimal relative to the Lorentz force so we’ll assume it is 0. This gives us a net force of tan(θ) =

~ + (~v × B)] ~ Fnet,y = m¨ y = q[E

(4.244)

The problem says nothing about a magnetic field existing (even though it should) so we’ll ~ = 0, giving assume B

~ = m¨ y = qE y¨ =

qV dm

qV d

(4.245) (4.246)

We know that the velocity in the x-direction is x˙ = L/t and the acceleration in the y-direction is y¨ = y/t. ˙ Solve for t in both equations and equate the two, to get

388

4.71. PGRE9677 #71

CHAPTER 4. PGRE9677 SOLUTIONS

L/x˙ = y/¨ ˙ y

(4.247)

Solve for velocity in the y-direction to get y˙ = L¨ y /x˙

(4.248)

We already solved for y¨ in Equation 4.246, so we can plug that in to get y˙ =

LqV dmx˙

(4.249)

then, plug 4.249 into 4.242 to get tan(θ) =

LqV dmxv ˙

(4.250)

and since x˙ = v, we can solve for θ to get θ = tan−1



LV q dmv 2



Correct Answer (A)

389

(4.251)

4.72. PGRE9677 #72

4.72

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #72

Recommended Solution This solution is far from rigorous but it is the quickest way to solve the problem. Negative feedback and positive feedback function similarly in an electronic circuit as it does in acoustics. Positive feedback of an audio wave involves an increase in amplitude for the wave. Think along the lines of placing a microphone too close to a speaker and making that high pitched squeal. Negative feedback, on the other hand, should cancel out some of the amplitude. From this, you should immediately know that (A) can’t be an aspect of negative feedback. Correct Answer (A)

390

4.73. PGRE9677 #73

4.73

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #73

Recommended Solution For a thermodynamic expansion, work done is Z Vf

P dV

W =

(4.252)

Vi

The problem gives us the ideal adiabatic expansion equation as PV γ = C

(4.253)

C = CV −γ (4.254) Vγ Making the substitution into the work equation and taking the integral should give us P =

Z Vf

W =C

V Vi

−γ

C dV = V 1−γ 1−γ 

Vf

(4.255) Vi

At this point, you can see that the denominator must contain a 1 − γ term and you can choose (C). However, if you want you can always substitute in C to get Pf V f − Pi V i 1−γ Correct Answer (C)

391

(4.256)

4.74. PGRE9677 #74

4.74

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #74

Recommended Solution Spontaneous events in a thermodynamic system always have positive value changes in entropy so get rid of (D) and (E). Additionally, since we know that the change in entropy is ∆S =

Z T2 dq T1

T

(4.257)

We are going to get a natural log component (unless one of our temperatures is 0 which is not the case) so eliminate (A). If you can’t get any farther than this, at least you go it down to 2 choices and you can guess. The next step you should take is to use Equation 4.257 to calculate the net change in entropy as the sum change in entropy of the 2 masses ∆Snet = ∆S1 + ∆S2

(4.258)

Because the masses are of the same size, they will both reach a temperature as an average of the two 500 K + 100 K = 300 K 2 then, using dq = mCdT and equation 4.259, we get ∆Snet = mC

Z 300 dT 100

T1

+

 Z 300 dT 500

T2

= mC[ln(3) + ln(3/5)] = mCln(9/5)

(4.259)

(4.260)

Correct Answer (B) Additional Note In the event that ETS didn’t give us objects with the same mass, it should be relatively straightforward to calculate the final heat of the system by the conservation of energy (heat) as Q1 + Q2 = 0.

392

4.75. PGRE9677 #75

4.75

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #75

Recommended Solution Fourier’s law of heat conduction gives ∂Q ∂t

= −k

I

~ ∇T · dA

(4.261)

S

∆T (4.262) ∆X Which tells us that heat transfer is proportional to the thermal conductivity of the material, k and the cross-sectional area of the material but inversely related to the length the heat transfers through. In this problem, the cross sectional area is the same for both, so the ratio is Q = −kA

QA (0.8 watt/m◦ C)/(4 mm) = = 16 QB (0.025 watt/m◦ C)/(2 mm) Correct Answer (D) 393

(4.263)

4.76. PGRE9677 #76

4.76

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #76

Recommended Solution Analysis of Gaussian wave packets are fascinating because some of the more interesting and familiar quantum mechanical laws fall out of them. In particular, the Heisenberg Uncertainty principle is one of those results ¯ h (4.264) 2 This is relevant because the uncertainty principle tells us that the wavepackets momentum can never be 0, meaning I is not possible. Eliminate (A), (C), (E). Now, compare II and III. If it’s true that the“width of the wave packet increases with time” then it isn’t possible for the statement “Amplitude of the wave packet remains constant with time” to also be true, since the wave packet stretching in time would alter the amplitude. From this, (D) can’t be true, leaving you with choice (B). ∆x ∆p ≥

Correct Answer (B)

394

4.77. PGRE9677 #77

4.77

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #77

Recommended Solution Recall that the expectation value for energy is given by the Hamiltonian operator as hH(t)i = hψ|H(t)|ψi

(4.265)

H = −JS1 · S2

(4.266)

S12 ψ1 = S1 (S1 + 1)ψ1

(4.267)

S22 ψ2 = S2 (S2 + 1)ψ2

(4.268)

We are given the Hamiltonian

and spin operators

We can use the polynomial identity a1 · a2 =

i 1h (a1 + a2 )2 − a21 − a22 2

to get 395

(4.269)

4.77. PGRE9677 #77

CHAPTER 4. PGRE9677 SOLUTIONS

J hψ1 |H(t)|ψ2 i = − [(S1 + S2 )2 − S12 − S22 ] 2 Substitute in the spin operators (Equations 4.267 and 4.268) to get hH(t)i = −

J [(S1 + S2 )(S1 + S2 + 1) − S1 (S1 + 1) − S2 (S2 + 1)] 2 Correct Answer (D)

396

(4.270)

(4.271)

4.78. PGRE9677 #78

4.78

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #78

Recommended Solution Semiconductors are useful devices because it is relatively straightforward to alter a semiconductors conductive properties by intentionally adding impurities to the semiconductor lattice. The process of adding these impurities is known as doping. Dopants in the lattice of an n-type semiconductor alter conductivity by donating their own weakly bound valence electrons to the material. This is precisely the description of (E). In general, you should try to remember that dopants in an n-type semiconductor always contribute electrons to the lattice rather than take from the lattice (those would be p-type semiconductors). Once you’ve concluded that the only solutions could be (D) or (E) it should seem reasonable that electrons that are donated aren’t going to get donated to a full valence shell. Correct Answer (E)

397

4.79. PGRE9677 #79

4.79

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #79

Recommended Solution In its most general form, heat capacity C is given by ∆Q C= (4.272) ∆T Which at the very least gives us our temperature dependence for the heat capacity. Specific to this problem, recall the energy level diagrams from thermodynamics. the heat capacity of an ideal gas is proportional to the sum of the degrees of freedom for each of the three energy levels. For a monotonic gas, particles will have three translational degrees of freedom corresponding to the three components of motion (~x, ~y , ~z) 



3 Cv = R (4.273) 2 For a diatomic molecule, we have to figure into the heat capacity the linear vibrational energy and the rotational energy 3 Cv = R + Rvib + Rrot (4.274) 2 From the energy level diagram (Figure 4.11), we can see that small changes in energy level correspond to translational motion. However, large quantities of energy are required for vibrational and rotational energy to play a part. From this, we get the low temperature (i.e. low energy) heat capacity as 3 Cv−low = R 2 and high temperature (i.e. high energy) heat capacity as 3 7 Cv−high = R + Rvib + Rrot = R 2 2 So the ratio of high temperature heat capacity to low temperature heat capacity is Cv−high /Cv−low =

398

7/2 R 7 = 3/2 R 3

(4.275)

(4.276)

(4.277)

4.79. PGRE9677 #79

CHAPTER 4. PGRE9677 SOLUTIONS

V1

Electronic Energy Levels

V0

Vibrational Energy Levels E n e r g y

R3 R2 R1

V1

R3 R2 R1

V0

Rotational Energy Levels

Figure 4.11: Energy level diagram and electron transitions Correct Answer (D)

399

4.80. PGRE9677 #80

4.80

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #80

Recommended Solution Immediately get rid of (A) and (E) because they suggest no dependence on the mass of either string. Next, consider the scenario in which the string on the right becomes infinitely massive (i.e. µr → ∞). When this occurs it won’t be possible for any amount of energy on the left string to create an amplitude on an infinitely massive string on the right and so amplitude should go to 0. Under this condition (B) will become 2, (D) will go become -1 so these can’t be correct. (C) is the only one which goes to 0 when µr → ∞. Correct Answer (C)

400

4.80. PGRE9677 #80

CHAPTER 4. PGRE9677 SOLUTIONS

Alternate Solution Consider the case when µl = µr . In this case, the two part string of different masses becomes a single string with one mass, call it µ, and the original amplitude of 1 will be maintained. From this (E) and (D) can be eliminated and (A) can be eliminated simply because it doesn’t acknowledge the dependency on string mass. Finally, eliminate (D) because the amplitude doesn’t go to 0 when the mass, µr goes to infinity. Correct Answer (C)

401

4.81. PGRE9677 #81

4.81

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #81

Recommended Solution This is one of the few problems I would recommend doing the math in full gritty detail. The number of beats between two waves comes from the difference in frequency between them. Beats can be observed (heard), for example, when two musical instruments are out of tune. To minimize the beats with a frequency of 73.416 Hz for D2 , we will need the harmonic multiplied by that frequency to be very close to 440 Hz. Of the harmonics given, 6 is the most reasonable both from the perspective of quick mental math and from the perspective that ETS likes to keep you from getting the correct answer by knowing only one of the pieces of information (i.e. there are two solutions with a harmonic of 6). Multiplying everything out completely, you should get (73.416 Hz)(6) = 440.49600 Hz ≈ 440.5 Hz

(4.278)

Since the number of beats is just the difference between the two frequencies, 440.5 Hz − 440.0 Hz = 0.05 Beats Correct Answer (B)

402

(4.279)

4.82. PGRE9677 #82

4.82

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #82

Recommended Solution

Figure 4.12: Reflection of light on a thin film The equation for constructive interference of a thin film (Figure 4.12 is 1 2nd = m + λ 2 



(4.280)

Plug in n = 1 for air our value for the wavelength as 488 nm, 1 2d = mλ + λ 2 

d=m

448 nm 2





+

448 nm 4

From this, we get

403

(4.281)



= m(244 nm) + 122 nm

(4.282)

4.82. PGRE9677 #82

CHAPTER 4. PGRE9677 SOLUTIONS

When m = 0: d = 122 nm When m = 1: d = 366 nm When m = 2: d = 610 nm Correct Answer (E)

404

4.83. PGRE9677 #83

4.83

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #83

Recommended Solution Note that when d goes to infinity, every peak and trough will be infinitely tall/short and you won’t even need a velocity to get the ball to free fall (i.e. when d → ∞ then v → 0). The only equation which fits the bill is (D). Correct Answer (D)

405

4.84. PGRE9677 #84

4.84

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #84

Recommended Solution Immediately get rid of (A) as it suggests that the normal mode has no dependence on either of the masses. Next, note that if we let the mass of one of our masses, say m2 go to infinity, then the other mass will oscillate about that mass as if it were connected to a stationary object. When this happens, dependence on mass m2 should disappear but dependence on m1 should remain. For (B) and (E), allowing m2 → ∞ forces the entire term, including m1 to disappear. Finally, comparing (C) and (D), get rid of (C) because it suggests that the normal mode has no dependence on the acceleration due to g or l, which is not true. Correct Answer (D)

406

4.85. PGRE9677 #85

4.85

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #85

Recommended Solution Taking the recommendation at the end of the problem, consider the limiting cases of M → 0 and M → ∞. Looking through the 5 options, you can immediately eliminate (D) and (E) because they both suggest that the mass of the string and the mass of the ring have no influence on the wavelength, which is not correct. Considering the first limiting case, when M → 0 then µ/M → ∞. In the case of (C), a sine function is going to limit the maximum and minimum values so we can eliminate (C). since (x) sin(x) cot(x) = tan1(x) = cos sin(x) and tan(x) = cos(x) , both can blow up to infinity if the bottom trig function goes to 0. Considering the second limiting case, when M → ∞ then µ/M → 0. It is also the case that when mass goes to infinity, the ring won’t move from any amount of force placed on the string so what we have is a fixed end on the ring side. This means that the only wavelengths possible are 407

4.85. PGRE9677 #85

CHAPTER 4. PGRE9677 SOLUTIONS

lengths of L = n2λ . This is the case because the fixed end on each side acts as a node and you have a bound standing wave. Checking this requirement with both (A) and (B), gives −1 (A) µ/M = 0 = cot( 2πL λ ) =⇒ cot (0) = π/2 + n π = −1 (B) µ/M = 0 = tan( 2πL λ ) =⇒ tan (0) = 0 + n π =

2πL λ

2πL λ

=⇒ L =

Of which, only (B) meets the necessary criteria. Correct Answer (B)

408

=⇒ L =

n 2

nλ 2

+

1 4

4.86. PGRE9677 #86

4.86

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #86

409

4.86. PGRE9677 #86

CHAPTER 4. PGRE9677 SOLUTIONS

Recommended Solution ~ will exhibit As a general rule, particles moving in an orthogonal direction to a magnetic field (B) cyclotron (helix shaped) motion with a direction of spin in agreement with the right hand rule. Of the choices, only (B) and (E) exhibit this phenomena and only (B) demonstrates actual helical motion. Correct Answer (B)

410

4.87. PGRE9677 #87

4.87

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #87

Recommended Solution Immediately eliminate (D) and (E) because they both suggest that charges in a magnetic field won’t cause the pith balls to move which is incorrect. Eliminate (B) because it doesn’t have a dependence on R and if R went to 0, the magnetic field would as well. Finally, consider that if we had a dependence on d as in (C) and let d → ∞, then angular momentum would also need to be infinite which would violate conservation of the momentum built up from the magnetic field. Correct Answer (A)

411

4.88. PGRE9677 #88

4.88

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #88

Recommended Solution Immediately eliminate all solutions that suggest that the magnetic field is anything other than 0 at the origin, specifically (E) and (D). There are a number of ways you can convince yourself

412

4.88. PGRE9677 #88

CHAPTER 4. PGRE9677 SOLUTIONS

this criteria must be true. For example, consider integrating the magnetic field generated over an infinitely small surface area which would give you no magnetic field at all. Alternatively, recall that a larger magnetic field vector is indicative of a larger magnetic field and that the magnetic field ~ = 0 (Figure 4.13). vectors get smaller as R gets smaller, until it approaches R = 0 when B

Figure 4.13: Magnetic field vectors decrease in magnitude as R → 0 Next, eliminate all solutions which don’t suggest that the magnetic field at point c and all radii larger than c is 0., i.e. (C) and (A). You can convince yourself that this must be the cases because the two cables have equivalent magnetic fields moving in opposite directions which will cancel each other out at and past point c. Correct Answer (B)

413

4.89. PGRE9677 #89

4.89

CHAPTER 4. PGRE9677 SOLUTIONS

PGRE9677 #89

Recommended Solution ~ In As soon as you see a charged particle in a magnetic field, think Lorentz force, F = q(~v × B). our particular problem the velocity vector and magnetic field vectors are orthogonal, so the cross ~ = vB. Since the object is rotating the net force on the particle should also be product of ~v × B 2 equal to the centripetal force, F = mRv . Set these two equations equal to one another and solve for momentum, p = mv. m v2 R

(4.283)

m v = qBR

(4.284)

qvB =

Using Pythagorean theorem, solve for the hypotenuse of the triangle drawn in the diagram, which also happens to be our radius R. You should get 414

4.89. PGRE9677 #89

CHAPTER 4. PGRE9677 SOLUTIONS

R2 = l2 + (R − s)2

(4.285)

R2 = l2 + R2 − 2sR + s2

(4.286)

l2 − 2sR + s2 = 0

(4.287)

expand the (R − s)2 term to get

Now, recall that s c, two events occur at the same time light-like: if |∆x/∆t| = 0, two events are connected by a signal that moves at the speed of light. Of the choices, the problem is describing a space-like interval which is (C). Correct Answer (C)

470

5.35. PGRE0177 #35

5.35

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #35

Recommended Solution According to our models for simple blackbody radiation, a blackbody’s power is proportional to the 4th power of its temperature P = κT 4

(5.107)

From this, the increase in power from tripling the temperature is P = κ(3T )4 = 81κT 4 which is (E). Correct Answer (E)

471

(5.108)

5.36. PGRE0177 #36

5.36

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #36

Recommended Solution An adiabatic expansion is not the same thing as an isothermal expansion and an isothermal expansion is the only type which maintains a constant temperature. Additionally, from the ideal gas equation we know that a change in volume should be accompanied by a change in temperature P ∆V = nR∆t Correct Answer (E)

472

(5.109)

5.37. PGRE0177 #37

5.37

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #37

Recommended Solution The quickest method to determine the thermodynamic work is to estimate the area under the P-V curve. The curve is roughly triangular shaped with base of 2 and height of 300. Calculate the area under the triangle as 1 1 P V = (2)(300) ≈ 300 (5.110) 2 2 this eliminates all but (B) and (D). We can then determine whether the work is negative or positive by the direction of the process. Recall that a clockwise process represents a heat engine and the work is positive. Alternatively, if the process is counter-clockwise, it’s a heat pump and the work is negative. This process is counter-clockwise so we choose (D). Correct Answer (D)

473

5.38. PGRE0177 #38

5.38

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #38

Recommended Solution Amplitude is maximized in an RLC circuit when we’ve reached the resonant frequency. This can be calculated by ω=√

1 LC

(5.111)

plug in your values to get ω = C =

1 LC 1 Lω 2 √

(5.112) (5.113)

1 (25 millihenries)(1 × 103 rad)2 1 = 7 2.5 × 10 millihenries-rad2 = 40 µF =

Correct Answer (D)

474

(5.114) (5.115) (5.116)

5.39. PGRE0177 #39

5.39

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #39

Recommended Solution At low frequencies and low energies, a capacitor and a resistor look like an impassible gap so these will act as a high pass filter (i.e. only high frequencies can pass) but an inductor will appear like just another bit of wire. At high frequencies and high energies, capacitors and resistors can be passed quite easily but passing through an inductor will generate a strong magnetic field and will impede the flow, making this a low pass filter. Looking through the choices, I. The inductor impedes high frequencies before it reaches the terminals so this can’t be a high pass filter II. The resistor allows the high energy to pass and the low frequencies will drop via the resistor. We know this is a high pass filter. III. The capacitor will allow high energies to pass through and low frequency voltage will drop. We know this is a high pass filter. IV. The high energy passes the resistor and passes with the capacitor without a drop in voltage. Without a drop in voltage, this can’t be a high pass filter. Correct Answer (D)

475

5.40. PGRE0177 #40

5.40

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #40

Recommended Solution After the switch is closed, we will start off with the maximum voltage and continually decrease as the resistor and inductor eat away at the initial voltage. From this, we can eliminate all options but (D) and (E). Next, compare the amount of time between the two remaining options to see that only half of the voltage dropping after 200 seconds is a vastly unrealistic voltage drop for any circuit, so we choose (D). Correct Answer (D)

476

5.41. PGRE0177 #41

5.41

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #41

Recommended Solution The existence of magnetic charge would be, in essence, the same thing as saying that magnetic monopoles exist. If this were the case, then II would not have to be changed to allow magnetic field lines to diverge completely and we can eliminate (A), (C) and (D). Next, consider that a magnetic monopole will also allow magnetic field lines to exist without curling back to its opposite pole, so III must change. Correct Answer (E)

477

5.42. PGRE0177 #42

5.42

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #42

Recommended Solution From Lenz’s law, we know that anytime a current is generated by a change in the magnetic or electric field, the generated emf will be such that it opposes this change in direction and magnitude. As the center ring moves towards ring A, this increases the field present and ring A will oppose this change with a current of opposite current, i.e. clockwise. Ring B, on the other hand, will have its field decreased so a current will be induced to oppose this decrease, which requires an anti-clockwise motion. Correct Answer (C)

478

5.43. PGRE0177 #43

5.43

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #43

Recommended Solution Start with the commutator relation [AB, C] = A[B, C] + [A, C]B

(5.117)

and apply it to the given commutator [Lx Ly , Lz ] = Lx [Ly , Lz ] + [Lx , Lz ]Ly

(5.118)

and replace the bracketed portions on the RHS with the commutation relations given in the problem (while recalling that the have the commutation relation in the opposite order from left to right simply changes the sign), to get

[Lx Ly , Lz ] = Lx [Ly , Lz ] + [Lx , Lz ]Ly

(5.119)

= Lx (i¯hLx ) + (−i¯hLy )Ly

(5.120)

=

i¯h(L2x

−

L2y )

Correct Answer (D)

479

(5.121)

5.44. PGRE0177 #44

5.44

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #44

Recommended Solution The problem gives us the energy equation as n2 π 2 ¯h2 (5.122) 2mL2 this tells us that the only difference between, say E1 and E2 will be the change in Energy from the squared quantum number, n = 1, 2, 3, . . .. So, we know that all allowed energies will be some squared integer multiple of E1 . Of those given, only (D) has a coefficient with a perfect square value (i.e. 32 = 9). En =

Correct Answer (D)

480

5.45. PGRE0177 #45

5.45

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #45

Recommended Solution First, recall that the expectation value for any operator is the sum of its terms, each one individually squared and multiplied by it’s respective eigenvalue. First, find the ”ket” values for |1i, |2i and 3i, 3 ¯hω 2 5 |2i = ¯hω 2 7 |3i = ¯hω 2 then multiply these by their respective coefficients, having been squared, to get |1i =

1 2 3 √ |1i − √ |2i + √ |3i 14 14 14          3 4 5 9 7 1 = ¯hω + ¯hω + ¯hω 14 2 14 2 14 2 86 = ¯hω 28 43 = ¯hω 14

|ψi =

Correct Answer (B)

481

(5.123) (5.124) (5.125)

(5.126) (5.127) (5.128) (5.129)

5.46. PGRE0177 #46

5.46

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #46

Recommended Solution Start with the de Broglie wavelength equation λ=

h p

(5.130)

and then recall the Schrdinger 1-D energy equation for a particle in a potential p2 + V (x) (5.131) 2m without doing much work, you can see that combining Equation 5.130 with Equation 5.131 will force you to have an inverse square root to substitute the p2 in Equation 2 into the p in Equation 1. This only matches option (E) so it must be the correct choice. E=

Correct Answer (E)

482

5.47. PGRE0177 #47

5.47

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #47

Recommended Solution First, eliminate choices (D) and (E) as both predict a negative change in entropy, which should not be true of a gas expanding outward to occupy a new open space. Next, recall from thermodynamics that entropy is defined as the natural log of the total number of states of a system S = kb ln(Ω)

(5.132)

where Ω is the number of states. For a volume divided into two parts, for n particles the number of states will be Ω = 2n

(5.133)

Plugging Equation 5.133 into Equation 5.132 and applying the rules of logarithms gives

S = kb ln(Ω)

(5.134)

n

= kb ln (2 )

(5.135)

= nkb ln(2)

(5.136)

which is given by choice (B). Correct Answer (B)

483

5.48. PGRE0177 #48

5.48

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #48

Recommended Solution You can initially eliminate all solutions that are less than 1, as we would expect the smaller and lighter N2 molecules to move faster than the larger and heavier O2 molecules. This step eliminates (A), (B), and (E). Between (C) and (D), you can either take a guess, or recall Graham’s law of effusion, which states Rate1 = Rate2

s

M2 M1

at which point, (C) becomes the obvious choice. Correct Answer (C)

484

(5.137)

5.49. PGRE0177 #49

5.49

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #49

Recommended Solution A Maxwell-Boltzmann system has the canonical partition function Z=

X

g e−Es /kb T

(5.138)

s

Where g is the degeneracy. Since we have two energies, we will need at least two terms in our energy equation so we can eliminate (A), (B) and (C). Next, plug everything in to see our coefficient of 2 appear, as in (E).

Z =

X

g e−Es /kb T

(5.139)

s

= 2e−/kT + 2e−2/kT h

= 2 e−/kT + e−2/kT Correct Answer (E)

485

(5.140) i

(5.141)

5.50. PGRE0177 #50

5.50

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #50

Recommended Solution Recall our equation relating frequency to wavelength, ν=

v λ

(5.142)

if our velocity is 3 vf = 0.97vo

(5.143)

so the initial velocity is ν0 λ v0 440 hz = λ v0 = (440 hz)(λ) νo =

(5.144) (5.145) (5.146)

then, compare this to the final velocity

νf

= = = =

vf λ (0.97v0 ) λ (0.97)(440 hz) λ 427 hz λ

Correct Answer (B) 486

(5.147) (5.148) (5.149) (5.150)

5.50. PGRE0177 #50

CHAPTER 5. PGRE0177 SOLUTIONS

Note: if you have any experience with wind instruments you know that a cold instrument always starts off a bit flat and the pitch of your instrument continually increases as it warms up. From this little fact, you should at least be able to determine that (D) and (E) can’t be correct.

487

5.51. PGRE0177 #51

5.51

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #51

Recommended Solution Consider a beam of unpolarized light headed towards you and you have three ideal polarizers in hand.

In the figure above, grey vectors indicate individual directions of oscillation, black vectors indicate the net polarization, greyed out areas indicate areas of absorbed polarization and white areas indicate unaffected polarization. If we wanted to completely eliminate all light using two of the polarizers, we could place them in series with a rotation of exactly π/2 between them, for example using the horizontal and vertical polarization. In this arrangement, anything that survived through the horizontal polarizer would be caught by the vertical polarizer and no light would be transmitted on the other side. However, imagine we were to place another polarizer between the horizontal and vertical polarizers such that this third polarizer is rotated π/4 or 45 with respect to the other two. As the light first passes through the horizontal polarizer, only half of the photons that hit the polarizer will pass through. These photons will continue to the 45 polarizer where, 488

5.51. PGRE0177 #51

CHAPTER 5. PGRE0177 SOLUTIONS

again, half of the remaining photons get absorbed and the other half pass through. After the photons pass through the 45 polarizer, the remaining photons will spread out from -22.5 to +67.5. This occurs because linearly polarized light will always completely fill a full 90 of angular spread, which I didn’t mention previously because the previous instances came out with a 90◦ spread. Finally, the remaining photons will pass through the vertical polarizer giving a final total of 1/8 the original number of photons

Correct Answer (B)

489

5.52. PGRE0177 #52

5.52

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #52

Recommended Solution Recall the three generic types of cubic crystals: Simple Cubic, Body Centered Cubic and Face Centered Cubic. Sum up the total area in each cubic to find Simple Cubic 1 Atom Body Centered Cubic 2 Atoms Face Centered Cubic 4 Atoms

490

5.52. PGRE0177 #52

(a) Simple Cubic

CHAPTER 5. PGRE0177 SOLUTIONS

(b) Body Centered Cubic

(c) Face Centered Cubic

Figure 5.6: 3 generic forms of cubic crystals Thus the primitive unit cell (simple cubic) contains half as much area, a3 /2 Correct Answer (C)

491

5.53. PGRE0177 #53

5.53

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #53

Recommended Solution Recall that semi-conductors, unlike regular conductors, conduct well at high temperatures and effectively not at all at extremely low temperatures. From this, we know that the resistivity at 0 should be very high and it should decrease as temperature increases. Only (B) matches this description. Correct Answer (B)

492

5.54. PGRE0177 #54

5.54

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #54

Recommended Solution Recall that impulse is Z t2

I=

F dt

(5.151)

t1

So we just want to find the area under this plot. If you don’t recall this equation, take note that we are given a y-axis in newtons and an x-axis in seconds. The only way to get these two units to end with kg · m/s is to take the area. Using the equation for the area under a triangle, you can quickly get I = 2 kg · m/s Correct Answer (C)

493

5.55. PGRE0177 #55

5.55

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #55

Recommended Solution We can quickly eliminate choices (B) and (D) based on the fact that they both suggest an asymmetrical set of values for an exceptionally symmetric problem. We also know that (A) can’t be right because it would violate conservation of momentum for a particle of mass m and velocity v0 to generate two particles with mass m and velocity v0 . Finally, since we know that horizontal momentum must be conserved and each piece gets half the horizontal momentum of the initial, then the addition of a vertical component of velocity requires, by the Pythagorean theorem, that the net velocity be greater than just the horizontal component. In other words 2 vnet = vx2 + vy2

Correct Answer (E)

494

(5.152)

5.56. PGRE0177 #56

5.56

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #56

Recommended Solution Start by finding the difference between the air density and the helium density to get the average density the balloon will experience ρavg = ρair − ρHe = 1.29 kg/m3 − 0.18 kg/m3 = 1.11 kg/m3

(5.153)

Now, note that the only way to get units of m3 from kg and kg/m3 is to do the following

V

m ρavg 300 kg = 1.11 kg/m3 = 270 m3

=

Correct Answer (D)

495

(5.154) (5.155) (5.156)

5.57. PGRE0177 #57

5.57

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #57

Recommended Solution Consider that if the density of the water were to increase, so to would the force and we don’t see this feature in (D) and (E). Next, eliminate (C) because the acceleration due to gravity,g, should not be a part of these calculations. Finally, compare units on (A) and (B) to get (A) ρv 2 A = (kg/m3 )(m2 /s2 )(m2 ) = kg m/s2 = N (B) ρvA/2 = (kg/m3 )(m/s)(m2 ) = kg/s and we choose (A). Correct Answer (A)

496

5.58. PGRE0177 #58

5.58

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #58

Recommended Solution The problem tells us that the electric field is pushing the proton in the +x-direction and, from Fleming’s left hand rule, we can quickly deduce that the magnetic field is pushing the proton in the -x-direction. Since we are told the proton isn’t deflected with a potential difference of V , we know that the forces must have balanced out exactly. On the second pass, however, the potential difference is doubled and, from the Lorentz force F = qvB, this will increase the force from the magnetic field but not for the electric field. In the second pass, the magnetic field force will be greater and will exceed the electric field force and the overall direction of deflection will be in the -x-direction. Correct Answer (B)

497

5.59. PGRE0177 #59

5.59

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #59

Recommended Solution The equation for a simple harmonic oscillator is m¨ x + kx = 0

(5.157)

which we get when L = m, C = 1/k and Q = x. Correct Answer (B)

Alternate Solution Start by eliminating any choices that suggest that Q = m, i.e. (C) and (E), as a change in mass over time is not a necessary condition for a simple harmonic oscillator. Next, since Q has to be x, from our previous elimination, we can reasonably conclude that 1/C should be 1/k so that the second term gives us Hooke’s law, i.e. kx, and you can eliminate (A) and (D). Correct Answer (B)

498

5.60. PGRE0177 #60

5.60

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #60

Recommended Solution Start by recognizing that the flux through the Gaussian surface should have some dependence on x, because if x blows up to infinity, so to should the flux. With (A) and (B) eliminated, next eliminate (C) because the area under consideration should be the area cut out by the entire circle minus the area cut out by x, not the area of the difference of the two. Finally, note that if x = 0, the area under consideration should just be that of a circle, which is πR2 as opposed to 2πR2 . Correct Answer (D)

499

5.61. PGRE0177 #61

5.61

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #61

Recommended Solution Considering the electric field first, when an electric field orthogonal to a conductor interacts with that conductor, the field (in a sense) disperses over it and the E field just to the left and to the ~ interacts with the right will be 0, from which we can eliminate (B), (D) and (E). Next, as the E ~ conductor, charges are moved around and this induces a magnetic field, so B 6= 0 and we choose (C). Correct Answer (C)

500

5.62. PGRE0177 #62

5.62

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #62

Recommended Solution Start with our cyclotron frequency equation Bq 2πm and then re-write Equation 5.158 to solve for m f=

m=

Bq 2πf

(5.158)

(5.159)

Now, plugging everything in gives us

m = = = = =

(π/4 teslas)(2e− ) 2π(1600 hz) e− (4)(1600 hz) 1.6 × 10−19 kg/s (4)(1600 hz) 1 × 10−22 kg 4 2.5 × 10−23 kg

Correct Answer (A)

501

(5.160) (5.161) (5.162) (5.163) (5.164)

5.63. PGRE0177 #63

5.63

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #63

Recommended Solution Ideally, we would be able to recall Wien’s displacement law b (5.165) T in which b is the Wien displacement law constant. Alternatively, you could quickly get exactly the same relationship with a bit of dimensional analysis by noticing that the only way to get a final unit of Kelvins would be to take the quotient of Wien’s displacement law constant with the maximum wavelength. Either way you do it, convert the peak wavelength of 2 µm to 2 × 10−6 m and solve λ=

T

2.9 × 10−3 m · K 2 × 10−6 m = 1, 500 K =

Correct Answer (D)

502

(5.166) (5.167)

5.64. PGRE0177 #64

5.64

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #64

Recommended Solution Options (C) and (D) are very much true and are frequently mentioned in a lot of pop physics books, television and other forms of media, so these two should be easily dismissed. (E) is also true because band spectrum, i.e. spectral lines so close together as to form a band, can’t appear when a single atom produces a single spectral line. Lastly, (B) is true because absorption spectroscopy and emission spectroscopy are essentially different ways of measuring the same thing, specifically wavelengths. Only (A) is left so that is our answer. Correct Answer (A)

503

5.65. PGRE0177 #65

5.65

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #65

Recommended Solution Start with the Taylor Series expansion of ex to get an approximation ex ≈ 1 + x e

hν/kT

= 1 + hν/kT

(5.168) (5.169)

Plug this approximation into the denominator of Einstein’s formula 

C = 3kNA 

= 3kNA

hν kT

2

hν kT

2

ehν/kT (1 + hν/kT − 1)2

(5.170)

ehν/kT (hν/kT )2

(5.171)

= 3kNA ehν/kT

(5.172)

Finally, as temperature blows up to infinity, ehν/kT = 1 and so we are left with C = 3kNA Correct Answer (D)

504

(5.173)

5.66. PGRE0177 #66

5.66

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #66

Recommended Solution Ignore the fact that this problem mentions anything about physics and treat it like a standard rate problem. If Jimmy can eat a whole apple pie in 24 minutes and Susie can eat a whole apple pie in 36 minutes, how long will it take to eat that same pie if both Jimmy and Susie are sharing a single pie? 24 ∗ 36 = 14.4 minutes 24 + 36 Correct Answer (D)

505

(5.174)

5.67. PGRE0177 #67

5.67

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #67

Recommended Solution First, eliminate (A) because we know that Uranium-238 can fission spontaneously. Next, eliminate (B) because, aside from it being a very vague claim that is not typical of a GRE answer, it is a strange conclusion to try and draw from the splitting of a Uranium atom that barely loses any mass/energy in the process. We can then eliminate (C) simply because it isn’t true that a nuclei with roughly half the particles of Uranium-238 would be equally massive. Finally, between (D) and (E), recall that binding energy generally increases as we move towards the most stable element, iron, and decreases as we move away. A change from Uranium-238 to a nuclei with A=120 would be a movement closer to iron, so we would likely so a binding energy of 8.5 MeV/nucleon rather than 6.7 MeV/nucleon. Correct Answer (E)

506

5.68. PGRE0177 #68

5.68

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #68

Recommended Solution Since an alpha particle is a helium atom, specifically He2+ with two protons and two neutrons, if lithium lost an alpha particle it would have lost one to many protons to become beryllium and (A) is eliminated. Losing an electron, positron or neutron wouldn’t alter the actual type of atom, so emitting these wouldn’t change anything and we can eliminate (B), (C) and (D). This leaves us (E). Correct Answer (E)

507

5.69. PGRE0177 #69

5.69

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #69

Recommended Solution From the description, we know that the light first interacts with the oil surface and the phase of the blue light is shifted 180◦ . Next, the light interacts with the glass surface and, because of the previous phase shift, we use the thin film interference equation 2nd = mλ

(5.175)

where m = 1, 2, 3, . . .. The thinnest oil film will be when m = 1, so we solve for d in Equation 5.175 and plug in our values mλ 2n 480 nm = 2(1.2) = 200 nm

d =

Correct Answer (B)

508

(5.176) (5.177) (5.178)

5.70. PGRE0177 #70

5.70

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #70

Recommended Solution Starting with the double slit equation ω sin(θ) = mλ

(5.179)

re-write Equation 5.179 to solve for the angle between fringes, θ, to get mλ (5.180) ω from Equation 5.180, we can easily see that doubling the frequency, ω, will result in a separation angle half as large as the initial 1.0 millimeter and we pick (B). sin(θ) =

Correct Answer (B)

509

5.71. PGRE0177 #71

5.71

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #71

Recommended Solution We can eliminate (E) right away because the proposed velocity is faster than light. Next, eliminate (A) and (B) because we know, from the Doppler effect, that an increase in wavelength corresponds to an object moving away from the observer. Lastly, we can choose between (C) and (D) by utilizing the doppler redshift equation

λ λ0 607.5 nm 121.5 nm

s

=

1 + v/c 1 − v/c

(5.181)

s

1 + v/c 1 − v/c 1 + v/c 25 = 1 − v/c 24 − 26(v/c) = 0 v = 12/13 c v = (12/13)c =

8

v = 2.8 × 10 m/s Correct Answer (D)

510

(5.182) (5.183) (5.184) (5.185) (5.186) (5.187)

5.72. PGRE0177 #72

5.72

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #72

Recommended Solution Right as the string breaks the block will be accelerating due to gravity and it will also be accelerating from the spring pulling down on the top block from the bottom block. Based on this, we know that the net acceleration of the top block must be larger than just that from acceleration, and we eliminate (A), (B), and (C). Lastly, summing the vertical forces on the top block, we get − ma = −mg − k∆x 2mg = k∆x which tells us that the acceleration should be 2g. Correct Answer (E) 511

(5.188) (5.189)

5.73. PGRE0177 #73

5.73

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #73

Recommended Solution In order for block B to not fall, the gravitational force downward must be in equilibrium with the frictional force upwards. Adding up the forces and setting the net force to 0, we get

Fnet = f − FG 0 = µFN − mg mg FN = µ (20 kg)(10 m/s2 ) = 0.50 = 400 N Correct Answer (D)

512

(5.190) (5.191) (5.192) (5.193) (5.194)

5.74. PGRE0177 #74

5.74

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #74

Recommended Solution The Lagrangian equation of motion can be calculated with d dt



d dt



∂L ∂ q˙

∂L =0 ∂q

(5.195)

= 2a¨ q

(5.196)

= 4bq 3

(5.197)



−

perform both derivatives to get ∂L ∂ q˙ ∂L ∂ q˙



plugging our results from Equation 5.197 into Equation 5.195 gives d dt



∂L ∂L − = 0 ∂ q˙ ∂q 2a¨ q − 4bq 3 = 0 

(5.198) (5.199) 3

2a¨ q = 4bq 2b 3 q¨ = q a Correct Answer (D)

513

(5.200) (5.201)

5.75. PGRE0177 #75

5.75

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #75

Recommended Solution We could approach this problem theoretically, but it will be quicker to just give the matrix a vector and see where it goes. to simplify things, use the vector ~v = h1, 0, 0i

(5.202)

multiplying everything out, we get a transformed vector √ v~0 = h1/2, 3/2, 0i which represents a 60◦ rotation clockwise about the z-axis. Correct Answer (E)

514

(5.203)

5.76. PGRE0177 #76

5.76

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #76

Recommended Solution The first of the five choices we can eliminate is (D), because electrons travel quite slowly in metals, making them exceptionally non-relativistic. We can also quickly eliminate (B) because there is no condition specified in the problem that would take the metal out of equilibrium. Next, eliminate (A) because electrons in a metal move freely but are generally more constrained in their degrees of freedom than free atoms. Finally, between (C) and (E), choose (C) because interactions between phonons and electrons are more technical than we would likely ever see on the GRE. Correct Answer (C)

515

5.77. PGRE0177 #77

5.77

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #77

Recommended Solution By Maxwell-Boltzmann statistics, we get the equation gi e−i /kT (5.204) Z kT and the energy are both given so plug them both into Equation 5.204 and solve to get Ni = N

gi e−i /kT Z (−0.1 eV)/(0.025 eV) gi e = N Z gi e−4 = N Z

Ni = N

which results in our ratio of e−4 . Correct Answer (E)

516

(5.205) (5.206) (5.207)

5.78. PGRE0177 #78

5.78

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #78

Recommended Solution Since we are talking about changing the arrangement of neutrinos and anti-neutrinos, both of which are leptons, we would be wise to check lepton number first. Doing so, we find the typical leptonic decay as µ− → e − L: 1 = 1 Le : 0 = 1 Lµ : 1 = 0

+ νˆe + νe − 1 + 1 − 1 +0 + 0 + 1

removing the anti-neutrino from our typical lepton decays results in µ− → e − L: 1 = 1 Le : 0 = 1 Lµ : 1 = 0

+ νe + 1 + 0 + 1

so lepton number is not conserved. Correct Answer (E)

517

5.79. PGRE0177 #79

5.79

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #79

Recommended Solution Start with the relativistic energy equation E 2 − (pc2 ) = (mc2 )2

(5.208)

we are given E = 10 GeV and p = 8 GeV/c so we plug these in and solve for m (10 GeV)2 − (8 GeV/c)2 c2 = m2 c4 2

2

(5.209)

2 4

(100 GeV ) − (64 GeV ) = m c 36 GeV

2

m

2

(5.210)

2 4

= m c

(5.211) 2

= 36 GeV /c 2

m = 6 Gev/c Correct Answer (D)

518

4

(5.212) (5.213)

5.80. PGRE0177 #80

5.80

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #80

Recommended Solution Start by eliminating option (E) because light will move slower than c when interacting with some medium, relativistic or not. Next, consider that when the tube and water are moving at 1/2c and moving in the same direction as the light, we would expect the light to pass through more quickly than if the tube was stationary. Recalling our equation for light speed in a medium, c 3 = c (5.214) n 4 so our answer should be larger than this and we eliminate (A) and (B). Finally, we can decide between (C) and (D) by considering the sum of our two relativistic velocities using v=

v0 = = = =

u+v 1 + vu c2 1/2c + 3/4c 1 + (1/2c)(3/4c) c2 5/4 c 11/8 10 c 11

Correct Answer (D)

519

(5.215) (5.216) (5.217) (5.218)

5.81. PGRE0177 #81

5.81

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #81

Recommended Solution Start with the equations for L2 and Lz L2 Ylm (θ, φ) = l(l + 1)¯h2 Ylm (θ, φ)

(5.219)

Lz Ylm (θ, φ) = m¯hYlm (θ, φ)

(5.220)

applying both eigenvalues to their respective operators, we get l(l + 1)¯h2 = 6¯h2 l = 2

(5.221) (5.222)

and m¯h = −¯h

(5.223)

m = −1

(5.224)

at which point we plug in our values for m and l into the original momentum eigenfunction and get Ylm (θ, φ) = Y2−1 (θ, φ) Correct Answer (B)

520

(5.225)

5.82. PGRE0177 #82

5.82

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #82

Recommended Solution For a two electron system, there are three triplet state configurations and one singlet state configuration. The single singlet configuration occurs when the total spin, S = S1 + S2 equals 0, which is only true when S1 = −S2 and can be written as 1 √ (|αi1 |βi2 − |αi2 |βi1 ) 2

(5.226)

we can then eliminate all choices with II as an option and, more to the point, all other configurations must be one of the three possible triplet configurations. Correct Answer (D)

521

5.83. PGRE0177 #83

5.83

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #83

Recommended Solution Depending on how well you recall your linear algebra, you might quickly recognize that the Pauli spin matrix given in this problem is a relatively typical transformation/rotation matrix. To demonstrate this, you can pick any arbitrary up and down spin vectors (I’ll be using 1,2,3 & 4 just for the purposes of distinguishing separate elements of the vectors). Start with spin up and down vectors

| ↑ i = (1, 2)

(5.227)

| ↓ i = (3, 4)

(5.228)

Now multiply each of those to the transformation matrix given in the problem to get the transformed vectors

| ↑ iσx = (2, 1)

(5.229)

| ↓ iσx = (4, 3)

(5.230)

522

5.83. PGRE0177 #83

CHAPTER 5. PGRE0177 SOLUTIONS

which shows that this Pauli matrix performs an orthogonal transformation on the vectors. Of the potential solutions, only (C) gives us the necessary result that swapping the order of the vectors will also swap the sign. Correct Answer (C)

523

5.84. PGRE0177 #84

5.84

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #84

Recommended Solution For a single electron transition, utilize our selection rules for the orbital and total angular quantum numbers

∆l = ±1

(5.231)

∆j = 0, ±1

(5.232)

we can eliminate transition A because ∆l = 0. For B and C, however, we have transitions B: ∆l = −1 and ∆j = −1 C: ∆l = −1 and ∆j = 0 both of which are allowed and so we choose (D). Correct Answer (D)

524

5.85. PGRE0177 #85

5.85

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #85

Recommended Solution The resistance of the wire is related to its length and area by R=

ρl a

(5.233)

this gives us two resistances ρ2L (5.234) A ρL R2 = (5.235) 2A which gives us a ratio for the resistances of R1 /R2 = 4/1. This ratio also represents, from V = IR, the ratio of voltages R1 =

V1 4 = V2 1

(5.236)

V1 + V2 = 7 volts

(5.237)

Finally, find the net voltage as

and use Equation 5.236 and Equation 5.237 to get V

= 1 + IR

(5.238)

= 1 V + 1.4 V

(5.239)

= 2.4 V

(5.240)

Correct Answer (A)

525

5.86. PGRE0177 #86

5.86

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #86

Recommended Solution Start with Ohm’s law and solve for current V R Then, we can find the induced emf from a changing magnetic field by I=

dφB | Z dt ~ · dA ~ = B

ε = N| φB

526

(5.241)

(5.242) (5.243)

5.86. PGRE0177 #86

CHAPTER 5. PGRE0177 SOLUTIONS

The area and, therefore, the magnetic flux changes during rotation so we find the magnetic flux as φB = Bπr2 sin(ωt)

(5.244)

ε = N Bπr2 ω cos(ωt)

(5.245)

and, therefore, the induced EMF is

Finally, substitute the EMF from Equation 5.245 into Equation 5.241 to get the final solution N Bπr2 ω cos(ωt) R (15 turns)(0.5 tesla)(0.01 m)2 (300 rad/sec) cos(ωt) = 9Ω = 25π cos(ωt)

I =

Correct Answer (E)

527

(5.246) (5.247) (5.248)

5.87. PGRE0177 #87

5.87

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #87

Recommended Solution For the left side of the diagram, the test charge falls inside the sphere meaning that the potential is constant and the field is 0. This means that the only force on the test charge will be the result of sphere Q pushing q to the left. Using Gauss’s law with a distance of 10d − d/2 = 19/2d, the net force is

F

= = =

1 qQ 4π0 r2 4qQ 4π0 (361d2 ) qQ 361π0 d2

Correct Answer (A)

528

(5.249) (5.250) (5.251)

5.88. PGRE0177 #88

5.88

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #88

Recommended Solution Eliminate (A) since a changing field must generate a magnetic field. Next, compare the units for each of the potential solutions to get (B): Akg = tesla ·s 2 (C): Akg = tesla ·s 2 (D): A·kg s2 m 6= tesla kg (E): A·s2 m 6= tesla We are left with (B) and (C) at which point you can guess or try to recall that the Biot-Savart law is proportional to 1/4π rather than 1/4π 2 Z

B=

µ0 I dl × rˆ 4π |r|2

Correct Answer (C)

529

(5.252)

5.89. PGRE0177 #89

5.89

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #89

Recommended Solution From conservation of angular momentum, Lf = L0

(5.253)

When the child is in the middle of the merry-go-round, he/she/it won’t contribute to the angular momentum, so the final momentum will be Lf = Iωf =

M R2 2

!

ωf

(5.254)

however, for the initial momentum we must use both the child at a distance R and the merrygo-round, 530

5.89. PGRE0177 #89

CHAPTER 5. PGRE0177 SOLUTIONS

!

Li = Iω0 =

M R2 + mR2 ω0 2

(5.255)

Plug Equations 5.254 and 5.255 into Equation 5.253 to get M R2 2

!

!

ωf

=

ωf

= = =

M R2 + mR2 ω0 2 2m ω0 + ω0 M 2 2.0 rad/sec + (2.0 rad/sec) 5 2.8 rad/sec

Correct Answer (E)

531

(5.256) (5.257) (5.258) (5.259)

5.90. PGRE0177 #90

5.90

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #90

Recommended Solution From simple common sense, with a shorter length in which oscillations can occur and a stronger spring constant in Figure 1, the period should be less than that of Figure 2. From this, we can eliminate all options that are 1 or greater, i.e. (C), (D) or (E). Between (A) and (B), recall that the period of a SHO can be written as r

T = 2π

m k

(5.260)

and s

T = 2π

l g

(5.261)

√ which tells us that doubling k and doubling l will both scale the period by 2. However, since each of these changes are happening in such a way as to give Figure 1 a longer period, the two scalings combine to give a total change in period of twice as much for figure 1 or, equivalently, T1 1 = T2 2 Correct Answer (A)

532

(5.262)

5.91. PGRE0177 #91

5.91

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #91

Recommended Solution At the top of the wedge, net energy is just the gravitational energy, Enet = mgh

(5.263)

all of this energy is converted to translational and rotational at the bottom of the wedge, 1 1 mgh = mv 2 + Iω 2 2 2 solving for I in Equation 5.264 gives, I=

2mghR2 − R2 m v2

and substituting the value given for v 2 533

(5.264)

(5.265)

5.91. PGRE0177 #91

CHAPTER 5. PGRE0177 SOLUTIONS

I = = =

2R2 mgh − R2 m v2 7 2 R m − R2 m 4 3 2 R m 4

Correct Answer (B)

534

(5.266) (5.267) (5.268)

5.92. PGRE0177 #92

5.92

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #92

Recommended Solution The Hamiltonian is the sum of kinetic and potential energy terms, as opposed to the Lagrangian which is the difference. The energy of a harmonic oscillator is 535

5.92. PGRE0177 #92

CHAPTER 5. PGRE0177 SOLUTIONS

1 V = k(∆l)2 2

(5.269)

1 p2 T = mv 2 = 2 2m

(5.270)

and the kinetic energy term is

adding the two of these gives us

H = T +V p21 p2 1 = + 2 + k(l − l0 )2 2m 2m 2 " # 2 1 p1 p22 2 + + k(l − l0 ) = 2 m m Correct Answer (E)

536

(5.271) (5.272) (5.273)

5.93. PGRE0177 #93

5.93

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #93

Recommended Solution If you do well with your physics history, you may recall that Bohr radius is defined as the smallest possible orbital distance for the hydrogen atom in its ground state, which is sufficient to choose a0 . Correct Answer (C)

Alternate Solution If you don’t recall the fact in the recommended solution, start by taking the squared wavefunction and multiplying it by a spherical shell "

dP

= =

√

#2

1 3/2 πa0

e

−r/a0

4πr2 dr

4 2 −2r/a0 r e dr a30

(5.274) (5.275)

differentiating dP in Equation 5.275 with respect to r and setting it equal to 0 allows us to solve for the minimum radius 2 2 −2r/a0 r e = 0 a0   r 2re−2r/a0 1 − = 0 a0

2re−2r/a0 −

537

(5.276) (5.277)

5.93. PGRE0177 #93

CHAPTER 5. PGRE0177 SOLUTIONS

at which point, it is easy to solve for the radius as r = a0 Correct Answer (C)

538

(5.278)

5.94. PGRE0177 #94

5.94

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #94

Recommended Solution From quantum mechanics, the first order energy shift for our specific hamiltonian is En(1) = hn0 |V (a + a† )|n0 i

(5.279)

multiply the Hamiltonian out to get 

a + a†

2

= a2 + aa† + a† a + a†2

(5.280)

at which point we can eliminate a2 and a†2 by orthogonality. Finally, solve for the energy with our remaining terms E = hn|(aa† + a† a|ni

(5.281)

= (2n + 1)V

(5.282)

= 5V

(5.283)

Correct Answer (E)

539

5.95. PGRE0177 #95

5.95

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #95

Recommended Solution Start by recalling the relative permitivity equation κ=

ε(ω) ε0

(5.284)

where ε(ω) and ε0 are the absolute permitivity and electric constant, respectively. Next, recalling that the Electric field is inversely proportional to ε(ω), which you could potentially realize from Coulomb’s law, we get

E ∝

1 ε(ω)

540

(5.285)

5.95. PGRE0177 #95

CHAPTER 5. PGRE0177 SOLUTIONS

∝

1 0 κ

(5.286)

E0 κ

(5.287)

Then applying E0 ∝ 1/ε0 , we get E=

Correct Answer (A)

541

5.96. PGRE0177 #96

5.96

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #96

Recommended Solution Recall the Larmor formula, which gives the power radiated by a charged object, P =

e2 a2 6π0 c3

(5.288)

since the sphere is not moving, just expanding at a stationary location, we get an acceleration, a = 0. Plugging this into Equation 1 clearly results in a total radiated power of zero as well. Correct Answer (E)

542

5.97. PGRE0177 #97

5.97

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #97

543

5.97. PGRE0177 #97

CHAPTER 5. PGRE0177 SOLUTIONS

Recommended Solution Consider the limiting case of θ = 0. In this case, there will be no difference between δθ0 and θ0 , which tells us that the solution should have some dependence on θ and, furthermore, that δθ0 = 0 when θ = 0. This eliminates (A), (B) and (C) due to their lack of theta dependence and (D) based on the its failure to have δθ0 = 0 when θ = 0. Note that taking the limit in (D) requires you to utilize L’hospitals rule to deal with the division of 0/0, at which point you get a pair of cosines which go to 1 at θ0 = 0 and θ = 0 and keeps δθ0 from going to 0. Correct Answer (E)

Alternate Solution Rather than applying a limiting case to the angle, θ, we can apply the limiting case n = 1. When n = 1, there is no transition to a new index of refraction and δθ0 should be equivalent to θ0 . We can immediately eliminate (D) because it lacks the dependence on n and we can next eliminate (A), (B) and (C) because their lack of θ0 dependence means we could never get δθ0 = θ0 when n = 1. Correct Answer (E)

544

5.98. PGRE0177 #98

5.98

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #98

Recommended Solution Examining both sums in the expression, it should become quite clear that whatever it represents, it will have units of energy. Looking at the units for the potential solutions, (A) Average energy is an energy! (B) The denominator of this particular expression is the partition function. You don’t really need to know that, however, provided you recognize that it is not an energy. (C) Absolutely not an energy. (D) Probability of a certain energy value is not, in itself, an energy value. (E) Entropy has units of J/K so it is not an energy. Correct Answer (A)

545

5.99. PGRE0177 #99

5.99

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #99

Recommended Solution Considering the problem qualitatively, after the collision occurs, we have three particles each with a mass m and some non-zero velocity. This means that our final energy, which by conservation of momentum must equal our initial energy, must be the sum of three particles with rest energy of mc2 (i.e. net rest energy will be 3mc2 ) and some kinetic energy. This tells us that the initial photon energy must be greater than just the final rest energies so we can eliminate (A), (B) and (C). Finally, equating the coefficients 4 and 5 in choices (D) and (E), respectively, to the Lorentz factor, γ, gives us a strong hint that higher values of γ will give us less realistic particle velocities (i.e. far too high) and we can choose (D). Correct Answer (D)

546

5.100. PGRE0177 #100

5.100

CHAPTER 5. PGRE0177 SOLUTIONS

PGRE0177 #100

Recommended Solution Recalling the acronym for the visible light spectrum, ROYGBIV, it should be clear that red light has a longer wavelength and lower energy than green light. Red light is given to us in this instance as λred = 632.82 nm so we know that the wavelength should be less than this and we can eliminate (D) and (E). Next, recall width of the visible spectrum is roughly between 380 nm and 750 nm and that green light is exactly in the middle of the spectrum. Thus, averaging the two wavelengths should give us a good approximation for the wavelength of green light 380 nm + 750 nm 1130 nm = = 565 nm 2 2 and this average is closest to (B). Correct Answer (B)

547

(5.289)

Chapter 6

Appendix 6.1

Equations

It would be very convenient if every question on these tests could be solved with nothing more than our intuitions and common sense. Sense this isn’t the case, we will need to do some calculations and, on this test, it isn’t feasible in the given time to re-derive most of the classical results. Below are the equations that had to be used at least once in my solutions.

6.1.1

Classical Mechanics

Kinematics: 1 ∆x = v0 t + at2 2 v 2 = v02 + 2ad v = +at v0 + vf ∆x = t 2 Newton’s Second Law: F = m¨ x Momentum: P = mx˙ Angular Momentum: L = Iω Torque: τ

= ~r × F~

τ

= Iα

548

6.1. EQUATIONS

CHAPTER 6. APPENDIX

Centripetal Acceleration: ac = ω 2 r = mrω 2 2π ω = T

F

Gravitational Potential Energy: U

= mgh m1 m2 = −G R = −W

U U Kinetic Energy:

Uk = Uk = Uk−rotational =

1 mv 2 2 p2 2m 1 2 Iω 2

Impulse: Z t2

I=

F dt t1

Parallel Axis Theorem: I = Icom + M h2 Work: Z

W =

F ·s

Moment of Inertia (Point Mass): I = M R2 Schwarzschild Radius: Re =

2GM c2

Lagrangian: L=T −V Lagrangian Equation of Motion: d dt



∂L ∂ q˙



549

−

∂L =0 ∂q

6.1. EQUATIONS

CHAPTER 6. APPENDIX

Hamiltonian: H =T +V Angular Frequency: ω =

2π T s

k m

ωSHO = Period of a SHO: r

T = 2π

m k

Rocket Mechanics: m

dm dv +u =0 dt dt

Kepler’s third law: T 2 ∝ βr3 Center of Mass: COM =

6.1.2

m1 x1 + m2 x2 mtot

Electricity & Magnetism

Maxwell’s Equations: ρ 0 ~ = 0 ∇·B ~ ~ = − ∂B ∇×E ∂t ~ = ∇·E

~ ~ = µ0 J~ + µ0 0 ∂ E ∇×B ∂t Coulombs Law: U

=

U

=

F

=

1 dQ 4π0 R 1 q1 q2 4π0 R 1 q1 q2 4π0 R2

Hooke’s Law: F = −kx 550

6.1. EQUATIONS

CHAPTER 6. APPENDIX

Faraday’s Law of Induction dφB dt

|ε| = N Magnetic Flux: Z Z

~ r, t) · dA B(~

φB = S

Malus’ Law: 1 I = c0 E02 cos2 (θ) 2 Gauss’ Law: ~ · dA ~= q E 0 Ampere’s Law ~ = µ0 i B 2πR Larmor Formula: q 2 a2 6π0 c3

P = Lorentz Force:



~ F = q ~v × B Hall Voltage: VH = −

IB dne

Magnetic Dipole Moment: Z

µ=

i dA

Relative Permitivity: κ=

551

ε(ω) ε0



6.1. EQUATIONS

6.1.3

CHAPTER 6. APPENDIX

Optics

Photoelectric Equation: Uk = hν − φ Snell’s Law: sin(θ1 ) n2 = sin(θ2 ) n1 n-slit constructive interference: d sin(θn ) = nλ n-slit destructive interference: 1 d sin(θn ) = n + λ 2 



Photon Energy: E = hν hc E = λ Thin Film Interference: 1 λ 2nd = m + 2 



Light speed through a medium: v=

1 µ

Traveling wave: 2π y(x, t) = A sin (x ± vt) λ 



Wave Propagation Velocity: v = fλ Drift velocity: vd =

i nqA

Compton Equation: λ0 − λ =

h (1 − cos(θ)) mp c

552

6.1. EQUATIONS

CHAPTER 6. APPENDIX

Bragg’s Law: 2d sin(θ) = nλ Optical Magnification (Convex Lens): M=

fobj feye

Rayleigh Criterion: sin(θ) = 1.22

λ d

Thin Lens Equation/Mirror Equation: 1 1 1 + = d0 di f

6.1.4

Thermodynamics & Stat Mech

Rydberg Formula: 1 1 1 = R∞ − 2 2 λ n1 n2 



Rydberg Formula for Hydrogen like atoms: 1 λH−like

= RZ 2



1 1 − n21 n22



Moseley’s Law: 1 λK−α

2

= R (Z − β)



1 1 − 2 2 n1 n2



Rydberg Energy: me e4 Z 2 2 n2 2¯  h 1 1 E = E0 − λ21 λ2   1 1 − E = n(13.6 eV ) λ21 λ22

E = −

Heat Capacity: C = C = C = CV

=

Cp =

∆Q ∆T ∂Q ∂T ∂S T    ∂T  ∂U ∂Q = ∂T V ∂T V     ∂Q ∂H = ∂T p ∂T p 553

6.1. EQUATIONS

CHAPTER 6. APPENDIX

Heat: Q = cm∆T First Law of Thermodynamics: dU

= dQ + dW

dU

= dQ − P dV

Ideal Gas Law: pV = nRT Thermodynamic Work: Z Vf

W =

P dV Vi

Entropy: ∆S =

Z T2 dq

T

T1

Fourier’s Law of Heat Conduction: ∂Q = −k ∂t

I

~ ∇T · dA

s

Mean free path: P (x) = nσdx Stefan-Boltzmann’s Law: j ∗ = σT 4 Bohr Energy: Etot =

−13.6 eV n2

Carnot Engine Efficiency: η=

W TC =1− QH TH

Fermi Velocity: s

vf =

2Ef m

Graham’s Law of Effusion Rate1 = Rate2

s

Wien’s Displacement Law λ=

554

b T

M2 M1

6.1. EQUATIONS

6.1.5

CHAPTER 6. APPENDIX

Quantum Mechanics

Particle Location (Probability): Z b

Pab =

|ψ(x)|2 dx

a

Infinite Square Well/Particle in a box: ψn (x, t) = Asin(kn x)e−iωn t Planck Length: s

lp =

G¯h c3

Expectation Value: hAi =

Z

hψ |A| ψi

Heisenberg Uncertainty Principle: ∆x∆p ≥ ∆E∆t ≥

¯ h 2 ¯h 2

Spin Operator: S12 ψ1 = S1 (S1 + 1)ψ1 Probability Current: ∂ψ ∂ψ ∗ ¯ ~ t) = h ψ∗ − ψ J(x, 2mi ∂x ∂x 



Quantum Harmonic Oscillator: 

En = h ¯ω n +

1 2



Wave Speed (de Broglie relations): vp = vp =

E p c2 v

Time-Independent Schrodinger Equation: E ψ(x) = −

¯2 2 h ∇ ψ(x) + V (x) ψ(x) 2m 555

6.1. EQUATIONS

CHAPTER 6. APPENDIX

Quantum Momentum Operator: ¯ h Pˆ = ∇ψ i Russel-Saunders Term Symbols: 2s+1

LJ

Total Angular Momentum Quantum Number: j =l+s Hermitian Operator: hAi =

Z

ˆ ψ ∗ (r)Aψ(r) dr

Canonical Partition Function: Z=

X

ge−Es /kb T

s

6.1.6

Special Relativity

Rest Energy: E = m0 c2 Lorentz Factor: γ=q

1 1−

v2 c2

Relativistic Energy: ER = γmc2 Relativistic Momentum: m0 v prel = γm0 v = q 2 1 − vc2 Relativistic Energy-Momentum: 

E 2 = mc2

2

+ (pc)2

Relativistic sum of velocities: u0 =

u+v 1 + vu c2

u =

u0 + v 0 0 0 1 + vcu2

556

6.1. EQUATIONS

CHAPTER 6. APPENDIX

Proper Time: ∆τ 2 = ∆t2 − ∆x2 ∆t2 = ∆τ 2 + ∆x2 Space-Time Interval: ∆S 2 = −(C∆t)2 + ∆x2 Length Contraction: s

L0 =

L v2 =L 1− 2 γ c

Time Dilation: ∆t ∆t0 = γ∆t = p 1 − v 2 /c2

6.1.7

Electronics

Ohm’s Law: V = IR Kirchoff ’s First Law: n X

I=0

k=1

Kirchoff ’s Second Law: n X

V =0

k=1

Current: i=

dq dt

Faraday’s law of induction: dφB ε= dt

Capacitance: C=

Q V

Frequency of an RLC Circuit: f=

1 √ 4π LC

557

6.1. EQUATIONS

CHAPTER 6. APPENDIX

Impedance Matching: ZS = ZL∗ Rg + jXg = Rg + jXl Capacitor Energy: 1 U = CV 2 2 Capacitor Discharge: V = V0 e−t/RC Resistance of a Wire: R=

ρl A

Biot-Savart Law: µ0 I dl × rˆ 4π |r|2

Z

B=

6.1.8

Special Topics

Acoustic Beats: beats = |f2 − f1 | Doppler Effect: "

f=

1 1±

#

vs ource vw ave

Cyclotron Resonance Frequency: ωc =

eB m

Frequency of a Pendulum: r

ω=

mgrcom I

Poisson Distribution Uncertainty Equation: σ u= √ N

558

f0

6.2. UNITS AND CONVERSIONS

6.2

CHAPTER 6. APPENDIX

Units and Conversions

6.2.1

Units

Current: A=

C s

P=

N m2

Pressure:

Energy: J=N·m=

kg · m2 = Pa · m3 = W · s s2

Power: W=

J N·m kg · m2 = = s s s3

Electric Charge: C=A·s=F·V Electric Potential: V=

W J J = = A A·s C

Capacitance: F=

J C = 2 V V

Resistance: Ω=

J V = A s · A2

Conductance: S = Ω−1 =

A V

Magnetic Flux Density: T=

V·s N Wb = = 2 m2 A·m m

Inductance: H=

m2 · kg Wb V·s 2 = A = A =Ω·s 2 s ·A

Entropy/Heat Capacity: J K 559

6.2. UNITS AND CONVERSIONS

6.2.2

CHAPTER 6. APPENDIX

Conversions

Volume: 1 cm3 = 1 mL 1 dm3 = 1 L Angstrom: 1˚ A = 0.1 nm = 1 × 10−9 m Micron: 1 µ = 1 × 10−6 m Electronvolt 1 eV = 1.7826 × 10−36 kg Kilogram per Cubic Meter: 1, 000

g kg kg =1 =1 3 m mL L

Revolutions: 1 rpm = 0.105

6.2.3

rad s

SI Prefixes Prefix peta tera giga mega kilo hecto deca BASE deci centi milli micro nano pico femto atto

Symbol P T G M k h da d c m µ n p f a

560

10n 1015 1012 109 106 103 102 101 101 = 1 10−1 10−2 10−3 10−6 10−9 10−12 10−15 10−18

Contact Information: [email protected]