Complete Solutions to the Physics GRE: PGRE0177

Complete Solutions to the Physics GRE

Exam #0177

Taylor Faucett

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BiBTeX: @Book{Faucett0177, author = {Taylor Faucett}, series = {Complete Solutions to the Physics GRE}, title = {Exam #0177}, pages = {36--38}, year = {2010}, edition = {first}, }

c 2010 Taylor Faucett

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Contents 1 Physics GRE Solutions 1.1 PGRE0177 #1 . . . 1.2 PGRE0177 #2 . . . 1.3 PGRE0177 #3 . . . 1.4 PGRE0177 #4 . . . 1.5 PGRE0177 #5 . . . 1.6 PGRE0177 #6 . . . 1.7 PGRE0177 #7 . . . 1.8 PGRE0177 #8 . . . 1.9 PGRE0177 #9 . . . 1.10 PGRE0177 #10 . . . 1.11 PGRE0177 #11 . . . 1.12 PGRE0177 #12 . . . 1.13 PGRE0177 #13 . . . 1.14 PGRE0177 #14 . . . 1.15 PGRE0177 #15 . . . 1.16 PGRE0177 #16 . . . 1.17 PGRE0177 #17 . . . 1.18 PGRE0177 #18 . . . 1.19 PGRE0177 #19 . . . 1.20 PGRE0177 #20 . . . 1.21 PGRE0177 #21 . . . 1.22 PGRE0177 #22 . . . 1.23 PGRE0177 #23 . . . 1.24 PGRE0177 #24 . . . 1.25 PGRE0177 #25 . . . 1.26 PGRE0177 #26 . . . 1.27 PGRE0177 #27 . . . 1.28 PGRE0177 #28 . . . 1.29 PGRE0177 #29 . . . 1.30 PGRE0177 #30 . . . 1.31 PGRE0177 #31 . . . 1.32 PGRE0177 #32 . . . 1.33 PGRE0177 #33 . . . 1.34 PGRE0177 #34 . . . 1.35 PGRE0177 #35 . . . 1.36 PGRE0177 #36 . . .

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3 3 4 6 7 8 9 10 11 12 13 14 15 17 18 19 21 22 23 24 25 26 28 29 30 31 33 35 36 37 38 39 40 41 42 43 44

CONTENTS

1.37 1.38 1.39 1.40 1.41 1.42 1.43 1.44 1.45 1.46 1.47 1.48 1.49 1.50 1.51 1.52 1.53 1.54 1.55 1.56 1.57 1.58 1.59 1.60 1.61 1.62 1.63 1.64 1.65 1.66 1.67 1.68 1.69 1.70 1.71 1.72 1.73 1.74 1.75 1.76 1.77 1.78 1.79 1.80 1.81 1.82 1.83 1.84

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

#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 . #82 . #83 . #84 .

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45 46 47 48 49 50 51 52 53 54 55 56 57 58 60 62 63 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 96

CONTENTS

1.85 PGRE0177 1.86 PGRE0177 1.87 PGRE0177 1.88 PGRE0177 1.89 PGRE0177 1.90 PGRE0177 1.91 PGRE0177 1.92 PGRE0177 1.93 PGRE0177 1.94 PGRE0177 1.95 PGRE0177 1.96 PGRE0177 1.97 PGRE0177 1.98 PGRE0177 1.99 PGRE0177 1.100PGRE0177

CONTENTS

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

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97 98 100 101 102 104 105 107 109 111 112 114 115 117 118 119

List of Figures 1.1 1.2 1.3 1.4 1.5 1.6

Electric potential at point P in relation to ring of radius R . . . . . . . . . . . . . . . P-V curve comparison of an Isothermal and Adiabtic process . . . . . . . . . . . . . Ray tracing diagram of a concave mirror with a longer focal length than object distance Tangential and centripetal acceleration of a particle constrained to a circle . . . . . . Relativistic space time diagram for time-like, space-like and light-like reference frames 3 generic forms of cubic crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

6 9 16 29 42 63

Chapter 1

Physics GRE Solutions 1.1

PGRE0177 #1

Recommended Solution For a moving pendulum in which we can neglect gravity, the only two forces on the the pendulum bob is centripetal acceleration due to its rotation and acceleration due to gravity, g. At point c, the centripetal acceleration should be pointing upwards and acceleration due to gravity must be pointing downwards. Sum these two forces to see that the net force on the pendulum bob can only point up or down, and we can eliminate (B) and (D). Next, check (A) to see that the net forces at point a and b should only have a horizontal acceleration in the left direction and, alternatively, a horizontal acceleration in the right direction for points d and e. Since this is not true in (A), we can eliminate it. Finally, recall that the centripetal acceleration for a pendulum will be minimized at its peaks, so the net acceleration should be pointing downward more than left or right, which more closely matches (C) than (E). Correct Answer (C)

3

1.2. PGRE0177 #2

1.2

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #2

Recommended Solution Start with the equation for friction, f = µFN

(1.1)

the normal force will be equal and opposite to the force due to the gravity, making Equation 1.1 f = µmg but since the net force in the horizontal will be rotational, f in 1.2 is then

4

(1.2)

1.2. PGRE0177 #2

CHAPTER 1. PHYSICS GRE SOLUTIONS

f mv 2 r v2 r

= µmg

(1.3)

= µmg

(1.4)

= µg

(1.5)

v2 µg

(1.6)

r = Now, convert velocity into rotational units, by

v = rω

(1.7)

rev = r 33.3 min rad ≈ πr s 

rad 2π rev





1 min 60 s



(1.8) (1.9)

Finally, substitute Equation 1.9 into Equation 1.6 and solve to get π2 r2 µg µg π2 3 π2 1 3

r = = ≈ ≈ which is closest to (D).

Correct Answer (D)

5

(1.10) (1.11) (1.12) (1.13)

1.3. PGRE0177 #3

1.3

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #3

Figure 1.1: Electric potential at point P in relation to ring of radius R

Recommended Solution Recall Kepler’s third law, ”The square of the orbital period of a planet is directly proportional to the cube of the semimajor axis of its orbit” or, equivalently T 2 ∝ R3 T

∝ R

3/2

which is option (D). Correct Answer (D)

6

(1.14) (1.15)

1.4. PGRE0177 #4

1.4

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #4

Recommended Solution Despite the change in energy, we can utilize the conservation of momentum to get the final and initial momentums pi = 2mvi

(1.16)

pf

(1.17)

= 3mvf

by conservation, pi = pf and we get 2mvi = 3mvf 2 vf = vi 3 Now, we need to find the initial and final energy of the system, as 1 (2m)vi2 = mvi2 2  2 1 3 2 2 Ef = (3m)vf2 = m vi2 = mvi2 2 2 3 3 At which it is clear that the difference between final and initial kinetic energy is 1/3. Ei =

Correct Answer (C)

7

(1.18) (1.19)

(1.20) (1.21)

1.5. PGRE0177 #5

1.5

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #5

Recommended Solution From the equipartion theorem, we know that the average energy for an ”ideal gas” with only translational degrees of freedom is 3 (1.22) hHi = kB T 2 where the 3 in our 3/2 comes from the 3 degrees of freedom for translational motion. For a harmonic oscillator, we must add to this 2 rotational degrees of freedom and 1 degree of freedom for its single dimension of oscillations. This brings the grand total to 6 degrees of freedom, making the average total energy 6 hHi = kB T = 3kB T 2 Correct Answer (D)

8

(1.23)

1.6. PGRE0177 #6

1.6

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #6

Recommended Solution The quickest solution to this problem is simply to be familiar with the P-V curves for adiabatic and isothermal processes. You will likely have seen these in a thermodynamics lab course,

Figure 1.2: P-V curve comparison of an Isothermal and Adiabtic process which is clearly (E). Correct Answer (E)

9

1.7. PGRE0177 #7

1.7

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #7

Recommended Solution From the diagram, we know that the two magnets have similar poles next to one another so we shouldnt have magnetic field lines from one to another, like in (A) (C) and (D). Next, we should get some repelling between the field lines which doesnt show up in (E), so we must select (B). Correct Answer (B)

10

1.8. PGRE0177 #8

1.8

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #8

Recommended Solution This specific problem is commonly used as one of the more simplistic, but by no means trivial, problems to introduce the method of image charges. Going through the entire proof during this exam would be a much more lengthy process than we can finish in a reasonable amount of time so we can really only solve this problem by knowing that a point charge will induce an exactly equal and opposite charge in the grounded plate, in fact choice (D). Correct Answer (D)

11

1.9. PGRE0177 #9

1.9

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #9

Recommended Solution With 5 charges set symmetrically and equidistant about the center, we should have 5 equal electric fields canceling out fields pointing in the opposite direction. Because of symmetry, the sum of all electric fields will equal 0. Correct Answer (A)

12

1.10. PGRE0177 #10

1.10

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #10

Recommended Solution Given two capacitors, C1 = 3 microfarad and C2 = 6 microfarad and potential difference V = 300 volt, the total energy can be found with 1 U = Ceq V 2 2 for capacitors in series, the equivalent capacitance is C1 C2 C1 + C2   (3 µF)(6 µF) = (3 µF) + (6 µF) 18 µF2 = 9 µF = 2 µF

Ceq =

(1.24)

(1.25) (1.26) (1.27) (1.28)

plug Equation 1.28 into Equation 1.24 and solve

U

1 (2 µF)(300 volt)2 2 = 0.09 =

Correct Answer (A)

13

(1.29) (1.30)

1.11. PGRE0177 #11

1.11

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #11

Recommended Solution We start with the thin lens equation 1 1 1 + = do di f

(1.31)

plug in the given values for do and f1 to solve for di 1 1 1 + = 40 cm di 20 cm 40 cm 1+ = 2 di di = 40 cm

(1.32) (1.33) (1.34)

since the first image is to the right of the second lens by 10 cm, for our next calculation we must use an object distance of d0 = −10 cm. Again, using the same equation as before, we get 1 1 1 + = −10 cm di 10 cm −10 cm 1+ = −1 di di = 5 cm which predicts a final image 5.0 cm to the right of the final lens. Correct Answer (A)

14

(1.35) (1.36) (1.37)

1.12. PGRE0177 #12

1.12

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #12

Recommended Solution Start with the mirror equation 1 1 + d0 di 1 di

= =

1 f 1 1 − ?? f d0

(1.38) (1.39)

Now, since we know from the image that the focal distance is larger than the object distance, we know that the RHS of Equation ?? must be negative and, therefore, the image distance is virtual and behind the mirror. Correct Answer (E)

Alternate Solution If we add an object at point O and do some ray tracing, we get Figure 1.3 15

1.12. PGRE0177 #12

CHAPTER 1. PHYSICS GRE SOLUTIONS

Figure 1.3: Ray tracing diagram of a concave mirror with a longer focal length than object distance Correct Answer (E)

16

1.13. PGRE0177 #13

1.13

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #13

Recommended Solution The Rayleigh Criterion can be used to give us the minimum resolution detail of a telescope, λ d so we can solve for d in Equation 1 and plug in our known values to get θ = 1.22

d = 1.22

λ θ

600 nm 3 × 10−5 rad = 2.5 cm

= 1.22

Correct Answer (B)

17

(1.40)

(1.41) (1.42) (1.43)

1.14. PGRE0177 #14

1.14

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #14

Recommended Solution From the description, we know that the detector is 100% efficient because exactly 50% of the samples are being detected when pressed up exactly on one of its two sides. Since this is the case, we just need to find out how much fewer gamma rays will hit the detector at a distance of 1 m away. Since the rays will disperse spherically, we want to take the ratio of the samples as it passes through a circular surface area (AC = πr2 ) to that of the surface area of a sphere (AS = 4πr2 ) at a distance of 1 m = 100 cm. The radius of the circular surface area is 4, so we let AC = 16π cm2 . Next, we can find the spherical surface area as AS = 4πr2

(1.44)

= 4π(100 cm) = 40, 000 cm

2

2

(1.45) (1.46)

Finally, take the ratio of the two areas to get AC AS

16π cm2 40, 000 cm2 = 4 × 10−4 =

Correct Answer (C)

18

(1.47) (1.48)

1.15. PGRE0177 #15

1.15

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #15

Recommended Solution Recall that precision is distinct from accuracy in that accuracy describes how close to the correct or true value a measurement is, while precision is a measurement of how closely grouped or how well a result can be reproduced. In other words, a group of measurements can be entirely incorrect but still be ”precise” if they are all extremely similar to one another. Of the plots given, (A) demonstrates the closest grouping of data points. 19

1.15. PGRE0177 #15

CHAPTER 1. PHYSICS GRE SOLUTIONS

Correct Answer (A)

20

1.16. PGRE0177 #16

1.16

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #16

Recommended Solution From a laboratory methods course, you probably did some problems on measurement uncertainty and came up with the uncertainty equation σ u= √ (1.49) N where σ is the standard deviation and N is the number of samples. Since our data set is discrete and completed over time, we can find the standard deviation via the Poisson distribution σ=

√

x ¯

(1.50)

where the average of our current data set appears to be, x ¯ = 2. This makes our standard √ deviation, σ = 2. Since we want to be within 1% of the average, we take 1% of 2 to get u = 0.02. √

N

2

σ u2 2 = 0.022 = 5000 =

Correct Answer (D)

21

(1.51) (1.52) (1.53)

1.17. PGRE0177 #17

1.17

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #17

Recommended Solution Your first approach to this problem should be to ensure that each of the possible solutions contains all 15 electrons for phosophorous. Checking this, by summing the subscripts, you’ll find 15 for each. This means that every single solution, except for the correct one, should have an incorrect progression from our standard energy level diagram.

From the energy level diagram above, it should be clear that the progression is 1s2 2s2 2p6 3s2 3p3 Correct Answer (B)

22

1.18. PGRE0177 #18

1.18

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #18

Recommended Solution In the problem, 79.0 eV is given as the ionization energy for both electrons which also tells us that the ionization for each electron individually must sum to this number. We also know that the first electron will be easier to pull from the atom because although it feels the same pull from the positively charged nucleus, it also has a negatively charged electron trying to push it away. Thus, we know that the first electron will have a lower ionization energy than half of the total. Since 24.6 eV is the only choice that meets this criteria, we choose (A). Correct Answer (A)

23

1.19. PGRE0177 #19

1.19

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #19

Recommended Solution The sun is powered by nuclear fusion of hydrogen atoms into helium atoms. Because the atomic mass of hydrogen is approximately 1 and the atomic mass of helium is roughly 4, it must be true, by conservation of energy, that 4 hydrogen atoms combine to make 1 helium atom. Correct Answer (B)

24

1.20. PGRE0177 #20

1.20

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #20

Recommended Solution Bremsstrahlung radiation refers to E&M radiation generated when a charged particle gets accelerated as the result of a collision with another charged particle, i.e choice (E). This is one of those problems that you either know, or you don’t. Unless you speak German, in which case you could break the Bremsstrahlung into its components, bremsen ”to brake” and Strahlung ”radiation”. Correct Answer (E)

25

1.21. PGRE0177 #21

1.21

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #21

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-α) is n=2 going to n=1, the second (Lyman-β) 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

!

(1.54)

For this problem we won’t need to compute anything, just compare λL and λB . Doing this for the shortest Lyman transition gives 1 1 1 = R 2− 2 λL 1 2 1 3 = R λL 4 λL = 4/(3R)



1 1 1 = R 2− 2 λB 2 3 1 5 = R λB 36 λB = 36/(5R)





(1.55) (1.56) (1.57)

and for the Balmer transition 

26

(1.58) (1.59) (1.60)

1.21. PGRE0177 #21

CHAPTER 1. PHYSICS GRE SOLUTIONS

making the ratio λL /λB =

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

Correct Answer (B)

27

(1.61)

1.22. PGRE0177 #22

1.22

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #22

Recommended Solution Recall our good friend Galileo, who demonstrated that an object in free fall in a vacuum will fall at the same acceleration regardless of its mass. In the problem given, the moon is our object, space is our vacuum and we conclude that the mass of the moon is unknowable with the given information. Correct Answer (B)

28

1.23. PGRE0177 #23

1.23

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #23

Recommended Solution In this problem, we have three vectors to consider. The first is the centripetal acceleration vector generated by the rotation of the particle. The second two are both tangential to the circle but one is a velocity vector, v = 10 m/s, and the other is a tangential acceleration, aT = 10 m/s2 . First, get the net acceleration by adding aT and aC

Figure 1.4: Tangential and centripetal acceleration of a particle constrained to a circle Since aT and aC are both 10 m/s2 , the angle between them must be 45◦ . This tells us that the net acceleration is always 45◦ from any tangential vectors and since the velocity vector is also a tangential vector, the angle between v and anet is 45◦ . Correct Answer (C)

29

1.24. PGRE0177 #24

1.24

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #24

Recommended Solution Start by realizing that without air resistance, the horizontal velocity will always be constant and positive, so vx vs. t must be plot II and we can eliminate (A) and (E). Next, in the y-direction we know that velocity must not be constant as it reaches some peak and its velocity becomes zero, before then changing direction and increasing its velocity in a negative direction. This description perfectly describes plot III and so we choose (C). Correct Answer (C)

30

1.25. PGRE0177 #25

1.25

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #25

Recommended Solution Start with the moment of inertia for a disk, I = 1/2mr2 , which is given in the front of your test booklet. Next, we can find the moment of inertia for the rest of the pennies using the parallel axis theorem, I = Ic om + ml2 again, the moment of inertia for a single penny is I = center of mass of l = 2r, so we get

Iop = = =

(1.62) 1/2mr2

but they have a radius from the

1 2 mr + ml2 2 1 2 mr + m(2r)2 2 9 2 mr 2

(1.63) (1.64) (1.65)

however, since we have 6 pennies, the sum of all 6 is just 54 2 mr 2 then add the moment of inertia of all 6 pennies with the inner penny to get, Iop,6 =

31

(1.66)

1.25. PGRE0177 #25

CHAPTER 1. PHYSICS GRE SOLUTIONS

Iip + Iop,6 =

54 2 1 2 55 2 mr + mr = mr 2 2 2

Correct Answer (E)

32

(1.67)

1.26. PGRE0177 #26

1.26

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #26

Recommended Solution Before the rod has moved, it only has potential energy of mgL (1.68) 2 where we use L/2 because that is the location of the center of mass. Next, recall the equation for rotational kinetic energy, which all of the energy is converted to at the bottom of the rods fall, U=

mgL 1 = Iω 2 2 2 The moment of inertia for a rod rotating at its tip is 1 I = mL2 3 substitute Equation 1.70 into Equation 1.69 and substitute ω = v/r to get mgL 2

= 33

Iω 2 2

(1.69)

(1.70)

(1.71)

1.26. PGRE0177 #26

CHAPTER 1. PHYSICS GRE SOLUTIONS mL2 v 2 6L2 v2 gL = 3 p v = 3gL =

Correct Answer (C)

34

(1.72) (1.73) (1.74)

1.27. PGRE0177 #27

1.27

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #27

Recommended Solution The Hermitian operator is defined as hAi =

Z

ˆ ψ ∗ (r)Aψ(r)dr

(1.75)

where Aˆ is the expectation value and is an observable. Therefore, Aˆ is, like all other observables, is real valued and we choose (A). Correct Answer (A)

35

1.28. PGRE0177 #28

1.28

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #28

Recommended Solution Recall the condition for orthogonality, hx|xi = 1

(1.76)

from this, we can solve for x by taking

hψ1 |ψ2 i = 0

(1.77)

(5 · 1) + (−3 · −5) + (2 · x) = 0

(1.78)

20 + 2x = 0

(1.79)

x = −10 Correct Answer (E)

36

(1.80)

1.29. PGRE0177 #29

1.29

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #29

Recommended Solution Recall that the expectation value of any state is, hAiψ =

X

aj |hψ|φj i|2

(1.81)

j

where aj is the eigen value. Thus, we square each term, multiply it by its eigenvalue and sum everything up to get ˆ = −1 + 1 + 2 hOi 6 2 3 = 1 Correct Answer (C)

37

(1.82) (1.83)

1.30. PGRE0177 #30

1.30

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #30

Recommended Solution Start by realizing that the wave function should decrease as the radius goes off to infinity, which is not true of option II and so we eliminate (B), (C) and (E). Next, as the radius goes to zero, the wave function should go to A, which is only true of I and we can choose (A). Note that even if you don’t recognize that the wave function should go to A when r → 0, you can at least be sure that it shouldn’t blow up to infinity like it does in option III. Correct Answer (A)

38

1.31. PGRE0177 #31

1.31

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #31

Recommended Solution One exceedingly useful piece of information to memorize is that the ground state energy of positronium is half that of the hydrogen atom, E0,H −13.6 eV = = −6.8 eV (1.84) 2 2 Then, using the Bohr model energy equation, substitute −13.6 eV with −6.8 eV and solve E0,pos =

1 1 − n21 n22   1 = −6.8 eV 1 − 9 = 6.0 eV 

E = −6.8 eV

Correct Answer (A)

39



(1.85) (1.86) (1.87)

1.32. PGRE0177 #32

1.32

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #32

Recommended Solution Start by equating the total relativistic energy, E = γm0 c2 , to the rest energy, E = m0 c2 , in the proportions given Enet = γm0 c2 = 2m0 c2

(1.88)

which clearly tells us that the lorentz factor is γ = 2. Next, solve for velocity in the Lorentz factor

2 =

p

1 1 − v 2 /c2

q

2 1 − v 2 /c2 = 1 3 v 2 /c2 = 4√ v =

3 c 2

(1.89) (1.90) (1.91) (1.92)

Finally, using our relativistic momentum and Lorentz factor, we can solve for p

p = γmo v √

3 = 2m0 c 2 √ = 3m0 c

Correct Answer (D)

40

(1.93) (1.94) (1.95)

1.33. PGRE0177 #33

1.33

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #33

Recommended Solution First, immediately eliminate (E) because a pion can’t achieve light speed. Next, notice that if you try to apply the classical concept of velocity (i.e. ∆x/∆T ), you get a ridiculously large number so relativity must be in full effect and (A) isn’t nearly fast enough. Now, start with our equation for the space time interval ∆s2 = ∆r2 − c2 ∆t2

(1.96)

in the pion’s frame, its change in position is 0 so we have ∆s2π = −c2 ∆t2

(1.97) −

2

8 2

= −(3 × 10 8) (1 × 10 )

(1.98)

= −9 m/s

(1.99)

Now, in the lab frame, we get the space time interval, ∆s2lab = (30 m)2 − c2 ∆t2lab ∆t2lab c2 ∆lab

= 909 √ 909 = √ c2 √ = 101 × 10−8

(1.100) (1.101) (1.102) (1.103)

Finally, using our standard velocity equation, plug in our values to get ∆Xlab ∆tlab 30 m = √ 101 × 10−8 ≈ 2.98 × 108

v =

Correct Answer (D)

41

(1.104) (1.105) (1.106)

1.34. PGRE0177 #34

1.34

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #34

Recommended Solution From the the displacement 4-vector, we get 3 general types of space-time intervals: time-like, space-like and light-like. These are typically drawn out on a space-time diagram as such

Figure 1.5: Relativistic space time diagram for time-like, space-like and light-like reference frames each interval is characterized by the following criteria time-like: if |∆x/∆t| < c, two events occur at the same location space-like: if |∆x/∆t| > 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)

42

1.35. PGRE0177 #35

1.35

CHAPTER 1. PHYSICS GRE 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

(1.107)

From this, the increase in power from tripling the temperature is P = κ(3T )4 = 81κT 4 which is (E). Correct Answer (E)

43

(1.108)

1.36. PGRE0177 #36

1.36

CHAPTER 1. PHYSICS GRE 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)

44

(1.109)

1.37. PGRE0177 #37

1.37

CHAPTER 1. PHYSICS GRE 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 (1.110) 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)

45

1.38. PGRE0177 #38

1.38

CHAPTER 1. PHYSICS GRE 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

(1.111)

plug in your values to get ω = C =

1 LC 1 Lω 2 √

(1.112) (1.113)

1 (25 millihenries)(1 × 103 rad)2 1 = 7 2.5 × 10 millihenries-rad2 = 40 µF =

Correct Answer (D)

46

(1.114) (1.115) (1.116)

1.39. PGRE0177 #39

1.39

CHAPTER 1. PHYSICS GRE 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)

47

1.40. PGRE0177 #40

1.40

CHAPTER 1. PHYSICS GRE 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)

48

1.41. PGRE0177 #41

1.41

CHAPTER 1. PHYSICS GRE 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)

49

1.42. PGRE0177 #42

1.42

CHAPTER 1. PHYSICS GRE 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)

50

1.43. PGRE0177 #43

1.43

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #43

Recommended Solution Start with the commutator relation [AB, C] = A[B, C] + [A, C]B

(1.117)

and apply it to the given commutator [Lx Ly , Lz ] = Lx [Ly , Lz ] + [Lx , Lz ]Ly

(1.118)

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

(1.119)

= Lx (i¯hLx ) + (−i¯hLy )Ly

(1.120)

=

i¯h(L2x

−

L2y )

Correct Answer (D)

51

(1.121)

1.44. PGRE0177 #44

1.44

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #44

Recommended Solution The problem gives us the energy equation as n2 π 2 ¯h2 (1.122) 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)

52

1.45. PGRE0177 #45

1.45

CHAPTER 1. PHYSICS GRE 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)

53

(1.123) (1.124) (1.125)

(1.126) (1.127) (1.128) (1.129)

1.46. PGRE0177 #46

1.46

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #46

Recommended Solution Start with the de Broglie wavelength equation λ=

h p

(1.130)

and then recall the Schrdinger 1-D energy equation for a particle in a potential p2 + V (x) (1.131) 2m without doing much work, you can see that combining Equation 1.130 with Equation 1.131 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)

54

1.47. PGRE0177 #47

1.47

CHAPTER 1. PHYSICS GRE 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)

(1.132)

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

(1.133)

Plugging Equation 1.133 into Equation 1.132 and applying the rules of logarithms gives

S = kb ln(Ω)

(1.134)

n

= kb ln (2 )

(1.135)

= nkb ln(2)

(1.136)

which is given by choice (B). Correct Answer (B)

55

1.48. PGRE0177 #48

1.48

CHAPTER 1. PHYSICS GRE 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)

56

(1.137)

1.49. PGRE0177 #49

1.49

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #49

Recommended Solution A Maxwell-Boltzmann system has the canonical partition function Z=

X

g e−Es /kb T

(1.138)

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

(1.139)

s

= 2e−/kT + 2e−2/kT h

= 2 e−/kT + e−2/kT Correct Answer (E)

57

(1.140) i

(1.141)

1.50. PGRE0177 #50

1.50

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #50

Recommended Solution Recall our equation relating frequency to wavelength, ν=

v λ

(1.142)

if our velocity is 3 vf = 0.97vo

(1.143)

so the initial velocity is ν0 λ v0 440 hz = λ v0 = (440 hz)(λ) νo =

(1.144) (1.145) (1.146)

then, compare this to the final velocity

νf

= = = =

vf λ (0.97v0 ) λ (0.97)(440 hz) λ 427 hz λ

Correct Answer (B) 58

(1.147) (1.148) (1.149) (1.150)

1.50. PGRE0177 #50

CHAPTER 1. PHYSICS GRE SOLUTIONS

Note: if you have any experience with wind instruments, which fortunately I do, you know from experience 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.

59

1.51. PGRE0177 #51

1.51

CHAPTER 1. PHYSICS GRE 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, 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 60

1.51. PGRE0177 #51

CHAPTER 1. PHYSICS GRE SOLUTIONS

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)

61

1.52. PGRE0177 #52

1.52

CHAPTER 1. PHYSICS GRE 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 Thus the primitive unit cell (simple cubic) contains half as much area, a3 /2 Correct Answer (C)

62

1.53. PGRE0177 #53

CHAPTER 1. PHYSICS GRE SOLUTIONS

(a) Simple Cubic

(b) Body Centered Cubic

(c) Face Centered Cubic

Figure 1.6: 3 generic forms of cubic crystals

1.53

PGRE0177 #53

63

1.53. PGRE0177 #53

CHAPTER 1. PHYSICS GRE SOLUTIONS

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)

64

1.54. PGRE0177 #54

1.54

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #54

Recommended Solution Recall that impulse is Z t2

I=

F dt

(1.151)

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)

65

1.55. PGRE0177 #55

1.55

CHAPTER 1. PHYSICS GRE 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)

66

(1.152)

1.56. PGRE0177 #56

1.56

CHAPTER 1. PHYSICS GRE 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

(1.153)

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)

67

(1.154) (1.155) (1.156)

1.57. PGRE0177 #57

1.57

CHAPTER 1. PHYSICS GRE 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)

68

1.58. PGRE0177 #58

1.58

CHAPTER 1. PHYSICS GRE 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)

69

1.59. PGRE0177 #59

1.59

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #59

Recommended Solution The equation for a simple harmonic oscillator is m¨ x + kx = 0

(1.157)

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)

70

1.60. PGRE0177 #60

1.60

CHAPTER 1. PHYSICS GRE 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)

71

1.61. PGRE0177 #61

1.61

CHAPTER 1. PHYSICS GRE 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)

72

1.62. PGRE0177 #62

1.62

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #62

Recommended Solution Start with our cyclotron frequency equation Bq 2πm and then re-write Equation 1.158 to solve for m f=

m=

Bq 2πf

(1.158)

(1.159)

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)

73

(1.160) (1.161) (1.162) (1.163) (1.164)

1.63. PGRE0177 #63

1.63

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #63

Recommended Solution Ideally, we would be able to recall Wien’s displacement law b (1.165) 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)

74

(1.166) (1.167)

1.64. PGRE0177 #64

1.64

CHAPTER 1. PHYSICS GRE 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)

75

1.65. PGRE0177 #65

1.65

CHAPTER 1. PHYSICS GRE 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

(1.168) (1.169)

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

(1.170)

ehν/kT (hν/kT )2

(1.171)

= 3kNA ehν/kT

(1.172)

Finally, as temperature blows up to infinity, ehν/kT = 1 and so we are left with C = 3kNA Correct Answer (D)

76

(1.173)

1.66. PGRE0177 #66

1.66

CHAPTER 1. PHYSICS GRE 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)

77

(1.174)

1.67. PGRE0177 #67

1.67

CHAPTER 1. PHYSICS GRE 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)

78

1.68. PGRE0177 #68

1.68

CHAPTER 1. PHYSICS GRE 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)

79

1.69. PGRE0177 #69

1.69

CHAPTER 1. PHYSICS GRE 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

(1.175)

where m = 1, 2, 3, . . .. The thinnest oil film will be when m = 1, so we solve for d in Equation 1.175 and plug in our values mλ 2n 480 nm = 2(1.2) = 200 nm

d =

Correct Answer (B)

80

(1.176) (1.177) (1.178)

1.70. PGRE0177 #70

1.70

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #70

Recommended Solution Starting with the double slit equation ω sin(θ) = mλ

(1.179)

re-write Equation 1.179 to solve for the angle between fringes, θ, to get mλ (1.180) ω from Equation 1.180, 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)

81

1.71. PGRE0177 #71

1.71

CHAPTER 1. PHYSICS GRE 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

(1.181)

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)

82

(1.182) (1.183) (1.184) (1.185) (1.186) (1.187)

1.72. PGRE0177 #72

1.72

CHAPTER 1. PHYSICS GRE 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) 83

(1.188) (1.189)

1.73. PGRE0177 #73

1.73

CHAPTER 1. PHYSICS GRE 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)

84

(1.190) (1.191) (1.192) (1.193) (1.194)

1.74. PGRE0177 #74

1.74

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #74

Recommended Solution The Lagrangian equation of motion can be calculated with d dt



d dt



∂L ∂ q˙

∂L =0 ∂q

(1.195)

= 2a¨ q

(1.196)

= 4bq 3

(1.197)



−

perform both derivatives to get ∂L ∂ q˙ ∂L ∂ q˙



plugging our results from Equation 1.197 into Equation 1.195 gives d dt



∂L ∂L − = 0 ∂ q˙ ∂q 2a¨ q − 4bq 3 = 0 

(1.198) (1.199) 3

2a¨ q = 4bq 2b 3 q¨ = q a Correct Answer (D)

85

(1.200) (1.201)

1.75. PGRE0177 #75

1.75

CHAPTER 1. PHYSICS GRE 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

(1.202)

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)

86

(1.203)

1.76. PGRE0177 #76

1.76

CHAPTER 1. PHYSICS GRE 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)

87

1.77. PGRE0177 #77

1.77

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #77

Recommended Solution By Maxwell-Boltzmann statistics, we get the equation gi e−i /kT (1.204) Z kT and the energy are both given so plug them both into Equation 1.204 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)

88

(1.205) (1.206) (1.207)

1.78. PGRE0177 #78

1.78

CHAPTER 1. PHYSICS GRE 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)

89

1.79. PGRE0177 #79

1.79

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #79

Recommended Solution Start with the relativistic energy equation E 2 − (pc2 ) = (mc2 )2

(1.208)

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

(1.209)

2 4

(100 GeV ) − (64 GeV ) = m c 36 GeV

2

m

2

(1.210)

2 4

= m c

(1.211) 2

= 36 GeV /c 2

m = 6 Gev/c Correct Answer (D)

90

4

(1.212) (1.213)

1.80. PGRE0177 #80

1.80

CHAPTER 1. PHYSICS GRE 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 (1.214) 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)

91

(1.215) (1.216) (1.217) (1.218)

1.81. PGRE0177 #81

1.81

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #81

Recommended Solution Start with the equations for L2 and Lz L2 Ylm (θ, φ) = l(l + 1)¯h2 Ylm (θ, φ)

(1.219)

Lz Ylm (θ, φ) = m¯hYlm (θ, φ)

(1.220)

applying both eigenvalues to their respective operators, we get l(l + 1)¯h2 = 6¯h2 l = 2

(1.221) (1.222)

and m¯h = −¯h

(1.223)

m = −1

(1.224)

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)

92

(1.225)

1.82. PGRE0177 #82

1.82

CHAPTER 1. PHYSICS GRE 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

(1.226)

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)

93

1.83. PGRE0177 #83

1.83

CHAPTER 1. PHYSICS GRE 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)

(1.227)

| ↓ i = (3, 4)

(1.228)

Now multiply each of those to the transformation matrix given in the problem to get the transformed vectors

| ↑ iσx = (2, 1)

(1.229)

| ↓ iσx = (4, 3)

(1.230)

94

1.83. PGRE0177 #83

CHAPTER 1. PHYSICS GRE 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)

95

1.84. PGRE0177 #84

1.84

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #84

Recommended Solution For a single electron transition, utilize our selection rules for the orbital and total angular quantum numbers

∆l = ±1

(1.231)

∆j = 0, ±1

(1.232)

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)

96

1.85. PGRE0177 #85

1.85

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #85

Recommended Solution The resistance of the wire is related to its length and area by R=

ρl a

(1.233)

this gives us two resistances ρ2L (1.234) A ρL R2 = (1.235) 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

(1.236)

V1 + V2 = 7 volts

(1.237)

Finally, find the net voltage as

and use Equation 1.236 and Equation 1.237 to get V

= 1 + IR

(1.238)

= 1 V + 1.4 V

(1.239)

= 2.4 V

(1.240)

Correct Answer (A)

97

1.86. PGRE0177 #86

1.86

CHAPTER 1. PHYSICS GRE 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

98

(1.241)

(1.242) (1.243)

1.86. PGRE0177 #86

CHAPTER 1. PHYSICS GRE SOLUTIONS

The area and, therefore, the magnetic flux changes during rotation so we find the magnetic flux as φB = Bπr2 sin(ωt)

(1.244)

ε = N Bπr2 ω cos(ωt)

(1.245)

and, therefore, the induced EMF is

Finally, substitute the EMF from Equation 1.245 into Equation 1.241 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)

99

(1.246) (1.247) (1.248)

1.87. PGRE0177 #87

1.87

CHAPTER 1. PHYSICS GRE 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)

100

(1.249) (1.250) (1.251)

1.88. PGRE0177 #88

1.88

CHAPTER 1. PHYSICS GRE 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)

101

(1.252)

1.89. PGRE0177 #89

1.89

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #89

Recommended Solution From conservation of angular momentum, Lf = L0

(1.253)

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

(1.254)

however, for the initial momentum we must use both the child at a distance R and the merrygo-round, 102

1.89. PGRE0177 #89

CHAPTER 1. PHYSICS GRE SOLUTIONS

!

Li = Iω0 =

M R2 + mR2 ω0 2

(1.255)

Plug Equations 1.254 and 1.255 into Equation 1.253 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)

103

(1.256) (1.257) (1.258) (1.259)

1.90. PGRE0177 #90

1.90

CHAPTER 1. PHYSICS GRE 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

(1.260)

and s

T = 2π

l g

(1.261)

√ 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)

104

(1.262)

1.91. PGRE0177 #91

1.91

CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #91

Recommended Solution At the top of the wedge, net energy is just the gravitational energy, Enet = mgh

(1.263)

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 1.264 gives, I=

2mghR2 − R2 m v2

and substituting the value given for v 2 105

(1.264)

(1.265)

1.91. PGRE0177 #91

CHAPTER 1. PHYSICS GRE SOLUTIONS

I = = =

2R2 mgh − R2 m v2 7 2 R m − R2 m 4 3 2 R m 4

Correct Answer (B)

106

(1.266) (1.267) (1.268)

1.92. PGRE0177 #92

1.92

CHAPTER 1. PHYSICS GRE 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 107

1.92. PGRE0177 #92

CHAPTER 1. PHYSICS GRE SOLUTIONS

1 V = k(∆l)2 2

(1.269)

1 p2 T = mv 2 = 2 2m

(1.270)

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)

108

(1.271) (1.272) (1.273)

1.93. PGRE0177 #93

1.93

CHAPTER 1. PHYSICS GRE 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

(1.274) (1.275)

differentiating dP in Equation 1.275 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 −

109

(1.276) (1.277)

1.93. PGRE0177 #93

CHAPTER 1. PHYSICS GRE SOLUTIONS

at which point, it is easy to solve for the radius as r = a0 Correct Answer (C)

110

(1.278)

1.94. PGRE0177 #94

1.94

CHAPTER 1. PHYSICS GRE 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

(1.279)

multiply the Hamiltonian out to get 

a + a†

2

= a2 + aa† + a† a + a†2

(1.280)

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

(1.281)

= (2n + 1)V

(1.282)

= 5V

(1.283)

Correct Answer (E)

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1.95. PGRE0177 #95

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CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #95

Recommended Solution Start by recalling the relative permitivity equation κ=

ε(ω) ε0

(1.284)

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)

112

(1.285)

1.95. PGRE0177 #95

CHAPTER 1. PHYSICS GRE SOLUTIONS

∝

1 0 κ

(1.286)

E0 κ

(1.287)

Then applying E0 ∝ 1/ε0 , we get E=

Correct Answer (A)

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1.96. PGRE0177 #96

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CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #96

Recommended Solution Recall the Larmor formula, which gives the power radiated by a charged object, P =

e2 a2 6π0 c3

(1.288)

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)

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1.97. PGRE0177 #97

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CHAPTER 1. PHYSICS GRE SOLUTIONS

PGRE0177 #97

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CHAPTER 1. PHYSICS GRE 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)

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1.98. PGRE0177 #98

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CHAPTER 1. PHYSICS GRE 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)

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1.99. PGRE0177 #99

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CHAPTER 1. PHYSICS GRE 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)

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1.100. PGRE0177 #100

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CHAPTER 1. PHYSICS GRE 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)

119

(1.289)