Walker4 ISM Ch32

Chapter 32: Nuclear Physics and Nuclear Radiation Answers to Even-Numbered Conceptual Questions 2.

The difference is that in an α decay only a single particle is emitted—the α particle—and it carries the energy released by the decay. In the case of β decay, two particles are emitted—the β particle (electron) and the corresponding antineutrino. These two particles can share the energy of decay in different amounts, which accounts for the range of observed energies for the β particles. (Of course, the antineutrinos are very difficult to detect.)

4.

Alpha particles, which can barely penetrate a sheet of paper, are very unlikely to expose film in a cardboard box. Beta particles, on the other hand, are able to penetrate a few millimeters of aluminum. Therefore, beta particles are more likely to expose the film than alpha particles.

6.

A change in isotope is simply a change in the number of neutrons in a nucleus. The electrons in the atom, however, respond only to the protons with their positive charge. Because electrons are responsible for chemical reactions, it follows that chemical properties are generally unaffected by a change in isotope.

8.

Above the N = Z line, a nucleus contains more neutrons than protons. This helps to make the nucleus stable, by spreading out the positive charge of the protons. If a nucleus were below the N = Z line, it would have more protons than neutrons, and electrostatic repulsion would blow the nucleus apart.

10.

No. Fossil dinosaur skeletons represent organic material—which is necessary for carbon-14 dating—but they are thousands of times too old for the technique to be practical.

12.

Yes. If the different isotopes have different decay rates—which is generally the case—they can still have the same activity if they are present in different amounts.

Solutions to Problems and Conceptual Exercises 1.

Picture the Problem: We are given three isotope designations and asked to calculate the atomic number (A), neutron number (N), and atomic mass (Z) for each isotope. Strategy: Use the designation of an isotope is ZA X , where A is the atomic mass, Z is the atomic number, and X is the chemical designation, to determine the atomic number and atomic mass of the three isotopes. Calculate the neutron number by subtracting the atomic number from the atomic mass. Solution: 1. (a) Write Z, N, and A for the isotope

238 92

U:

Z = 92 N = A − Z = 238 − 92 = 146 A = 238

2. (b) Write Z, N, and A for the isotope

239 94

Z = 94

Pu:

N = A − Z = 239 − 94 = 145 A = 239

3. (c) Write Z, N, and A for the isotope

144 60

Z = 60

Nd:

N = A − Z = 144 − 60 = 84 A = 144

Insight: Note that for each of these isotopes the neutron number is about 1½ times the atomic number. The neutron number is about equal to the atomic number for small isotopes and about 1½ times the atomic number for large isotopes. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 1

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation 2.

Picture the Problem: We are given three isotope designations and asked to calculate the atomic number (A), neutron number (N), and atomic mass (Z) for each isotope. Strategy: Use the designation of an isotope is ZA X , where A is the atomic mass, Z is the atomic number, and X is the chemical designation, to determine the atomic number and atomic mass of the three isotopes. Calculate the neutron number by subtracting the atomic number from the atomic mass. Solution: 1. (a) Write Z, N, and A for the isotope

202 80

Z = 80

Hg:

N = A − Z = 202 − 80 = 122 A = 202

2. (b) Write Z, N, and A for the isotope

220 86

Z = 86

Rn:

N = A − Z = 220 − 86 = 134 A = 220

3. (c) Write Z, N, and A for the isotope

93 41

Z = 41

Nb:

N = A − Z = 93 − 41 = 52 A = 93

Insight: Note that for each of these isotopes, the neutron number is about 1½ times the atomic number. The neutron number is about equal to the atomic number for small isotopes and about 1½ times the atomic number for large isotopes.

3.

Picture the Problem: The nuclear radius increases as the atomic mass number increases. Strategy: Insert the mass number (A) from each isotope into equation 32-4 to calculate the nuclear radius. Solution: 1. (a) Calculate the radius of 2. (b) Calculate the radius of Insight:

4.

197 79

60 27

197 79

r = (1.2 × 10 −15 m ) A1/3 = (1.2 fm )(197 )

1/3

Au (A = 197):

r = (1.2 fm )( 60 )

1/3

Co (A = 60):

Au has over three times the number of nucleons as

60 27

= 7.0 fm

= 4.7 fm

Co , yet its radius in not even twice as large.

Picture the Problem: The number of neutrons in a chlorine nucleus can be determined from its nuclear radius. Strategy: Use equation 32-4 to calculate the mass number of the chlorine atom. The periodic table shows that the atomic number of chlorine is 17. Subtract the atomic number from the atomic mass to calculate the neutron number. 3

r3

⎛ 4.0 × 10−15 m ⎞ =⎜ ⎟ = 37 −15 ⎝ 1.2 × 10 m ⎠

Solution: 1. Solve equation 32-4 for the atomic mass:

A=

2. Solve equation 32-1 for the neutron number:

A = Z + N ⇒ N = A − Z = 37 − 17 = 20

(1.2 ×10

−15

m)

3

Insight: At best we can say there are between 19 and 21 neutrons because the radius is only given to two significant figures. For instance, note that adding one additional neutron to make A = 38 also gives a radius of 4.0 × 10−15 m.

5.

Picture the Problem: The nuclear densities of thorium-238 and an alpha particle (helium-4) can be compared. Strategy: The nuclear density is the mass of the nucleus divided by its volume. Calculate the mass by multiplying the mass number by the mass of one nucleon. The volume is the volume of a sphere whose radius is given by equation 33-4. Solution: 1. (a) Write the density as mass over volume:

ρ=

M Am = 4 3 V 3πr

2. Use equation 33-4 to write the radius and simplify:

ρ=

3 Am

( )

4π r0 A

1 3

3

=

3 Am 3m = 3 4π r0 A 4π r03

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 2

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation 3. Insert the constants to calculate the nuclear density of 22890Th :

ρ=

3 (1.67 × 10−27 kg )

4π (1.2 × 10

−15

m)

3

= 2.3 × 1017 kg/m3

4. (b) The nuclear density of an alpha particle will be the same as the nuclear density of thorium because the nuclear density is independent of the mass number.

ρ = 2.3 × 1017 kg/m3

5. (c) Write the nuclear density of an alpha particle:

Insight: The density of all nuclei is approximately 2.3×1017 kg/m3, regardless of the identity of the isotope.

6.

Picture the Problem: An alpha particle with initial kinetic energy approaches a stationary gold nucleus and comes to rest due to the electrostatic repulsion between the two particles. Strategy: We want to know the required kinetic energy such that the alpha particle comes to rest a distance d = 22.5 fm from the gold nucleus. Set the initial kinetic energy equal to the electrostatic potential energy (equation 20-8) at a distance of 22.5 fm. Solution: 1. (a) Set the initial kinetic energy equal to the electrostatic potential energy, noting that Z = 79 and q = 2e:

K i = Ef =

k ( Ze ) q d

(8.99 ×10 =

= 1.62 × 10 −12 J ×

N ⋅ m 2 /C2 ) ( 79 )( 2 ) (1.60 × 10−19 C )

9

2

22.5 × 10−15 m

1 MeV = 10.1 MeV 1.60 × 10−13 J

2. (b) The kinetic energy is reduced by a factor of 4. As a result, the alpha particle is unable to get as close to the gold nucleus. Because the potential energy is inversely proportional to distance, the distance of closest approach is increased by a multiplicative factor of 4. Insight: Likewise, doubling the initial speed will decrease the distance of closest approach by a factor of 4.

7.

Picture the Problem: An alpha particle with initial kinetic energy approaches a stationary gold nucleus and comes to rest due to the electrostatic repulsion between the two particles. Strategy: Use equation 7-6 to calculate the velocity from the kinetic energy. Set the initial kinetic energy equal to the electrostatic potential energy (equation 20-8) at the distance of closest approach and solve for the distance. Finally, replace the nuclear charge of gold in the electrostatic potential energy equation with the charge on copper to calculate the distance of closest approach to the copper nucleus. The mass of the alpha particle is the mass of 42 He minus the mass of two electrons, or mα = mHe-4 − 2me = 4.002603 u − ( 2 )( 0.000549 u ) = 4.001505 u ( 931.5 MeV/c 2 1 u ) = 3727 MeV/c 2 . 1 2 mv ⇒ v = 2

Solution: 1. (a) Solve equation 7-6 for v:

K=

2. Insert the numerical values:

v=

3. (b) Set the kinetic energy equal to the electrostatic energy and solve for the distance:

K i = Ef = d=

2 ( 0.85 MeV )

=c

3727 MeV/c 2 k ( Ze ) q d

(8.99 ×10

9

2K m

1.7 = 0.021c = 6.4 × 106 m/s 3727

⇒ d=

N ⋅ m /C 2

2

k ( Ze ) q Ef

) ( 79 )( 2 ) (1.60 ×10

0.85 MeV × 1.60 × 10−13 J/MeV

−19

C)

2

= 0.27 pm

4. (c) Because Z = 29 for copper and Z = 79 for gold, the repulsive force the alpha particle experiences is much less when it approaches the copper nucleus. Thus, the distance of closest approach would be less than that found in part (b). Insight: Verify for yourself that the distance of closest approach for copper is 0.098 pm for a 0.85 MeV alpha particle.

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 3

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

8.

Picture the Problem: A marble that has same density as a nucleus would be so massive that a relatively small number of them would have a mass equal to the mass of Earth. Strategy: Solve equation 15-1 for the mass in terms of the density and volume. Set the volume of the marble equal to the volume of a sphere and solve for the mass. Finally, divide the mass of Earth by the mass of the marble to determine the number of marbles equivalent to the mass of Earth. Solution: 1. (a) Solve equation 15-1 for m:

3 m = ρV = ( 2.3 × 1017 kg m3 ) ⎡ 43 π ( 0.015 m ) ⎤ = 3.3 × 1012 kg ⎣ ⎦

2. (b) Divide the mass of Earth by the mass of the marble:

N=

M E 5.97 × 1024 kg = = 1.8 × 1012 marbles m 3.3 × 1012 kg

Insight: Although 1.8×1012 seems like a large number of marbles, if the marbles were spread out only one-marble deep they would cover an area less than half the size of Rhode Island.

9.

Picture the Problem: The nuclear radius increases as the atomic mass number increases. Strategy: Use equation 32-4 to calculate the radius of the phosphorous nucleus with A = 30. To calculate the number of nucleons that would form a nucleus that is twice as large, multiply the radius of the phosphorous nucleus by two and solve equation 32-4 for the number of nucleons. r = (1.2 × 10 −15 m ) A1/3 = (1.2 fm )( 30 )

1/3

Solution: 1. (a) Set A = 30 in equation 32-4:

= 3.7 fm

2. (b) Because the mass number is proportional to the radius cubed, doubling the radius requires 23 = 8 times the mass number. So, A = 8 ( 30 ) = 240 . r = (1.2 fm ) A1 3 = 2rP = 2 ( 3.7 fm ) = 7.4 fm

3. (c) Set r = 2rP and solve for A:

A1/3 = ( 7.4 fm ) (1.2 fm ) = 6.2 A = ( 6.2 ) = 240 3

Insight: Isotopes with 240 nuclei are some of the largest elements, such as neptunium, plutonium, and americium.

10. Picture the Problem: The nuclear radius increases as the atomic mass number increases. Strategy: We want to calculate the number of nucleons that would be contained in a nucleus that is twice the radius of an alpha particle. Equation 32-4 indicates that the radius is proportional to the cube root of the atomic mass number A. Therefore, in order to double the radius, you must increase A by the factor 23 = 8. A = 8 AHe = 8 ( 4 ) = 32

Solution: 1. (a) Multiply the mass number of helium by 8:

32 15

2. (b) Write the symbol for phosphorus with A = 32:

P

Insight: Helium is one of the smallest elements. Doubling its radius produces an element the size of phosphorous. As noted in problem 9, some of the largest isotopes are only twice the radius of phosphorous. Therefore, the radii of known nuclei vary by only about a factor of four from the smallest to the largest.

11. Picture the Problem: A uranium nucleus (A = 236) splits into two nuclei of the same size (A = 118). We want to calculate the radius of the uranium nucleus and the radii of the two daughter nuclei. Strategy: Use equation 32-4 to calculate the radius of each nucleus. Solution: 1. (a) When the volume is reduced by a factor of 2, the radius becomes

1 2 ≈ 0.8 times the initial radius.

3

Therefore, the new radius is more than one-half the original radius. r = (1.2 × 10−15 m ) A1/3 = (1.2 fm )( 236 )

1/3

2. (b) Insert A = 236 into equation 32-4: 3. (c) Insert A =

1 2

( 236 ) = 118

r = (1.2 fm )(118 )

1/3

into equation 32-4:

= 7.4 fm

= 5.9 fm

Insight: To have the daughter nuclei with ½ the radius of the uranium nucleus, it would have to break into eight nuclei of equal atomic number. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 4

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

12. Picture the Problem: A large number of nucleons are required to create a nucleus that has a weight of one pound. Strategy: Because the weight is given as one pound, first convert the weight to newtons. Divide the weight by the acceleration of gravity to calculate the mass. Calculate the atomic mass number by dividing the mass by the mass of one nucleon. Insert the atomic mass number into equation 32-4 to calculate the radius.

W 1 lb ( 4.45 N/lb ) = = 0.454 kg g 9.81 m/s 2

Solution: 1. (a) Calculate the mass in kilograms:

M =

2. Divide by the mass of one nucleon:

A=

3. (b) Insert the atomic mass into equation 32-4:

r = (1.2 × 10−15 m )( 3 × 1026 )

M 0.454 kg = = 3× 1026 nucleons m 1.67 × 10−27 kg 1/3

= 800 nm

Insight: The nucleus would be about the size of a microscopic bacterium, but would weigh as much as a pint of milk!

13. Picture the Problem: A nucleus undergoes α decay. Strategy: The radius of a nucleus depends upon the total number of nucleons A that it contains (equation 32-4). Alpha decay involves the emission of four nucleons from the nucleus. Use these facts to answer the conceptual question. Solution: 1. (a) In α decay, the radius of the daughter nucleus is less than that of the original nucleus, because the daughter nucleus contains four fewer nucleons. 2. (b) The best explanation is II. When the nucleus undergoes decay it ejects two neutrons and two protons. This decreases the number of nucleons in the nucleus, and therefore its radius will decrease. Statements I and III are each false. Insight: Statement III is true for beta decay (β) or gamma decay (γ), but not alpha decay.

14. Picture the Problem: A nucleus undergoes β decay. Strategy: The radius of a nucleus depends upon the total number of nucleons A that it contains (equation 32-4). Beta decay involves the emission of an electron from the nucleus, but the number of nucleons remains unchanged. Use these facts to answer the conceptual question. Solution: 1. (a) In β decay the radius of the daughter nucleus is the same as that of the original nucleus, because the daughter nucleus contains the same number of nucleons. 2. (b) The best explanation is III. When a nucleus emits a β particle a neutron is converted to a proton, but the number

of nucleons is unchanged. As a result, the radius of the daughter nucleus is the same as that of the original nucleus. Statements I and II are each false. Insight: In reference to statement I, it is possible for a nucleus to gain an electron in a process called electron capture. However, even in that process the number of nucleons remains the same, and so does the radius of the nucleus.

15. Picture the Problem: A nucleus undergoes α, β, or γ decay. Strategy: The chemical identity of an atom is determined by the number of protons in its nucleus, so that any nuclear decay process that changes the number of protons in the nucleus will result in the production of a new element. Solution: Both alpha (α) and beta (β) decay result in a new element, because the atomic number Z changes. For example, in alpha decay the atomic number of a nucleus changes from Z to Z – 2. In beta decay, the atomic number changes from Z to Z + 1. In contrast, gamma (γ) decay is simply a release of energy with no change in atomic number. Insight: While it is possible to make gold out of thallium or platinum by means of alpha decay or beta decay, respectively, the alchemist’s dream of producing large quantities of gold remains elusive because of the long time and large expense it would require to produce pure gold by either radioactive decay or nuclear reaction processes. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 5

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

16. Picture the Problem: We are given part of a nuclear reaction and are asked to balance the equation by determining the missing constituent. Strategy: Set the sum of the atomic numbers on each side of the equation equal in order to calculate the unknown atomic number. From the atomic number determine the element. Set the sum of the atomic masses on each side of the equation equal to calculate the unknown atomic mass. Using the atomic mass and atomic number complete the equation. Solution: 1. Calculate the unknown atomic number:

3 + 1 = 2 + Z ⇒ Z = 2, which is helium, or He

2. Calculate the unknown atomic mass:

7 +1 = 4 + A ⇒ A = 4

3. Complete the nuclear reaction:

7 3

Li + 11 H → 42 He +

4 2

He

Insight: The atomic mass number and the charge are always conserved in a nuclear reaction.

17. Picture the Problem: We are given part of a nuclear reaction and are asked to balance the equation by determining the missing constituent. Strategy: Set the sum of the atomic numbers on each side of the equation equal in order to calculate the unknown atomic number. From the atomic number determine the element. Set the sum of the atomic masses on each side of the equation equal to calculate the unknown atomic mass. Using the atomic mass and atomic number complete the equation. Solution: 1. Calculate the unknown atomic number:

90 = 88 + Z ⇒ Z = 2, which is helium, or He.

2. Calculate the unknown atomic mass:

234 = 230 + A ⇒ A = 4

3. Complete the nuclear reaction:

234 90

Th →

230 88

Ra +

4 2

He

Insight: In this reaction, the thorium underwent an alpha decay to produce radium.

18. Picture the Problem: We are given part of a nuclear reaction and are asked to balance the equation by determining the missing constituent. Strategy: Set the sum of the atomic numbers on each side of the equation equal in order to calculate the unknown atomic number. Although it is not expressly written, the electron has an atomic number Z = −1. The neutrino does not carry a charge so it has an atomic number Z = 0. From the atomic number determine the element. Set the sum of the atomic masses on each side of the equation equal to calculate the unknown atomic mass. The atomic mass number for the electron and the neutrino is zero. Using the atomic mass and atomic number complete the equation. Solution: 1. Calculate the unknown atomic number:

Z = 7 + ( −1) + 0 = 6, which is carbon, or C.

2. Calculate the unknown atomic mass:

A = 14 + 0 = 14

3. Complete the nuclear reaction:

14 6

C → 147 N + e – + υ

Insight: In this reaction, carbon-14 decays by beta-minus decay to nitrogen-14. In the process, an electron anti-neutrino is also produced.

19. Picture the Problem: We are given a nuclear decay series that includes α and β decay processes. Strategy: Compare the daughter product of each decay event with the parent isotope. If the number of protons has decreased by two and the atomic mass has decreased by 4 amu, an alpha decay has occurred. If the number of protons increases by one while the atomic mass has remained the same, a beta decay has occurred. Solution: The 14 decays in this series are as follows: α decay; β decay; β decay; α decay; α decay; α decay; α decay; β decay; β decay; α decay; α decay; β decay; α decay; β decay. Insight: A gamma decay would not change either the number of protons or the atomic mass. A gamma decay only involves the removal of energy from the nucleus by means of a high-energy photon.

20. Picture the Problem: We are given part of a nuclear reaction and are asked to balance the equation by determining the missing constituent, and to calculate the amount of energy released in the reaction. Strategy: Set the sum of the atomic numbers on each side of the equation equal in order to calculate the unknown atomic number. Set the sum of the atomic masses on each side of the equation equal to calculate the unknown atomic mass. From the atomic mass and atomic number determine the missing particles in the equation. To calculate the energy released, calculate the change in mass during the reaction. Multiply this change in mass by the speed of light squared to Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 6

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

calculate the energy released. Solution: 1. Find the unknown atomic number:

1 = 2 + Z ⇒ Z = −1

2. Calculate the unknown atomic mass:

4 = 4+ A ⇒ A = 0

The first particle is an electron because Z = −1 and A = 0. Beta-minus decay also produces an anti-neutrino.

H → 23 He + e – + υ

3. Complete the nuclear reaction:

3 1

4. Look up the masses of hydrogen and helium in Appendix F and take their difference:

mH = 3.016049 u

5. Multiply by c2 and convert to eV:

mHe = 3.016029 u

Δm = mHe − mH = 3.016029 u − 3.016049 u = − 0.000020 u ⎛ 931.5 MeV/c 2 ⎞ 2 E = Δm c 2 = 0.000020 u ⎜ ⎟ c = 19 keV 1u ⎝ ⎠

Insight: The masses in Appendix F include the mass of the electrons in each atom. The mass of hydrogen includes one electron, which is present before the decay. The mass of the helium atom includes two electrons. In this problem, the two electrons are the electron from the hydrogen atom and the electron from the beta decay. The mass of the neutrino is small enough that it can be ignored.

21. Picture the Problem: Two isotopes undergo radioactive alpha decay. Strategy: Determine the daughter isotope for each parent isotope by subtracting the atomic number of the alpha particle (Z = 2) from the atomic number of the parent. Also subtract the atomic mass of the alpha particle (A = 4) from the atomic mass of the parent. Determine the daughter isotope from these results and write the reaction equation. To calculate the energy released, subtract the masses of the daughter products (including the alpha particle) from the parent mass, where the masses are given in Appendix F. Multiply the mass difference by c2 and convert the answers to MeV. Solution: 1. (a) Determine the daughter isotope:

Z = 84 − 2 = 82, which is lead, or Pb. A = 212 − 4 = 208

Po →

208 82

Pb + 42 He

2. Write the reaction equation:

212 84

3. Calculate the mass difference:

mi = 211.988852 u mf = 207.97664 u + 4.002603 u = 211.97924 u Δm = mf − mi = 211.97924 u − 211.988852 u = − 0.00961 u

4. Multiply by c2 and convert to MeV: 5. (b) Determine the daughter isotope: 6. Write the reaction equation: 7. Calculate the mass difference:

⎛ 931.5 MeV/c 2 E = Δm c 2 = 0.00961 u ⎜ 1u ⎝

⎞ 2 ⎟ c = 8.95 MeV ⎠

Z = 94 − 2 = 92, which is uranium, or U. A = 239 − 4 = 235 239 94

4 Pu → 235 92 U + 2 He

mi = 239.052158 u mf = 235.043925 u + 4.002603 u = 239.046528 u Δm = mf − mi = 239.046528 u − 239.052158 u = − 0.005630 u

8. Multiply by c2 and convert to MeV:

⎛ 931.5 MeV/c 2 ⎞ 2 E = Δm c 2 = 0.005630 u ⎜ ⎟ c = 5.244 MeV 1u ⎝ ⎠

Insight: The masses in Appendix F include the mass of the electrons in the atom. In this problem the parent atoms had two more electrons than the daughter isotopes. These electrons were accounted for by using the mass of helium (which contains two electrons) rather than the mass of the alpha particle (helium nucleus). Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 7

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

22. Picture the Problem: Two isotopes undergo radioactive β − decay. Strategy: Determine the daughter isotope for each parent isotope by subtracting the atomic number of the β − particle (Z = −1) from the atomic number of the parent. The atomic mass number of the daughter is equal to the atomic mass number of the parent because the atomic mass number of the β − particle is zero. Determine the daughter isotope from these results and write the reaction equation. Include the creation of an anti-neutrino in the reaction. To calculate the energy released, subtract the masses of the daughter products from the parent mass, where the masses are given in Appendix F. Multiply the mass difference by c2 and convert the answers to MeV. Solution: 1. (a) Determine the daughter isotope:

Z = 16 − ( −1) = 17, which is chlorine, or Cl.

2. Write the reaction equation:

35 16

A = 35

3. Calculate the mass difference:

35 S → 17 Cl + e – + υ

mi = 34.969033 u mf = 34.968853 u

Δm = 34.968853 u − 34.969033 u = − 0.000180 u

4. Multiply by c2 and convert to MeV:

⎛ 931.5 MeV/c 2 E = Δm c 2 = 0.000180 u ⎜ 1u ⎝

⎞ 2 ⎟ c = 0.168 MeV ⎠

5. (b) Determine the daughter isotope:

Z = 82 − ( −1) = 83, which is bismuth, or Bi. A = 212

6. Write the reaction equation: 7. Calculate the mass difference:

212 82

Pb →

212 83

Bi + e – + υ

mi = 211.99188 u mf = 211.991272 u

Δm = 211.991272 u − 211.99188 u = − 0.00061 u

8. Multiply by c2 and convert to MeV:

⎛ 931.5 MeV/c 2 ⎞ 2 E = Δm c 2 = 0.00061 u ⎜ ⎟ c = 0.57 MeV 1u ⎝ ⎠

Insight: The masses in Appendix F include the mass of the electrons in the atom. The mass of the parent includes one fewer electron than the daughter product. The extra electron included in the daughter mass from Appendix F accounts for the mass of the beta particle when calculating the mass difference of the entire reaction. The mass of the neutrino is small enough that it can be ignored.

23. Picture the Problem: Two isotopes undergo radioactive β + decay. Strategy: Determine the daughter isotope for each parent isotope by subtracting the atomic number of the β + particle (Z = 1) from the atomic number of the parent. The atomic mass number of the daughter is equal to the atomic mass number of the parent because the atomic mass number of the β + particle is zero. Determine the daughter isotope from these results and write the reaction equation. Include the production of a neutrino in the decay. To calculate the energy released, subtract the mass of the daughter products (including the mass of two electrons) from the parent mass. The masses are given in Appendix F. Multiply the mass difference by c2 and convert the answers to MeV. Solution: 1. (a) Determine the daughter isotope:

Z = 9 − 1 = 8, which is oxygen, or O. A = 18

2. Write the reaction equation:

18 9

3. Calculate the mass difference:

mi = 18.000938 u

F → 188 O + e + + υ

mf = 17.999159 u + 2 ( 5.49 × 10−4 u ) Δm = 18.000257 u − 18.000938 u = − 6.81× 10−4 u Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 8

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation 4. Multiply by c2 and convert to MeV:

⎛ 931.5 MeV/c 2 ⎞ 2 E = Δm c 2 = 6.81× 10−4 u ⎜ ⎟ c = 0.634 MeV 1u ⎝ ⎠

5. (b) Determine the daughter isotope:

Z = 11 − 1 = 10, which is neon, or Ne, and A = 22

6. Write the reaction equation:

22 11

Na →

22 10

Ne + e + + υ

mi = 21.994435 u

7. Calculate the mass difference:

mf = 21.991384 u + 2 ( 5.49 × 10−4 u ) Δm = 21.992482 u − 21.994435 u = − 0.001953 u ⎛ 931.5 MeV/c 2 E = Δm c 2 = 0.001953 u ⎜ 1u ⎝

8. Multiply by c2 and convert to MeV:

⎞ 2 ⎟ c = 1.819 MeV ⎠

Insight: The masses in Appendix F include the mass of the electrons in the atom. The mass of the parent includes one more electron than the daughter product, but that electron still orbits the nucleus in the daughter product after the nuclear decay. The mass of the extra electron in the daughter, as well as the mass of the β + particle, must be taken in to account when calculating the mass difference. The mass of the neutrino is small enough that it can be ignored.

24. Picture the Problem:

211 82

Pb undergoes radioactive β − decay to

211 83

Bi .

Strategy: To calculate the energy released, subtract the mass of the 211 83 Bi from the mass of the 2 are given in Appendix F. Multiply the mass difference by c and convert the answer to MeV.

211 82

Pb , where the masses

mi = 210.98874 u, mf = 210.98726 u

Solution: 1. Calculate the mass difference:

Δm = 210.98726 u − 210.98874 u = − 0.00148 u ⎛ 931.5 MeV/c 2 E = Δm c 2 = 0.00148 u ⎜ 1u ⎝

2. Multiply by c2 and convert to MeV:

Insight: The masses in Appendix F include the mass of the electrons in the atom. The mass of 211 83

211 82

⎞ 2 ⎟ c = 1.38 MeV ⎠ Pb includes one fewer

211 83

electron than Bi . The extra electron included in the mass of Bi accounts for the mass of the beta particle when calculating the mass difference of the entire reaction. The mass of the neutrino is small enough that it can be ignored. 25. Picture the Problem:

66 28

Ni undergoes radioactive β − decay.

Strategy: Determine the daughter isotope by subtracting the atomic number of the β − particle (Z = −1) from the atomic number of nickel (A = 28). The atomic mass number is unchanged in the reaction because the atomic mass number of the β − particle is zero. Determine the daughter isotope from these results and write the reaction equation. The maximum kinetic energy of the released electron will be equal to the total energy released in the reaction. To 2 calculate the energy released, subtract the mass of the daughter product from the mass of 66 28 Ni . Multiply the result by c and convert the answer to MeV. Solution: 1. (a) Determine the daughter isotope:

Z = 28 − ( −1) = 29, which is copper, or Cu. A = 66 Ni →

66 29

Cu + e − + υ

2. Write the reaction equation:

66 28

3. (b) Calculate the energy released:

⎛ 931.5 MeV/c 2 E = Δm c 2 = ( 65.9291 − 65.9289 u ) ⎜ 1u ⎝

⎞ 2 ⎟ c = 0.2 MeV ⎠

Insight: Almost all of the released energy will be in the kinetic energy of the electron because the mass of the copper is much greater than the mass of the electron. The neutrino was first hypothesized because the kinetic energy of the electron is generally smaller than the energy released by β − decays. The only possible explanation is for another (undetected) particle to carry away some of the energy; we now know this particle is a neutrino. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

26. Picture the Problem: Carbon-14 undergoes beta decay with a half-life of 5730 y. Strategy: Consider the random nature of radioactive decay when answering the conceptual questions. Solution: 1. (a) Yes, it is possible for a particular nucleus in a sample of carbon-14 to decay after only 1 s has passed because of the entirely random nature of radioactive decay. The half-life only describes the average time required for half of a large number of nuclei to decay. 2. (b) Yes, it is possible for a particular nucleus in a sample of carbon-14 to decay after 10,000 y have passed because of the entirely random nature of radioactive decay. Insight: All we can say for sure is that half of the initial carbon-14 nuclei (on average) will have decayed after 5730 years have passed.

27. Picture the Problem: Carbon-14 undergoes beta decay with a half-life of 5730 y. Strategy: Consider the assumptions involved in carbon-14 dating when answering the conceptual question. Solution: When we assign ages based on carbon-14 dating, we assume that the ratio of carbon-14 to carbon-12 in the atmosphere has remained constant. If the ratio had been smaller 10,000 years ago, the initial decay rate of carbon-14 would have been smaller as well. As a result, less time would be required for the decay rate to decrease to its current value. Therefore, the true age of an object would be less than the age we have assigned to it. Insight: The atmospheric ratio of carbon-14 to carbon-12 can be affected by changes in the flux of cosmic radiation.

28. Picture the Problem: After two days only one-quarter of an isolated radioactive sample is still radioactive. Strategy: Use the definition of radioactive half-life to answer the conceptual question. Solution: A radioactive sample will decrease by a factor of two in one half-life, and by a factor of four in two half-lives. Therefore, this sample has been in the closed container for two half-lives. It follows that its half-life is one day. Insight: The atmospheric ratio of carbon-14 to carbon-12 can be affected by changes in the flux of cosmic radiation.

29. Picture the Problem: Radioactive decay of radon occurs at a rate characterized by the decay constant and the half-life. Strategy: Solve equation 32-10 for the decay constant. Solution: Solve equation 32-10 for λ:

T1 2 =

ln 2

λ

⇒ λ=

ln 2 ln 2 = = 0.181 d −1 T1 2 3.82 d

Insight: Because the half-life is measured in days in this example, the decay constant is written in units of d−1. It can also be written as 2.1× 10−6 s −1 .

30. Picture the Problem: Radioactive decay occurs at a rate characterized by the decay constant and the half-life. Strategy: Use equation 32-10 to write the half-life from the decay constant.

ln 2 = 78 s 8.9 × 10−3 s −1 Insight: An examination of the half-lives in Appendix F indicates that this substance is probably thallium-210. T1/ 2 =

Solution: Solve equation 32-10:

ln 2

λ

=

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32 – 10

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

31. Picture the Problem: The number of radioactive nuclei in a sample decays to a sixteenth of its initial value within a specified period of time. Strategy: In each half-life the number of nuclei drops by a factor of two. Calculate how many half-lives are necessary to drop to a sixteenth of the initial number of nuclei. Divide the total time by the number of half-lives to calculate the half-life. n

Solution: 1. Calculate the number of half-lives:

1 ⎛1⎞ ⇒ n=4 ⎜ ⎟ = ⎝ 2 ⎠ 16

2. Divide the time by the number of half-lives:

T1 2 =

T 18 d = = 4.5 d n 4

Insight: Note that after each half-life the number of radioactive nuclei decreases by one half. Therefore, after 4.5 days 1/2 are left, after 9.0 days 1/4 are left, after 13.5 days 1/8 are left, and after 18 days 1/16 remain.

32. Picture the Problem: Radioactive decay of oxygen-15 occurs at a rate characterized by a constant half-life. Strategy: Use equation 32-10 to calculate the decay constant from the half-life. Solve equation 32-9 for the time when N = 10−4 N0. Solution: 1. Solve equation 32-10 for λ:

λ=

ln 2 ln 2 = = 0.00568 s −1 T1/ 2 122 s

N = 10− 4 N 0 = N 0 e − λ t

2. Set N = 10−4 N 0 in equation 32-9 and simplify:

eλ t = 104 t=

3. Take the natural log of both sides and solve for t:

ln104

λ

=

ln104 = 1620 s = 27.0 min 0.00568 s −1

Insight: Oxygen-15 is commonly used as a tracer in medical tests. After about 30 minutes, the radioactive decay of the isotope has dropped below detectable levels.

33. Picture the Problem: The activity of a technetium-99 sample depends upon the decay constant and the number of radioactive nuclei it contains. Strategy: Use equation 32-10 to calculate the decay constant from the half-life. Then solve equation 32-11 for the necessary nuclei. Solution: 1. (a) Calculate λ:

λ=

ln 2 ln 2 = = 0.115 h −1 T1/ 2 6.05 h

2. (b) Solve equation 32-11 for the number of nuclei:

N=

R

Insight: This amount of pure

22 43

λ

=

1.50 × 10−6 Ci ⎛ 3600 s ⎞ ⎛ 3.7 × 1010 s -1 ⎞ 9 ⎟ = 1.7 × 10 nuclei ⎜ ⎟⎜ 0.1146 h -1 ⎝ h ⎠ ⎝ 1 Ci ⎠

Tc would have a mass of only 2.8 × 10−13 g .

34. Picture the Problem: After the number of radioactive nuclei has decreased, so has the activity of the sample. Strategy: Use equation 32-12 to calculate the activity from the initial activity and the decay constant.

R = R0 e − λt = (1.50 μ Ci ) e

Solution: Insert t = 2.05 h into equation 32-12:

(

)

− 0.115 h −1 ( 2.05 h )

= 1.20 μ Ci

Insight: If the doctor wants an activity of 1.50 µCi at the time of injection (2.05 hours after preparation) the initial activity would need to be 1.9 µCi.

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32 – 11

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

35. Picture the Problem: The radioactive decay of carbon-14 can be used to determine the age of fossilized remains. Strategy: Use equation 32-10 to calculate the decay constant for carbon-14 from the half-life (5730 y). Then solve equation 32-9 for the time, setting N = 0.0925 N 0 . ln 2 ln 2 = = 1.210 × 10− 4 y −1 T1/ 2 5730 y

Solution: 1. Calculate the decay constant:

λ=

2. Set N = 0.0925 N 0 in equation 32-9 and simplify:

0.0925 N 0 = N 0 e − λ t ⇒ 0.0925 = e− λ t

3. Take the natural log of both sides and solve for t:

t=−

ln 0.0925

λ

=−

ln 0.0925 = 1.97 × 104 y 1.210 × 10− 4 y −1

Insight: According to the carbon dating, it has been 19,700 years since the grass in the basket stopped growing.

36. Picture the Problem: The radioactive decay of carbon-14 can be used to determine the age of fossilized remains. Strategy: Use equation 32-10 to calculate the decay constant for carbon-14 from the half-life (5730 y). Then solve equation 32-9 for the time, setting N = 0.150 N 0 . ln 2 ln 2 = = 1.210 × 10− 4 y −1 T1/ 2 5730 y

Solution: 1. Calculate the decay constant:

λ=

2. Set N = 0.150 N 0 in equation 32-9 and simplify:

0.150 N 0 = N 0 e− λ t ⇒ 0.150 = e − λ t

3. Take the natural log of both sides and solve for t:

t=−

ln 0.150

λ

=−

ln 0.150 = 1.57 × 104 y 1.210 × 10− 4 y −1

Insight: According to the carbon dating it has been 15,700 years since the saber-toothed tiger died.

37. Picture the Problem: The radioactive decay of carbon-14 can be used to determine the age of fossilized remains. Strategy: Use equation 32-10 to calculate the decay constant for carbon-14 from the half-life (5730 y). Then solve equation 32-9 for the time, setting N = 0.175 N 0 . ln 2 ln 2 = = 1.210 × 10− 4 y −1 T1/ 2 5730 y

Solution: 1. Calculate the decay constant:

λ=

2. Set N = 0.175 N 0 in equation 32-9 and simplify:

0.175 N 0 = N 0 e − λ t ⇒ 0.175 = e− λ t

3. Take the natural log of both sides and solve for t:

t=−

ln 0.175

λ

=−

ln 0.175 = 1.44 × 10 4 y 1.210 × 10− 4 y −1

Insight: According to the carbon dating it has been 14,400 years since the tree died and was burned in the fire pit.

38. Picture the Problem: The activity of a gold-198 sample depends upon its decay constant and the number of radioactive nuclei that it contains. Strategy: First use equation 32-10 to calculate the decay constant for gold-198, and then use equation 32-11 to calculate the number of nuclei necessary to produce the activity. Finally divide the number of nuclei by Avogadro’s number and multiply the result by the molar mass (198 g/mole). Solution: 1. Calculate the decay constant:

λ=

ln 2 ln 2 = = 0.2567 d −1 T1/ 2 2.70 d

2. Solve equation 32-11 for N:

N=

R

3. Calculate the mass:

m=M

λ

=

225 Ci ⎛ 86, 400 s ⎞ ⎛ 3.7 × 1010 s −1 ⎞ 18 ⎟ = 2.80 × 10 ⎟⎜ −1 ⎜ 0.2567 d ⎝ d 1 Ci ⎠⎝ ⎠

N 2.80 × 1018 = (198 g ) = 0.92 mg NA 6.022 ×1023

Insight: The required mass is less than 1 milligram. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 12

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

39. Picture the Problem: We are given the half-life of

241 95

Am and want to calculate how long it takes for the activity to

drop to 1 525 of the initial activity. Strategy: Use equation 32-10 to calculate the decay constant for americium-241. Then calculate the time using equation 32-13. Solution: 1. Calculate the decay constant:

λ=

2. Set R = R0 / 525 in equation 32-13:

t=

ln 2 ln 2 = = 0.00160 y −1 T1/ 2 432 y 1

λ

ln

R0 432 y ⎛ R0 ⎞ 432 y = ( ln 525) = 3.90 ×103 y ⎜ ln ⎟= R ln 2 ⎝ R0 / 525 ⎠ ln 2

Insight: The structure that the smoke detector was protecting probably will not be around in 3900 years.

40. Picture the Problem: The decay constant of 90 38 Sr can be used to find the fractional amount of radioactive nuclei that remains after 50.0 years, 60.0 years, and 70.0 years. Strategy: Use equation 32-10 to find the decay constant of Sr-90 from the half-life given in Appendix F. Then solve equation 32-9 for the fraction remaining as a function of time and convert the fraction to a percentage. ln 2 ln 2 = = 0.02407 y −1 T1/ 2 28.8 y

Solution: 1. Calculate the decay constant using the half-life of 28.8 years:

λ=

2. Solve equation 32-9 for the fraction of remaining nuclei:

N − ( 0.2407 y-1 ) t = e − λt = e N0

3. (a) Set t = 50.0 y:

e

4. (b) Set t = 50.0 y:

e

5. (c) Set t = 50.0 y:

e

(

)

(

)

(

)

− 0.2407 y−1 ( 50.0 y )

− 0.2407 y −1 ( 60.0 y )

− 0.2407 y−1 ( 70.0 y )

= 0.300 = 30.0% = 0.236 = 23.6% = 0.185 = 18.5%

Insight: The health risk posed by strontium-90 is not the percent remaining, but the possibility that the decaying Sr-90 may damage cells. The evaluation of that risk involves an understanding of the total activity of the Sr-90 and to what extent it is concentrated near the most sensitive tissues.

41. Picture the Problem: The nuclear binding energy of gold-197 is the amount of energy that must be added to completely separate the nucleus into protons and neutrons. Strategy: In the gold nucleus there are 79 protons and 197 − 79 = 118 neutrons. To calculate the binding energy, subtract the mass of the 197 79 Au nucleus from the combined mass of the 79 protons and 118 neutrons. The pertinent

masses are given in Appendix F. We use the mass of hydrogen as the mass of the proton because the listed mass of gold-197 includes the mass of the electrons in that atom. Multiply the mass difference by c2 and convert the result to MeV. Solution: 1. Calculate the mass of the protons and neutrons:

mf = 79 (1.007825 u ) + 118 (1.008665 u ) = 198.640645 u

2. Subtract the mass of the gold atom:

Δm = mf − mi = 198.640645 u − 196.96654 u = 1.67411 u

3. Convert the mass to energy:

⎛ 931.5 MeV/c 2 E = Δmc 2 = 1.67411 u ⎜ 1u ⎝

⎞ 2 ⎟ c = 1559 MeV ⎠

Insight: This binding energy is greater than the rest energy of one of the nucleons. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 13

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

42. Picture the Problem: The nuclear binding energy of lithium-7 is the amount of energy that must be added to completely separate the nucleus into protons and neutrons. Strategy: In the lithium nucleus there are 3 protons and 7 − 3 = 4 neutrons. To calculate the binding energy, subtract the mass of the Li-7 nucleus from the combined mass of the 3 protons and 4 neutrons. The pertinent masses are given in Appendix F. We use the mass of hydrogen for the mass of the proton because the listed mass of lithium-7 includes the mass of the electrons in that atom. Multiply the mass difference by c2 and convert the result to MeV. Solution: 1. Calculate the mass of the protons and neutrons:

mf = 3 (1.007825 u ) + 4 (1.008665 u ) = 7.058135 u

2. Subtract the mass of the gold atom:

Δm = mf − mi = 7.058135 u − 7.016003 u = 0.042132 u

3. Convert the mass to energy:

⎛ 931.5 MeV/c 2 ⎞ 2 E = Δmc 2 = 0.042132 u ⎜ ⎟ c = 39.25 MeV 1u ⎝ ⎠

Insight: Dividing the binding energy by the number of nucleons gives 5.6 MeV/nucleon, as shown in Figure 32-9.

43. Picture the Problem: The average binding energy per nucleon is the amount of energy needed to completely separate a nucleus into protons and neutrons, divided by the number of nucleons. Strategy: The number of protons in the nucleus is given by the atomic number, Z. The number of neutrons is the atomic mass less the atomic number, or N = A − Z. To calculate the binding energy, subtract the mass of the nucleus from the combined mass of the protons and neutrons. The pertinent masses are given in Appendix F. We use the mass of hydrogen for the mass of the proton because the listed masses of the iron and uranium isotopes in Appendix F include the mass of the electrons in those atoms. Multiply the mass difference by c2 and convert the result to MeV. Divide the result by the atomic mass to calculate the binding energy per nucleon. Solution: 1. (a) For 56 26 Fe , calculate the mass of the protons and neutrons:

mf = 26 (1.007825 u ) + 30 (1.008665 u ) = 56.463400 u

2. Subtract from the mass of Fe-56:

Δm = mi − mf = 56.463400 u − 55.934939 u = 0.528461 u

3. Convert the mass to energy:

⎛ 931.5 MeV/c 2 E = Δm c 2 = 0.528461 u ⎜ 1u ⎝

4. Divide by the number of nucleons:

E 492.26 MeV = = 8.790 MeV/nucleon A 56

5. (b) For

238 92

U , calculate the mass

of the protons and neutrons:

⎞ 2 ⎟ c = 492.3 MeV ⎠

mf = 92 (1.007825 u ) + 146 (1.008665 u ) = 239.984990 u

6. Subtract from the mass of U-238:

Δm = mi − mf = 239.984990 u − 238.050786 u = 1.934204 u

7. Convert the mass to energy:

⎛ 931.5 MeV/c 2 ⎞ 2 E = Δm c 2 = 1.934204 u ⎜ ⎟ c = 1801.7 MeV 1u ⎝ ⎠

8. Divide by the number of nucleons:

E 1801.7 MeV = = 7.570 MeV/nucleon A 238

Insight: The binding energy per nucleon of Fe-56 is greater than the binding energy per nucleon of U-238. Therefore, it is energetically favorable for U-238 to break up into smaller nuclei, such as Fe-56.

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32 – 14

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

44. Picture the Problem: The average binding energy per nucleon is the amount of energy needed to completely separate a nucleus into protons and neutrons, divided by the number of protons and neutrons in the nucleus. Strategy: The number of protons in the nucleus is given by the atomic number, Z. The number of neutrons is the atomic mass less the atomic number, or N = A − Z. To calculate the binding energy, subtract the mass of the nucleus from the combined mass of the protons and neutrons. The pertinent masses are given in Appendix F. We use the mass of hydrogen for the mass of the proton because the listed masses of the helium and zinc isotopes in Appendix F include the mass of the electrons in those atoms. Multiply the mass difference by c2 and convert the result to MeV. Divide the result by the atomic mass to calculate the binding energy per nucleon. Solution: 1. (a) For 42 He , calculate the mass of the protons and neutrons:

mf = 2 (1.007825 u ) + 2 (1.008665 u ) = 4.032980 u

2. Subtract from the mass of 42 He:

Δm = mi − mf = 4.032980 u − 4.002603 u = 0.030377 u

3. Convert the mass to energy:

⎛ 931.5 MeV/c 2 E = Δm c 2 = 0.030377 u ⎜ 1u ⎝

4. Divide by the number of nucleons:

E 28.296 MeV = = 7.074 MeV/nucleon 4 A

64 5. (b) For 30 Zn , calculate the mass of the protons and neutrons:

mf = 30 (1.007825 u ) + 34 (1.008665 u ) = 64.529360 u

6. Subtract from the mass of

64 30

⎞ 2 ⎟ c = 28.296 MeV ⎠

Δm = mi − mf = 64.529360 u − 63.929145 u = 0.600215 u

Zn:

7. Convert the mass to energy:

⎛ 931.5 MeV/c 2 E = Δm c 2 = 0.600215 u ⎜ 1u ⎝

8. Divide by the number of nucleons:

E 559.10 = = 8.736 MeV/nucleon 64 A

⎞ 2 ⎟ c = 559.10 MeV ⎠

Insight: Because the binding energy per nucleon for 42 He is less than the binding energy per nucleon of energetically favorable for the small nuclei of helium to fuse together to make larger nuclei.

45. Picture the Problem: Energy must be added to an

16 8

64 30

Zn it is

O atom in order to eject one neutron, resulting in 158 O + 10 n .

Strategy: Calculate the difference in mass between the oxygen-16 atom and the sum of the masses of an O-15 atom and a neutron. Multiply the mass difference by c2 and convert the result to MeV. Solution: 1. Calculate the mass of 2. Subtract the mass of the

16 8

O:

3. Convert the mass to energy:

15 8

O + 10 n:

mf = 15.003065 u + 1.008665 u = 16.011730 u Δm = 16.011730 u − 15.994915 u = 0.016815 u ⎛ 931.5 MeV/c 2 E = Δm c 2 = 0.016815 u ⎜ 1u ⎝

⎞ 2 ⎟ c = 15.66 MeV ⎠

Insight: A neutron cannot spontaneously be ejected from 168 O unless 15.66 MeV is added to the nucleus.

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32 – 15

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

46. Picture the Problem: We can determine whether the proton or the neutron is more tightly bound to the 168 O nucleus by calculating the energy necessary to free a proton and the energy necessary to free a neutron from the nucleus. Strategy: To calculate the energy necessary to remove a proton from the nucleus, multiply the mass difference between the 168 O and the 157 N + 11H by c2 and converting the result to MeV. Calculate the mass difference from the masses given

in Appendix F. Calculate the energy necessary to remove a neutron from the nucleus, by the same process, only using the masses of 158 O + 10 n . Solution: 1. (a) Calculate the mass of 2. Subtract the mass of the

16 8

15 7

N + 11H:

mf = 15.000109 u + 1.007825 u = 16.007934 u Δm = 16.007934 u − 15.994915 u = 0.013019 u

O:

⎛ 931.5 MeV/c 2 ⎞ 2 E = Δm c 2 = 0.013019 u ⎜ ⎟ c = 12.13 MeV 1u ⎝ ⎠

3. Convert the mass to energy: 4. (b) Calculate the mass of 5. Subtract the mass of the

16 8

15 8

mf = 15.003065 u + 1.008665 u = 16.011730 u

O + 10 n:

Δm = 16.011730 u − 15.994915 u = 0.016815 u

O:

⎛ 931.5 MeV/c 2 ⎞ 2 E = Δm c 2 = 0.016815 u ⎜ ⎟ c = 15.66 MeV 1u ⎝ ⎠

6. Convert the mass to energy:

7. (c) The neutron is more tightly bound than the proton because the proton is repulsed by the other protons in the nucleus. The fact that the energy required to remove the neutron is more than that required to remove the proton is verification that the neutron is more tightly bound. Insight: In the upper atmosphere, where oxygen molecules are bombarded with cosmic radiation, the production of 15 15 7 N is much more prevalent than the production of 8 O.

47. Picture the Problem: For the fission reaction 01 n + of neutrons b that are produced.

235 92

101 1 U → 132 50 Sn + 42 Mo + b 0 n , we are asked to calculate the number

Strategy: Balance the reaction equation by adding the correct number of neutrons so that the sum of the atomic numbers and the sum of the atomic masses of both sides of the equation are equal. 0 + 92 = 50 + 42 + b ( 0 )

Solution: 1. Verify that the atomic number is balanced:

92 = 92 1 + 235 = 132 + 101 + b (1)

2. Calculate b by balancing the atomic mass:

b = 1 + 235 − 132 − 101 = 3 neutrons

Insight: The number of ejected neutrons does not contribute to the balancing of the atomic number because the atomic number of a neutron is zero.

48. Picture the Problem: For the fission reaction 01 n + daughter product and the energy released.

235 92

1 U → 133 51 Sb + ? + 5 0 n , we are asked to calculate the missing

Strategy: Balance the atomic number and atomic mass on both sides of the reaction equation to identify the missing product. Calculate the difference in mass between the parent products 01 n + 235 92 U and the daughter products 133 51

Sb + ? + 5 01 n . Convert this mass difference to energy to determine the amount of energy released in the reaction.

Solution: 1. Balance the atomic number:

0 + 92 = 51 + Z + 5 ( 0 ) Z = 92 − 51 = 41, which is niobium or Nb.

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James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation 2. Balance the atomic mass:

1 + 235 = 133 + A + 5 (1) A = 1 + 235 − 133 − 5 = 98 98 41

Nb .

3. Determine the missing isotope:

The missing daughter product is

4. Calculate the initial mass:

mi = 1.008665 u + 235.043925 u = 236.052590 u

5. Calculate the final mass:

mf = 132.915237 u + 97.910331 u + 5 (1.008665 u )

6. Calculate the mass difference:

= 235.868893 u Δm = 235.868893 u − 236.052590 u = − 0.183697 u

7. Convert the difference into the energy released:

⎛ 931.5 MeV/c 2 ⎞ 2 E = Δm c 2 = 0.183697 u ⎜ ⎟ c = 171.1 MeV 1u ⎝ ⎠

Insight: The energy released by the fission is carried away primarily by the neutrons as kinetic energy. 1 235 88 136 49. Picture the Problem: For the fission reaction 0 n + 92 U → 38Sr + 54 Xe + neutrons , we are asked to calculate the number of neutrons produced and the energy released. Strategy: Balance the atomic number and atomic mass on both sides of the reaction equation to identify the missing product. Calculate the difference in mass between the parent products 01 n + 235 92 U and the daughter

136 products 88 38 Sr + 54 Xe + neutrons. Convert this mass difference to energy to determine the amount of energy released in

the reaction. Solution: 1. Balance the atomic mass to determine the number of neutrons:

1 + 235 = 88 + 136 + n (1) n = 1 + 235 − 88 − 136 = 12 n+

235 92

136 1 U → 88 38 Sr + 54 Xe + 12 0 n

2. Write out the complete reaction:

1 0

3. Calculate the initial mass:

mi = 1.008665 u + 235.043925 u = 236.052590 u

4. Calculate the final mass:

mf = 87.905625 u + 135.90722 u + 12(1.008665 u) = 235.91683 u

5. Calculate the mass difference:

Δm = 235.91683 u − 236.052590 u = − 0.13576 u

6. Convert the difference into energy released:

⎛ 931.5 MeV/c 2 ⎞ 2 E = Δm c 2 = 0.13576 u ⎜ ⎟ c = 126.5 MeV 1u ⎝ ⎠

Insight: This released energy is primarily distributed as kinetic energy of the neutrons.

50. Picture the Problem: We are asked to calculate the amount of gasoline that would have to be burned in order to release the same energy as that released when 1.0-lb of 235 92 U undergoes nuclear fission. Strategy: Calculate the amount of energy released when 1-lb of 235 92 U undergoes fission by multiplying the energy released in each fission event by the number of nuclei. To calculate the number of nuclei, divide the total mass by the mass of one nuclei, 235.0 u. Convert both the pound and u to kilograms. Divide the released energy by the energy released by one gallon of gas in order to find the equivalent number of gallons of gas. Solution: 1. Calculate the energy released by the uranium:

E=

2. Divide by the energy released in one gallon of gas:

N=

mU ⎞ 1.0 lb ⎛ 1 kg ⎞ ⎛ 1u 6 EU = ⎟ (173 × 10 eV ) ⎜ ⎟⎜ −27 MU 235.0 u ⎝ 2.2 lb ⎠ ⎝ 1.66 × 10 kg ⎠

= 2.015 × 1032 eV (1.60 × 10−19 J/eV ) = 3.2 × 1013 J E Egasoline

=

3.2 × 1013 J = 1.6 × 105 gallons of gas 2.0 × 108 J/gal

Insight: The energy released by each nuclear fission event is on the order of a million times greater than the energy released by each chemical reaction when gasoline is burned. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

51. Picture the Problem: Each fission event of

235 92

U nuclei releases 173 MeV of energy.

Strategy: Divide the total energy by the energy per fission to calculate the number of nuclei needed. Multiply the mass of one nucleus by the number of nuclei to calculate the total mass. E 8.4 × 1019 J ⎛ 1 MeV ⎞ 30 = ⎜ ⎟ = 3.035 × 10 EU 173 MeV ⎝ 1.6 × 10−13 J ⎠

Solution: 1. Calculate the number of 235 92 U nuclei:

N=

2. Multiply by the mass of one nuclei:

m = Nmu = 3.035 × 1030 ( 235.0 u ) (1.66 × 10 −27 kg/u ) = 1.2 × 106 kg

Insight: This mass would occupy a volume of only 62 m3, the size of a small classroom. However, it would not be wise to store it all in one room because of the danger of reaching the critical mass!

52. Picture the Problem: Each fission event of

235 92

U nuclei releases 173 MeV of energy.

Strategy: Divide the power output by the energy per reaction to calculate the required reaction rate. Solution: Calculate the reaction rate:

R=

P 150 × 106 W ⎛ eV ⎞ 18 −1 = ⎜ ⎟ = 5.4 × 10 s EU 173 × 106 eV ⎝ 1.60 × 10−19 J ⎠

Insight: This rate corresponds to the fission of 2.1 mg of

235 92

U per second.

53. Picture the Problem: Energy is released by the nuclear reaction: 12 H + 12 H → 13 H + 11H. Strategy: Calculate the mass difference between the initial two deuterium atoms and the resulting tritium and hydrogen, using the masses in Appendix F. Multiply the mass difference by c2 and convert the resulting energy to MeV. Solution: 1. Calculate the mass of 12 H + 12 H:

mi = 2(2.014102 u) = 4.028204 u

2. Calculate the mass of 13 H + 11H:

mf = 3.016049 u + 1.007825 u = 4.023874 u

3. Find the difference in mass:

Δm = 4.023874 − 4.028204 u = − 0.004330 u

4. Convert the difference into the energy released:

⎛ 931.5 MeV/c 2 ⎞ 2 E = Δm c 2 = 0.004330 u ⎜ ⎟ c = 4.033 MeV 1u ⎝ ⎠

Insight: The deuterium is stable, but the tritium produced in this reaction has a half-life of 12.33 y.

54. Picture the Problem: Energy is released by the nuclear reaction: 11 H + 10 n → 12 H. Strategy: Calculate the mass difference between the initial two atoms and the resulting deuterium, using the masses in Appendix F. Multiply the mass difference by c2 and convert the resulting energy to MeV. Solution: 1. Calculate the mass of 11 H + 10 n: 2. Look up the mass of

2 1

H:

mi = 1.007825 u + 1.008665 u = 2.016490 u mf = 2.014102 u

3. Find the difference in mass:

Δm = 2.014102 u − 2.016490 u = − 0.002388 u

4. Convert the difference into the energy released:

⎛ 931.5 MeV/c 2 E = Δm c 2 = 0.002388 u ⎜ 1u ⎝

⎞ 2 ⎟ c = 2.224 MeV ⎠

Insight: Once the deuterium is created it can be fused with another deuterium atom to release more energy in the helium fusion cycle.

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James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

55. Picture the Problem: Energy is released by the nuclear reaction: 11 H + 12 H → 32 He + γ . Strategy: Calculate the mass difference between the initial two atoms and the resulting helium-3, using the masses in Appendix F. Multiply the mass difference by c2 and convert the resulting energy to MeV. Solution: 1. Calculate the mass of 11 H + 12 H: 2. Look up the mass of

3 2

He:

mi = 1.007825 u + 2.014102 u = 3.021927 u mf = 3.016029 u

3. Find the difference in mass:

Δm = 3.016029 u − 3.021927 u = −0.005898 u

4. Convert the difference into the energy released:

⎛ 931.5 MeV/c 2 E = Δm c 2 = 0.005898 u ⎜ 1u ⎝

⎞ 2 ⎟ c = 5.494 MeV ⎠

Insight: When helium-3 is produced from the two atoms, a gamma ray must carry away some of the energy in order to conserve both momentum and energy.

56. Picture the Problem: We are asked to find the missing daughter product and the energy released for the fusion reaction, 12 H + 13 H → ? + 10 n . Strategy: Balance the atomic number and atomic mass on both sides of the reaction equation to identify the missing product. Calculate the difference in mass between the parent products (12 H + 13 H) and the daughter products (? + 10 n). Convert this mass difference to energy to determine the amount of energy released in the reaction. Divide the power output by the energy per reaction to calculate the required number of reactions per second. Solution: 1. (a) Balance the atomic number:

1+1 = Z + 0 Z = 2, which is helium, or He.

2. Balance the atomic mass:

2 + 3 = A +1 ⇒ A = 4

3. Write the fusion reaction:

2 1

4. Calculate the initial mass:

mi = 2.014102 u + 3.016049 u = 5.030151 u

5. Calculate the final mass:

mf = 4.002603 u + 1.008665 u = 5.011268 u

6. Calculate the mass difference:

Δm = 5.011268 u − 5.030151 u = − 0.018883 u

7. Convert the difference into the energy released:

⎛ 931.5 MeV/c 2 E = Δm c 2 = 0.018883 u ⎜ 1u ⎝

8. (b) Divide the power by the energy per reaction:

R=

H + 31 H → 42 He + 01 n

⎞ 2 ⎟ c = 17.589 MeV ⎠

P 25 × 106 W ⎛ 1 eV ⎞ 18 −1 = ⎜ ⎟ = 8.9 × 10 s −19 6 E 17.589 × 10 eV ⎝ 1.60 × 10 J ⎠

Insight: At this rate, it would take 19 hours to produce one mole (4 grams) of helium-4.

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James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

57. Picture the Problem: The Sun converts a portion of its mass into energy by means of nuclear fusion. Strategy: Divide the power output of the Sun by the speed of light squared to calculate the mass per second converted into energy. Calculate the total mass converted to energy by multiplying the mass conversion rate by the age of the Sun. Finally, divide the converted mass by the original mass of the Sun to find the percentage that was converted to energy. Solution: 1. (a) Calculate the rate of mass conversion:

3.90 × 10 26 W Δm PSun = 2 = = 4.33 × 109 kg/s 2 8 Δt c 3.00 10 m/s × ( )

2. (b) Multiply the mass conversion rate by the age of the Sun:

ΔM =

3. Divide the converted mass by the original mass:

ΔM 6.16 × 1026 kg = × 100 = 0.0307% M + ΔM 2.00 × 1030 kg + 6.16 × 1026 kg

⎛ 3.16 × 107 s ⎞ Δm T = ( 4.33 × 109 kg/s )( 4.50 × 109 y ) ⎜ ⎟ Δt 1y ⎝ ⎠ 26 = 6.16 × 10 kg

Insight: Over its entire life span, the Sun converts a tiny amount of its mass into energy, even though it converts mass at the astonishing rate of 4.33 million metric tons per second!

58. Picture the Problem: The amount of tissue damage caused by a dose of radiation depends upon the type of particle that is absorbed by the tissue. Strategy: Use equation 32-18 to set the dose in rem of the protons equal to the dose in rem of the alpha particles. Solve for the dose in rad of the protons. Solution: 1. Set the doses in rem equal:

( Dose in rem )α = ( Dose in rem )p ( Dose in rad )α RBEα = ( Dose in rad )p RBEp

2. Solve for ( Dose in rad )p :

( Dose in rad )p = ( Dose in rad )α

RBEα 20 = ( 55 rad ) = 110 rad RBEp 10

Insight: The body can be exposed to twice the radiation from protons as from alpha particles because the absorption of proton radiation is half that of the alpha radiation.

59. Picture the Problem: The amount of tissue damage caused by a dose of radiation depends upon the type of radiation that is absorbed by the tissue. Strategy: Use equation 32-18 to set the dose in rem of the X-rays equal to the dose in rem of the heavy ion particles. Solve for the dose in rad of the X-rays. The RBEs are listed in Table 32-3. Solution: 1. Set the doses in rem equal:

( Dose in rem )HI = ( Dose in rem )X-ray ( Dose in rad )HI RBEHI = ( Dose in rad )X-ray RBEX-ray

2. Solve for ( Dose in rad )X-ray :

( Dose in rad )X-ray = ( Dose in rad )HI

RBEHI RBEHI

20 = 1.0 × 103 rad 1 Insight: The body can be exposed to twenty times the radiation from X-rays as from heavy ions because the absorption of X-ray radiation is one-twentieth that of the heavy ion radiation. = 50 rad

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32 – 20

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

60. Picture the Problem: A 78-kg person absorbs the energy from 52 mrem of particles that have an RBE of 15. Strategy: Divide the dose in rem by the RBE in order to convert it into a dose in rad. Use equation 32-16 to convert the rad dosage to units of J/kg and then multiply by the person’s mass in order to find the total energy absorbed. 52 × 10−3 rem = 3.5 × 10 −3 rad 15

Solution: 1. (a) Convert the dose to rad:

Dose in rad =

2. Multiply by the mass and convert to joules:

⎛ 0.01 J/kg ⎞ E = ( Dose in rad ) m = ( 3.5 × 10−3 rad ) ⎜ ⎟ 78 kg = 2.7 mJ ⎝ 1 rad ⎠

3. (b) Because the energy absorbed is inversely proportional to the RBE as long as the dose in rem remains constant, the energy absorbed will decrease if the RBE is increased. Insight: If the RBE were increased to 20, the energy absorbed would be reduced to 2.0 mJ as long as the dose remained 52 mrem. The new radiation is more effective in creating biological damage, so a smaller amount of energy (in rad) is needed to produce the same biologically equivalent dose (in rem).

61. Picture the Problem: A 0.17-kg cancerous growth receives a 225-rad dose of radiation. Strategy: Multiply the radiation dose by the mass of the tumor in order to calculate the absorbed energy in joules. To find the change in temperature, solve equation 16-13 for the temperature change, where C = 4186 J kg ⋅ K . Solution: 1. (a) Calculate the absorbed energy:

⎛ 0.01 J ⎞ E = ( Dose in rad ) m = ( 225 rad )( 0.17 kg ) ⎜ ⎟ = 0.38 J ⎝ rad ⎠

2. (b) Solve equation 16-13 for the temperature:

ΔT =

Q E 0.3825 J = = = 0.54 mK mc mc 0.17 kg ( 4186 J/kgK )

Insight: The radiation does not cause significant heating of the tumor.

62. Picture the Problem: A 72-kg patient receives a radiation dose of 32 mrad over their entire body from alpha particles that have a relative biological effectiveness (RBE) of 13. Strategy: Use equation 32-18 to calculate the dosage in rem. Use equation 32-16 to convert the rad dosage to units of J/kg and then multiply by the patient’s mass in order to find the total energy absorbed. Solution: 1. (a) Calculate the rem dosage:

Dose in rem = ( Dose in rad )( RBE ) = ( 0.032 rad )(13) = 0.42 rem

2. (b) Multiply the dosage by the patient’s mass:

⎛ 0.01 J/kg ⎞ E = ( Dose in rad ) m = ( 0.032 rad )( 72 kg ) ⎜ ⎟ = 0.023 J ⎝ 1 rad ⎠

Insight: The received dose is about twice that received from inhaling radon over the course of a year (see Table 32-4).

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32 – 21

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

63. Picture the Problem: A patient receives a dose of radiation by ingesting a radioactive pharmaceutical containing

32 15

P.

Strategy: In order to calculate the number of electrons emitted over the seven days, we must calculate the total number of nuclei that decay. This will be the difference between the number of nuclei initially present and the number remaining after the seven-day period. First, use the half-life of the phosphorus to calculate the decay constant. Then use equation 32-11 and the decay constant to determine the initial number of nuclei. Subtract from the initial number of nuclei the number present after seven days, which can be calculated using equation 32-9. Multiply the number of electrons by the energy in each electron to find the total energy absorbed. Divide the energy absorbed by the mass of the tissue to determine the dosage in rad. Multiply the dosage by the RBE to calculate the dosage in rem. ln 2 ln 2 = = 0.04854 d −1 T1/ 2 14.28 d

Solution: 1. (a) Calculate the decay constant:

λ=

2. Calculate the initial number of nuclei:

N0 =

3. Calculate the final number of nuclei:

N = N 0 e − λ t = ( 2.39 × 1012 ) e

4. Calculate the number of electrons:

ne = N 0 − N = 2.39 × 1012 − 1.70 × 1012 = 6.9 × 1011 electrons

5. (b) Find the energy absorbed:

E = ne Ee = ( 6.9 ×1011 ) ( 705 keV ) (1.6 × 10−16 J/keV ) = 0.078 J

6. (c) Determine the dose in rad:

( Dose in rad ) =

7. Calculate the dose in rem:

dose in rem = ( dose in rad ) RBE = ( 62 rad )(1.50 ) = 93 rem

R0

λ

=

1.34 × 106 s −1 ⎛ 86, 400 s ⎞ 12 ⎜ ⎟ = 2.39 × 10 nuclei d 0.04854 d −1 ⎝ ⎠

(

)

− 0.04854 d −1 ( 7.00 d )

= 1.70 ×1012 nuclei

E 0.078 J ⎛ 1 rad ⎞ = ⎜ ⎟ = 62 rad m 0.125 kg ⎝ 0.01 J/kg ⎠

Insight: The total dosage the patient will receive if all of the original nuclei decay is 323 rem, close to the lethal dose of 500 rem. As it is, 93 rem is sufficient to damage the patient’s blood-forming tissues.

64. Picture the Problem: An α particle (charge +2e ) and a β particle (charge − e ) deflect in opposite directions when they pass through a magnetic field. Strategy: Recall the radius of curvature of a charged particle in a uniform magnetic field, r =

mv (equation 22-3) qB

when answering the conceptual question. Solution: In general, the amount of deflection is inversely proportional to the radius of curvature; that is, a large radius implies very little deflection. Recall, however, that the radius of curvature in a magnetic field (equation 22-3) is directly proportional to a particle’s mass and inversely proportional to its charge. Therefore, the α particle—which has twice the charge but roughly 8000 times the mass—has a larger radius of curvature by a factor of 4000. We conclude that the β particle deflects by the greater amount. Insight: We can also answer this question in terms of acceleration: F = qvB = ma . The greater the acceleration a of the particle, the greater its deflection in a certain period of time. We can therefore write the ratio: aβ qβ vB mβ qβ mα qβ 7295mβ = = = = 3647 ⇒ aβ = 3647aα The β particle will deflect much more than the α. aα qα vB mα qα mβ 2qβ mβ

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32 – 22

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

65. Picture the Problem: Radioactive samples A and B have equal half-lives, but the initial activity of sample A is twice that of sample B. Strategy: Use the principles involved in the description of radioactive decay to answer the conceptual question. Solution: The two samples have equal half-lives, therefore their numbers decrease with time by the same factor, e − λ t . It follows that the activity of sample A will always be twice that of sample B. We conclude that the ratio of the activity of sample A to that of sample B after two half-lives have elapsed is 2.00. Insight: If sample A had a shorter half-life than sample B, after awhile the activity of sample A would decrease and fall below the activity of sample B.

66. Picture the Problem: The initial activity of radioactive sample A is twice that of sample B. After two half-lives of sample A have elapsed, the two samples have the same activity. Strategy: Use the principles involved in the description of radioactive decay to answer the conceptual question. Solution: In two half-lives, the activity of sample A will be reduced to one-quarter its initial value. The initial activity of sample B was half that of sample A, but after two half-lives of sample A its activity is now one-quarter the initial activity of sample A—that is, the two samples have the same activity. It follows that the activity of sample B decreased by a factor of two in the same time that the activity of sample A decreased by a factor of four. Therefore, the ratio of the half-life of B to the half-life of A must be 2.00 (the half-life of sample B is twice as long as that of sample A). Insight: If sample A had the same half-life as sample B, the population of each would decrease with time by the same factor ( e − λ t ) and the activity of sample A would always remain twice that of sample B.

67. Picture the Problem: A coal-burning power plant and a nuclear power plant each consume fuel in order to produce a given amount of electricity. Strategy: Compare the energy yield per kilogram of coal and the energy yield per kilogram of uranium in order to determine the relative masses of the fuels used by each power plant. Solution: The amount of coal burned in a conventional power plant is much greater than the amount of uranium consumed in a nuclear power plant for a given amount of energy production. The reason is that in a coal-powered plant, energy is released as a result of chemical reactions. In a nuclear-powered plant the reactions occur within the nucleus, and therefore they release millions of times more energy than comparable chemical reactions. As a result, the amount of coal that is burned is much greater than the amount of uranium that is consumed. Insight: For instance, if we compare mid-grade bituminous coal in a 38% efficient power plant with a 32% efficient nuclear power plant that uses 3% enriched 235U fuel, we discover that in order to produce 8760 GWh (one year of continuous operation of a 1.0 GW power plant) the coal-burning plant must consume 3.15 million tons of coal while the nuclear plant needs only 38.7 tons of uranium fuel. Keep in mind that the mass of the uranium fuel decreases a little bit (about 1.10 kg) as some of its mass is converted into energy, while the “coal” mass actually increases as it is reacted with oxygen to produce ash, CO2, NOx, SOx, H2O, and various other compounds.

68. Picture the Problem: We are asked to determine the number of protons and neutrons in several isotopes. Strategy: Calculate the number of neutrons by subtracting the atomic number from the atomic mass (N = A − Z). The number of protons is equal to the atomic number (Z). Solution: 1. (a) Find the number of neutrons and protons in 2. (b) Calculate the number of neutrons and protons in

211 82

232 90

Pb:

Th:

neutrons = 232 − 90 = 142 protons = 90 neutrons = 211 − 82 = 129 protons = 82

neutrons = 60 − 27 = 33 protons = 27 Insight: Note that the ratio of neutrons to protons is about 1.5 for Th-232 and Pb-211, but is smaller (about 1.2) for the smaller isotope, Co-60. 3. (c) Calculate the number of neutrons and protons in

60 27

Co:

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James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

69. Picture the Problem: Three different decay processes produce three different daughter nuclei. Strategy: Balance the decay equations for the atomic number and the atomic mass. Determine the chemical symbol of the daughter isotope from the atomic number. Solution: 1. (a) Find the value of A 4 A and Z in 210 82 Pb → Z X + 2 He:

A = 210 − 4 = 206 Z = 82 − 2 = 80, which is mercury, or Hg.

2. Identify the daughter isotope:

A Z

3. (c) Find the value of A and Z A − in 239 92 U → Z X + e + υ :

A = 239 − 0 = 239 Z = 92 − ( −1) = 93, which is neptunium (Np).

4. Identify the daughter isotope:

A Z

5. (c) Calculate the value of A and Z in 116 C → ZA X + e + + υ :

A = 11 − 0 = 11 Z = 6 − 1 = 5, which is boron, or B.

6. Identify the daughter isotope:

A Z

X=

X=

206 80

239 93

X=

X=

X = 115 X =

206 80

239 93

11 5

Hg

Np

B

Insight: The total charge and total mass number is always conserved for any radioactive decay process.

70. Picture the Problem: The dose in rad that is required to give a cancerous tumor a certain dose in rem depends upon the relative biological effectiveness (RBE) of the alpha radiation. Strategy: Use equation 32-18 to calculate the dose in rad from the given dose in rem and the RBE found in Table 32-3. Solution: Solve equation 32-18 for the dose in rad: Dose in rem = Dose in rad × RBE Dose in rem 3800 rem Dose in rad = = RBE 10 to 20 = 190 to 380 rad Insight: Because the RBE for alpha particles lies in the range 10 to 20, the dose in rad cannot be fixed to a single value.

71. Picture the Problem: The dose in rem that is received by a patient depends upon the dose in rad and the relative biological effectiveness (RBE) of the gamma radiation. Strategy: Use equation 32-18 to calculate the dosage in rem. The RBE for gamma rays is given in Table 32-3. Dose in rem = Dose in rad × RBE = ( 260 )(1) = 260 rem

Solution: Apply equation 32-18 directly:

Insight: The dose in rem is the same as the dose in rad because the RBE for gamma rays is one.

72. Picture the Problem: The two radioactive decay series that begin 232 208 90Th and end with 82 Pb are shown in the figure at the right. Strategy: Use the representations of alpha decay, A A− 4 4 A A − to Z X → Z − 2 Y + 2 He and beta minus decay, Z X → Z +1 Y + e determine the daughter nuclei of each decay process illustrated in the figure. Solution: The nine intermediary nuclei on the upper series are as follows: 228 228 228 224 220 216 212 212 212 88 Ra, 89 Ac, 90Th, 88 Ra, 86 Rn, 84 Po, 82 Pb, 83 Bi, and 84 Po.

The tenth intermediary, on the lower (left) series, is

208 81

Tl.

Insight: A decay chain like this one can produce a wide variety of daughter isotopes in a rock. The different half-lives of each process produce the daughter products in specific ratios. The measurement of the ratios of these daughter isotopes can therefore give a reasonable estimate to the elapsed time from which the 23290Th was originally produced or inserted into the rock. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

73. Picture the Problem: The radioactive decay of potassium-40 can be used to determine the age of a Moon rock. Strategy: Solve equation 32-9 for the age of the rock, where the decay constant is calculated from the half-life. Then use equation 32-9, setting N 0 = 19.5% (the amount of potassium present today) and solving for the time t when N = 10.0% of the potassium-40 remains. ln 2 ln 2 = = 5.78 × 10−10 y −1 T1 2 1.20 × 109 y

Solution: 1. (a) Calculate the decay constant of K-40:

λ=

2. Solve equation 32-9 for the time:

N = N 0 e − λt t=−

3. (b) Calculate the time until only 10% remains:

t=−

⎛ N ⎞ 1 ln ⎜ ln ( 0.195 ) = 2.83 × 109 y ⎟=− −10 −1 5.78 × 10 y λ ⎝ N0 ⎠ 1

⎛ N ⎞ 1 ⎛ 0.100 ⎞ 9 ln ⎜ ln ⎜ ⎟=− ⎟ = 1.16 × 10 y −10 −1 N 5.776 10 y 0.195 × λ ⎝ 0⎠ ⎝ ⎠ 1

Insight: Part (b) of this problem could also have been calculated setting N 0 = 100% and calculating the total time for 90.0% of the potassium to decay. The remaining time required for 90.0% of the nuclei to decay would then be the total time minus the time that was found in part (a) and that has already elapsed for 80.5% of the nuclei to decay.

74. Picture the Problem: The mantle of a gas lantern contains a small amount of radioactive

232 90

Th .

Strategy: Calculate the activity by multiplying the decay constant by the number of nuclei present. The decay constant is calculated from equation 32-10. Determine the number of nuclei present by dividing the mass of the thorium-232 by the mass on one thorium atom (found in Appendix F). Solution: 1. Calculate the decay constant of Th-232:

λ=

ln 2 ln 2 1y ⎛ ⎞ −18 −1 = ⎜ ⎟ = 1.563 × 10 s 10 7 T1 2 1.405 × 10 y ⎝ 3.156 × 10 s ⎠

2. Calculate the number of Th-232 nuclei:

N=

⎞ M ( 0.325 g ) ⎛ 1u 20 = ⎜ ⎟ = 8.43 × 10 atoms m th 232.04 u ⎝ 1.66 × 10−24 g ⎠

3. Find the activity:

R = λ N = (1.563 × 10−18 s −1 )( 8.43 × 1020 ) = 1.32 × 103 Bq

4. (b) The activity would be reduced by a factor of 2, because it is inversely proportional to half-life. Insight: If the half-life were doubled, the decay constant would become 2.467 × 10−11 y −1 and the activity would become 658 Bq.

75. Picture the Problem: An unknown isotope produces the same nucleus by β − decay that

214 84

Po produces by α-decay.

Strategy: Determine the daughter product in alpha decay of polonium-214 by balancing the atomic number and atomic mass in the decay equation. Write an equation for β − decay that produces the same daughter product and balance that equation in order to determine the parent isotope. Solution: 1. Calculate the value of A 4 A and Z in 214 84 Po → Z X + 2 He:

A = 214 − 4 = 210 Z = 84 − 2 = 82, which is lead, or Pb

2. Identify the daughter isotope:

A Z

3. Calculate the value of A and Z 0 − in: ZA A → 210 82 Pb + −1 e + υ :

A = 210 + 0 = 210 Z = 82 + ( −1) = 81, which is thallium, or Tl

4. Identify the parent isotope:

A Z

X=

A=

210 82

210 81

Pb

Tl

Insight: A third radioactive decay that produces lead-210 is β + -decay from 210 83 Bi. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 25

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

76. Picture the Problem: The maximum Coulomb force, the electrostatic potential energy, and the initial kinetic energy of an alpha particle can be determined from the distance of its closest approach between a stationary nickel nucleus. Strategy: Use equation 19-5 to find the Coulomb force, where the charge on the alpha particle is 2e and the charge on the nickel nucleus is 28e. Calculate the potential energy using equation 20-8, and then find the initial kinetic energy by setting it equal to the maximum electrostatic potential energy. F=

Solution: 1. (a) Calculate the Coulomb force:

kqQ k ( 2e )( 28e ) = r2 d2

(8.99 ×10 = U=

2. (b) Find the electrostatic potential energy:

9

Nm 2 /C 2 ) ( 56 ) (1.60 × 10 −19 C )

(15 ×10

−15

m)

2

= 57 N

2

kqQ k ( 2e )( 28e ) = r d

(8.99 ×10 =

9

Nm 2 /C2 ) ( 56 ) (1.60 × 10−19 C )

2

15 × 10−15 m ⎛ 1 MeV ⎞ U = 0.86 pJ ⎜ ⎟ = 5.4 MeV −1 ⎝ 1.60 × 10 pJ ⎠ K i = U = 5.4 MeV

3. (c) Determine the initial kinetic energy:

Insight: Decreasing the separation distance by a factor of 2 will quadruple the Coulomb force and double the potential energy, requiring that the alpha particle has twice as much initial kinetic energy.

77. Picture the Problem: The activity in curies of a sample of pure

226 88

Ra can be determined from its mass.

Strategy: Calculate the decay constant from the half-life of 226 88 Ra as listed in Appendix F. Determine the number of nuclei by dividing the total mass of the sample by the mass of one nucleus. Finally, multiply the decay constant by the number of nuclei to calculate the activity in becquerels, and convert the answer to curies. Solution: 1. Calculate λ from the half-life:

λ=

ln 2 ln 2 = = 4.33 × 10−4 y −1 T1 2 1.60 × 103 y

2. Determine the number of nuclei:

N=

⎞ 1u M ⎛ 0.0017 kg ⎞ ⎛ 21 =⎜ ⎟ = 4.5 × 10 ⎟⎜ −27 m ⎝ 226.025406 u ⎠ ⎝ 1.66 ×10 kg ⎠

3. Calculate the activity:

1y ⎛ ⎞ R = λ N = ( 4.33 × 10− 4 y −1 )( 4.5 ×1021 ) ⎜ ⎟ 7 × 3.16 10 s ⎝ ⎠ ⎛ ⎞ 1 Ci = 6.2 × 1010 Bq ⎜ ⎟ = 1.7 Ci 10 ⎝ 3.7 × 10 Bq ⎠

Insight: The unit of curie is based on the number of decays per second of a 1-gram sample of pure 226 88 Ra (refer to the discussion surrounding equation 32-6). Therefore, a 1.7-gram sample should have an activity of 1.7 Ci.

78. Picture the Problem: We are asked to compare the half-lives of two samples. Sample A starts out with four times as many nuclei, but after two days the two samples contain the same number of radioactive nuclei. Strategy: Use equation 32-9 to set the number of nuclei of the two samples equal after 2.00 days have elapsed. The initial number of nuclei is equal to 4N 0 for sample A and N 0 for sample B. Solve the resulting equation for the ratio of the half-lives. Solution: 1. (a) Because nuclei of type A are decaying at a faster rate than type B, type B has the longer half-life. NA = NB

2. (b) Set N A = N B after two days and simplify:

( 4 N0 ) e

− λA t

= ( N 0 ) e − λB t

4 = e(

λA − λB ) t

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James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

(λ

3. Take the natural log of both sides and solve for the decay constant of sample B:

A

− λ B ) t = ln 4

λB = λA −

ln 4 t

ln 2 ln 4 ln 2 ln 4 − = − = 0.693 d −1 T1 2, A t 0.500 d 2.00 d

4. Write the decay constant of sample A in terms if its half-life:

λB =

5. Solve for the half-life of sample B:

T1 2, B =

ln 2

λB

=

ln 2 = 1.00 d 0.693 d −1

Insight: The half-life of sample B is double the half-life of sample A, and in the two-day period sample A decayed for 4 half-lives but sample B decayed for only 2 half-lives.

79. Picture the Problem: The range of radii of stable nuclei can be determined from the range of mass numbers for stable nuclei. Strategy: Use equation 32-4 to calculate the maximum and minimum radii. Set the surface area equal to the surface area of a sphere and calculate the ratio. Repeat for the volume of a sphere. rmin = (1.2 × 10−15 m ) (1)

1/3

Solution: 1. (a) Calculate the maximum and minimum radii:

= 1.2 × 10−15 m

rmax = (1.2 × 10−15 m ) ( 209 )

1/3

= 7.1× 10−15 m

2. Write the range of radii:

1.2 ×10−15 m ≤ r ≤ 7.1× 10−15 m

3. (b) Calculate the ratio of surface areas:

S max 4π rmax 2 ⎛ rmax ⎞ ⎛ 7.1× 10−15 m ⎞ = =⎜ ⎟ =⎜ ⎟ = 35.2 Smin 4π rmin 2 ⎝ rmin ⎠ ⎝ 1.2 × 10−15 m ⎠

4. (c) Calculate the ratio of volumes:

Vmax 43 π rmax 3 ⎛ rmax ⎞ ⎛ 7.1× 10−15 m ⎞ = 4 =⎜ ⎟ =⎜ ⎟ = 209 −15 3 Vmin ⎝ rmin ⎠ ⎝ 1.2 × 10 m ⎠ 3 π rmin

2

2

3

3

Insight: Note that the ratio of the surface areas is the ratio of the mass numbers raised to the 2/3 power and that the ratio of the volumes is equal to the ratio of the mass numbers.

80. Picture the Problem: The density of a neutron star is the same as the density of a nucleus, so that a neutron star is very much smaller than our Sun even though its mass is comparable. Strategy: Set the volume of the neutron star equal to the mass divided by the density. Set the volume equal to the volume of a sphere and solve for the radius. 4 = π r3 ρ 3

m

Solution: 1. Solve equation 15-1 for V:

V=

2. Solve the expression for the radius:

⎛ 3m ⎞ r =⎜ ⎟ ⎝ 4πρ ⎠

1/3

⎡ 3 ( 0.50 ) ( 2.00 × 1030 kg ) ⎤ ⎥ =⎢ ⎢⎣ 4π ( 2.3 × 1017 kg/m3 ) ⎥⎦

1/3

= 10 km

Insight: Neutrons stars are extremely small, dense objects. If the entire Earth were this dense, its radius would be only 184 m.

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James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

81. Picture the Problem: The age of a carbon-containing specimen can be determined from its mass and activity. Strategy: Equation 32-14 shows that the activity of a natural sample of carbon has an activity of 0.231 Bq per gram. Find the activity per gram of the mummy sample, and use it together with equation 32-13 and the initial activity per gram in order to determine the age of the sample. The decay constant for carbon-14 is 1.21× 10− 4 y −1 . Solution: 1. (a) Find the activity per gram of the carbon in the mummy:

R1g =

2. (b) Use equation 32-13 to find the age of the sample:

t=

1

λ

R 1.38 Bq = = 0.1765 Bq/g m 7.82 g ln

R0,1g R1g

=

⎛ 0.231 Bq ⎞ 1 ln ⎜ ⎟ = 2220 y −4 −1 1.21× 10 y ⎝ 0.1765 Bq ⎠

Insight: The activity of the one-gram sample is over 75% the initial activity, so the sample is younger than one half-life.

82. Picture the Problem: We are asked to calculate the number of nuclear fissions that would operate a light bulb for 2.5 days given that the conversion of energy is 32% efficient. Strategy: Multiply the power dissipated by the time elapsed to calculate the energy consumed by the light bulb. Divide this energy by the energy harvested from one fission reaction to find the number of reactions required to light the bulb. Finally, multiply the number of reactions by the mass of one uranium atom to calculate the total required mass. Solution: 1. (a) Calculate the energy consumed:

E = Pbulb t = (120 W )( 2.5 d )( 86, 400 s/d ) = 26 MJ

2. Divide the energy by the energy extracted from one reaction:

N reaction =

E 25.9 MJ 1 MeV ⎛ ⎞ = ⎜ −19 ε Efission ( 0.32 )( 212 MeV ) ⎝ 1.60 ×10 MJ ⎟⎠

= 2.39 × 1018 reactions = 2.4 × 1018 reactions

3. (b) Multiply the number of reactions by the mass of uranium:

m = N reaction M U = ( 2.39 ×1018 ) ( 235.043925 u ) (1.66 ×10 −27 kg/u ) = 9.3 × 10 −7 kg = 0.00093 g = 0.93 mg

Insight: This mass is miniscule when compared with the 2.5 kg of coal (assuming 27 MJ of energy released per 1.0 kg of bituminous coal and a 38% conversion efficiency) that would need to be burned to produce the same energy.

83. Picture the Problem: Energy is released when three helium atoms fuse to create one carbon-12 nucleus. Strategy: Subtract the mass of the carbon atom from the mass of three helium atoms (the nuclei of which are alpha particles) to calculate the mass difference. The masses in Appendix F include the mass of six electrons with the carbon atom and two electrons each with the helium atoms, so the electrons all subtract out. Multiply the mass difference by the speed of light squared to calculate the energy released. Solution: 1. (a) Some of the initial mass of the alpha particles is converted to energy. Therefore, the mass of carbon-12 is less than the mass of the three alpha particles. 2. (b) Calculate the change in mass:

Δm = 3 ( 4.002603 u ) − 12.000000 u = 0.007809 u

3. Convert the mass difference to energy :

⎛ 931.5 MeV/c 2 E = Δm c 2 = ( 0.007809 u ) ⎜ 1u ⎝

⎞ 2 ⎟ c = 7.274 MeV ⎠

Insight: Older, red-giant stars have consumed their primary fuel by fusing hydrogen into helium, and subsequently they fuse helium nuclei (alpha particles) into carbon as outlined in this problem. As mentioned in Section 32-6, hydrogen fusion releases 27 MeV per cycle, so we can see that helium fusion produces less energy for the star.

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James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

84. Picture the Problem: A heavy dose of gamma rays would be sufficient to melt a block of water. Strategy: The dosage is the energy per unit mass. The latent heat of fusion is also measured in energy per unit mass, so that we need only convert the latent heat of fusion for water into a dosage equivalent in rad. Solution: Convert the latent heat to rad:

( Dose in rad ) = L = ( 33.5 × 104

⎛ 1 rad ⎞ 6 J/kg ) ⎜ ⎟ = 33.5 × 10 rad ⎝ 0.01 J/kg ⎠

Insight: A very large dose of radiation is necessary to melt the ice. However, the small doses that are used to irradiate food can kill off any bacteria present without significantly affecting the temperature of the food.

85. Picture the Problem: We are asked to calculate the dosage of radiation necessary to heat water by one degree. Strategy: Use equation 16-13 to calculate the heat needed to raise the temperature of water by 1.0 C°. Divide this heat by the mass and convert the units into a dosage equivalent in rad. Solution: 1. (a) Calculate the heat needed:

Q = mcw ΔT = (1.00 kg )( 4186 J/kg ⋅ K )(1 C° ) = 4186 J

2. Divide the heat by the mass and convert to rad:

⎛ 4186 J ⎞ ⎛ 1.0 rad ⎞ 5 ⎜ ⎟⎜ ⎟ = 4.2 ×10 rad ⎝ 1.0 kg ⎠ ⎝ 0.01 J/kg ⎠

3. (b) The dosage in rad is the energy per unit mass required to heat the water. The dosage will stay the same. Insight: A very large dose of radiation is necessary to heat water. However, the small doses used to irradiate food can kill off any bacteria present without significantly affecting the temperature of the food.

86. Picture the Problem: A patient absorbs radiation when she undergoes a chest X-ray. Strategy: Use equation 32-18 to write the dose in rad and then convert the dose from rad to joules per kilogram. Multiply the dose by the mass of the patient in order to calculate the energy absorbed. dose in rem ⎛ 35 mrem ⎞ =⎜ ⎟ = 41.2 mrad RBE ⎝ 0.85 ⎠

Solution: 1. Calculate the dose in rad:

dose in rad =

2. Convert to J/kg:

⎛ 0.01 J/kg ⎞ dose = ( 0.0412 rad ) ⎜ ⎟ = 0.412 mJ/kg ⎝ rad ⎠

3. Multiply by the mass exposed to the X-rays:

E = dose ( m ) = ( 0.412 mJ/kg ) ( 14 ⋅ 72 kg ) = 7.4 mJ

Insight: This energy is insufficient to cause any measurable heating of the body.

87. Picture the Problem: A

226 88

Ra nucleus recoils after it emits a 0.186-MeV photon.

Strategy: Use the conservation of momentum to calculate the speed of the recoiling nucleus. The momentum of the photon is its energy divided by the speed of light (equation 30-11). Solution: Set pRa = pγ and solve

for the speed of the nucleus:

E c E 1 1u ⎛ ⎞ ⎛ 0.186 MeV ⎞ v= = ⎜ ⎟ 2 ⎟⎜ mc ( 226.025406 u ) ⎝ 931.5 MeV/c ⎠ ⎝ c ⎠

mv =

= ( 8.83 × 10−7 ) c = 264 m/s

Insight: The recoil speed of the nucleus is larger than the thermal speed (use equation 17-13 to find that vrms = 182 m/s at T = 300 K if the radium atom were in the gas phase) of the radium.

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32 – 29

James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation

88. Picture the Problem: All of the energy released in one hour by the alpha decay from a 50.0-g sample of to heat a 4.75-kg container of water.

239 94

Pu is used

Strategy: First calculate mass difference of the nuclear decay reaction in order to find the amount of energy released by each decay event. Then use equation 32-9 to calculate the number of nuclei that decay in one hour. Multiply the number of decays in one hour by the energy released per decay to calculate the total heat absorbed by the water. Finally, solve equation 16-13 for the change in temperature of the water when it absorbs the heat from the alpha decays. mi = 239.052158 u

Solution: 1. Calculate the mass difference 235 4 in the reaction 239 94 Pu → 92 U + 2 He:

mf = 235.043925 u + 4.002603 u = 239.046528 u Δm = 239.046528 u − 239.052158 u = − 0.005630 u

2. Convert the mass difference to energy released per decay:

⎛ 931.5 MeV/c 2 ⎞ 2 E = Δm c 2 = ( 0.005630 u ) ⎜ ⎟ c = 5.244 MeV 1u ⎝ ⎠

3. Calculate the decay constant:

λ=

4. Determine the initial number of nuclei:

N0 =

ln 2 ln 2 ⎛ 1 y ⎞⎛ 1 d ⎞ −9 −1 = ⎜ ⎟⎜ ⎟ = 3.28 ×10 h 4 T1/ 2 ( 2.41× 10 y ) ⎝ 365.25 d ⎠ ⎝ 24 h ⎠

⎞ m 0.0500 kg ⎛ 1u 23 = ⎜ ⎟ = 1.26 × 10 M 239.052158 u ⎝ 1.66 × 10−27 kg ⎠

n = number of decays = N 0 − N = N 0 − N 0 e − λ t = N 0 (1 − e − λ t )

5. Determine the number of decays in one hour:

(

n = (1.26 × 1023 ) 1 − e

(

)

− 3.28×10−9 h −1 (1.00 h )

) = 4.13 ×10

14

decays

6. Multiply the decays by the energy per decay:

⎛ 1.60 × 10−13 J ⎞ E = ( 4.13 × 1014 decays ) ( 5.244 MeV/decay ) ⎜ ⎟ MeV ⎝ ⎠ = 347 J

7. Solve equation 16-13 for the temperature change:

ΔT =

Q 347 J = = 0.0175 K mw cw ( 4.75 kg )( 4186 J/kg ⋅ K )

Insight: The temperature increase in the water is barely measurable. Alpha decay produces a tiny amount of thermal energy when compared with nuclear fission reactions.

89. Picture the Problem: The energy from alpha decay in a sphere of 235 92 U heats up the sphere, until the rate at which energy is produced in the sphere equals the rate at which the energy is radiated away through blackbody radiation. Strategy: Determine the mass of the sphere and the number of nuclei present from the density and radius of the sphere. Calculate the decay constant from the half-life (equation 32-10) and the activity from the number of nuclei and the decay constant (equation 32-11). Use the mass difference for each reaction to find the energy released per decay. Multiply the energy released by the activity to calculate the power output of the sphere. Set this power output equal to the power radiated by blackbody radiation (equation 16-19) and solve for the temperature of the sphere. Solution: 1. Calculate the mass of the sphere:

m = ρV = ρ ( 34 π r 3 ) = (18.95 g/cm3 ) 43 π ( 2.25 cm ) = 0.9042 kg

2. Determine the number of nuclei present:

N=

⎞ M ⎛ 0.9042 kg ⎞ ⎛ 1u 24 =⎜ ⎟⎜ ⎟ = 2.317 × 10 nuclei −27 m ⎝ 235.04 u ⎠ ⎝ 1.6605 ×10 kg ⎠

3. Calculate the decay constant:

λ=

⎞⎛ ln 2 ⎛ ln 2 1y ⎞ −17 −1 =⎜ ⎟⎜ ⎟ = 3.121× 10 s 8 T1/ 2 ⎝ 7.038 × 10 y ⎠ ⎝ 3.156 × 107 s ⎠

4. Use equation 32-11 to write the activity:

R = λ N = ( 3.121× 10−17 s −1 )( 2.317 × 1024 ) = 7.230 × 107 s −1

3

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James S. Walker, Physics, 4th Edition

Chapter 32: Nuclear Physics and Nuclear Radiation 5. Calculate the mass difference 231 4 for the decay 235 92 U → 90 Th + 2 He:

mi = 235.043925 u mf = 231.036297 u + 4.002603 u = 235.0389 u Δm = 235.0389 u − 235.043925 u = − 0.005025 u

6. Convert Δm to released energy:

⎛ 931.5 MeV/c 2 E = Δm c 2 = ( 0.005025 u ) ⎜ 1u ⎝

⎞ 2 ⎟ c = 4.681 MeV ⎠

7. Determine the power released:

P = ER = ( 4.681 MeV ) ( 7.230 × 107 s −1 )(1.60 × 10−13 J/MeV ) = 5.42 × 10−5 J/s = 54.2 μ W

8. Set the power equal to the power emitted by blackbody radiation:

P = eσ A (T 4 − Ts4 ) ⇒ T = T=

4

( 5.67 ×10

4

P + Ts4 eσ A

5.42 × 10 −5 W

−8

W m ⋅ K ) 4π ( 0.0200 m ) 2

4

2

+ ( 293 K )

4

= 293.0019 K

9. Calculate the change in temperature:

ΔT = T − Ts = 293.0019 − 293.0000 K = 1.9 mK

Insight: This temperature difference would be difficult to detect. The alpha decay considered here does not lead to the tremendous heating that occurs during the nuclear fission of uranium-235 (see Section 32-5).

90. Picture the Problem: The activity of an iodine-131 sample depends upon the decay constant and the number of radioactive nuclei it contains. Strategy: Find the decay constant from the half-life using equation 32-10. Solution: Solve equation 32-10 for the decay constant:

λ=

ln 2 ln 2 = = 9.98 × 10−7 s −1 T1/ 2 8.04 d × 86,400 s/d

Insight: The longer the half-life, the smaller the decay constant λ.

91. Picture the Problem: The activity of an iodine-131 sample depends upon the decay constant and the number of radioactive nuclei it contains. Strategy: Use the decay constant of 9.98×10−7 s−1 from the previous question, together with the sample size in order to find the activity of the sample. Solution. Calculate the activity:

⎛ ⎞ 1 Ci = 1.2 Ci R = λ N = ( 9.98 × 10−7 s −1 )( 4.5 × 1016 ) ⎜ 10 −1 ⎟ ⎝ 3.7 × 10 s ⎠

Insight: If the half-life were decreased to 4.02 d, more nuclei must decay per second in order that the same number of nuclei can decay in half the time. The activity would increase to 2.4 Ci.

92. Picture the Problem: The activity of an iodine-131 sample depends upon the decay constant and the number of radioactive nuclei it contains. ⎛ ln 2 ⎞ Strategy: Use equations 32-10 and 32-11 to find the dependence of R on T1 2 : R = λ N = ⎜ N . Use this expression ⎜ T1 2 ⎟⎟ ⎝ ⎠ to answer the conceptual question.

Solution: If the half-life T1/ 2 of iodine-131 were only half of its actual value, the activity R of the sample would be increased to twice its initial value because R is inversely proportional to T1/ 2 . Insight: A shorter half-life means that more nuclei decay per second, leading to a higher activity of the sample. Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32 – 31