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Responses to Questions

1. (a) n+13756Ba → ?+γ

Conserve nucleon number: 1 137+ = +A 0. Thus, A = 138. Conserve charge: 0 56+ = +Z 0. Thus, Z = 56. This is 13856Ba, or barium-138.

(b) n+13756Ba → 13755Cs ?+

Conserve nucleon number: 1 137 137+ = +A. Thus, A = 1. Conserve charge: 0 56 55+ = +Z. Thus, Z = 1. This is 11H, or p (a proton).

(c) d+21H → 42He ?+

Conserve nucleon number: 2 2 4+ = +A. Thus, A = 0. Conserve charge: 1 1 2+ = +Z. Thus, Z = 0. This is γ, or a gamma ray (a photon).

(d) α+19779Au → ? d+

Conserve nucleon number: 4 197+ = +A 2. Thus, A = 199. Conserve charge: 2 79+ = +Z 1. Thus, Z = 80. This is 19980Hg, or mercury-199.

2. The reaction is 2211Na+21H → 32He ?.+ The reactants have 12 protons and 24 nucleons, so they must have 12 neutrons. Thus the products must have 12 protons and 12 neutrons as well. Since the alpha has 2 protons and 2 neutrons, the other product must have 10 protons and 10 neutrons. That nucleus is

20

10Ne. Thus, the full equation is 2211Na+21H → 24He+1020Ne .

3. Neutrons are good projectiles for producing nuclear reactions because they are neutral and they are massive. If you want a particle to hit the nucleus with a lot of energy, then a more massive particle is the better choice. A light electron would not be as effective. Using a positively charged projectile like an alpha or a proton means that the projectile will have to overcome the large electrical repulsion from the positively charged nucleus. Neutrons can penetrate directly to the nucleus and cause nuclear reactions.

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4. For spontaneous radioactive decay reactions, Q>0, answer (b). Otherwise, more energy must be added to make the reaction proceed.

5. The thermal energy from nuclear fission appears in the kinetic energy of the fission products (daughter nuclei and neutrons). In other words, the fission products are moving very fast (especially the neutrons, due to the conservation of momentum).

6. (a) Yes, since the multiplication factor is greater than 1 (f = 1.5), a chain reaction can be sustained. (b) The difference would be that the chain reaction would proceed more slowly, and to make sure the chain reaction continued, you would need to be very careful about leakage of neutrons to the

surroundings.

7. Uranium can’t be enriched by chemical means because chemical reactions occur similarly with all of the isotopes of a given element. The number of neutrons does not influence the chemistry, which is primarily due to the valence electrons. Thus, trying to enrich uranium by chemical means, which means trying to increase the percentage of 23592U in the sample compared with 23892U, is impossible. 8. First of all, the neutron can get close to the nucleus at such slow speeds due to the fact that it is neutral

and will not be electrically repelled by either the electron cloud or the protons in the nucleus. Then, once it hits a nucleus, it is held due to the strong nuclear force. It adds more energy than just kinetic energy to the nucleus due to E mc= 2. This extra amount of energy leaves the nucleus in an excited state. To decay back to a lower energy state, the nucleus may undergo fission.

9. For a nuclear chain reaction to occur in a block of porous uranium, the neutrons being emitted by the decays must be slowed down. If the neutrons are too fast, then they will pass through the block of uranium without interacting, effectively prohibiting a chain reaction. Water contains a much higher density of protons and neutrons than does air, and those protons and neutrons will slow down (moderate) the neutrons, enabling them to take part in nuclear reactions. Thus, if water is filling all of the porous cavities, then the water will slow the neutrons, allowing them to be captured by other uranium nuclei, and allow the chain reaction to continue, which might lead to an explosion.

10. Ordinary water does not moderate, or slow down, neutrons as well as heavy water; more neutrons will also be lost to absorption in ordinary water. However, if the uranium in a reactor is highly enriched, then there will be many fissionable nuclei available in the fuel rods. It will be likely that the few moderated neutrons will be absorbed by a fissionable nucleus, and it will be possible for a chain reaction to occur.

11. A useful fission reaction is one that is self-sustaining. The neutrons released from an initial fission process can go on to initiate further fission reactions, creating a self-sustaining reaction. If no neutrons were released, then the process would end after a single reaction and not be very useful.

12. Heavy nuclei decay because they are neutron-rich, especially after neutron capture. After fission, the smaller daughter nuclei will still be neutron-rich and relatively unstable and will emit neutrons in order to move to a more stable configuration. Lighter nuclei are generally more stable with approximately equal numbers of protons and neutrons; heavier nuclei need additional neutrons in order to decrease Coulomb repulsion between the protons.

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accidental leakage and contamination. In the electric generating portion of the system, waste heat must be given off to the surroundings, and keeping all of the radioactive portions of the energy generation process as far away from this “release point” as possible is a major safety concern.

14. Fission is the process in which a larger nucleus splits into two or more fragments, roughly equal in size, releasing energy. Fusion is the process in which smaller nuclei combine to form larger nuclei, also releasing energy.

15. Fossil fuel power plants are less expensive to construct, and the technology is well known. However, the mining of coal is dangerous and can be environmentally destructive, the transportation of oil can be damaging to the environment through spills, the production of power from both coal and oil

contributes to air pollution and the release of greenhouse gases into the environment, and there is a limited supply of both coal and oil. Fission power plants produce no greenhouse gases and virtually no air pollution, and the technology is well known. However, they are expensive to build, they produce thermal pollution and radioactive waste, and when accidents occur they tend to be very destructive. Uranium is also dangerous to mine. Fusion power plants produce very little radioactive waste and virtually no air pollution or greenhouse gases. Unfortunately, the technology for large-scale sustainable power production is not yet known, and the pilot plants are very expensive to build.

16. Gamma particles penetrate better than beta particles because they are neutral and have no mass. Thus, gamma particles do not interact with matter as easily or as often as beta particles, allowing them to better penetrate matter.

17. The large amount of mass of a star creates an enormous gravitational attraction, which causes the gas to be compressed to a very high density. This high density creates very high pressure and high temperature. The high temperatures give the gas particles a large amount of kinetic energy, which allows them to overcome the Coulomb repulsion and then fuse when they collide. These conditions at the center of the Sun and other stars make the fusion process possible.

18. Stars, which include our Sun, maintain confinement of the fusion plasma with gravity. The huge amount of mass in a star creates an enormous gravitational attraction on the gas molecules, and this attractive force overcomes the outward repulsive forces from electrostatics and radiation pressure.

19. Alpha particles are relatively large and are generally emitted with relatively low kinetic energies. They are not able to penetrate the skin, so they are not very destructive or dangerous as long as they stay outside the body. If alpha emitters are ingested or inhaled, however, then the protective layer of skin is bypassed, and the alpha particles, which are charged, can do tremendous amounts of damage to the lungs and other delicate internal tissues due to ionizing effects. Thus, there are strong rules against eating and drinking around alpha emitters, and the machining of such materials, which produces fine dust particles that could be inhaled, can be done only in sealed conditions.

20. The absorbed dose measures the amount of energy deposited per unit mass of absorbing material and is measured in grays (Gy) or rads. The gray is the SI unit and is 1 J/kg. A rad is 0.01 Gy. The effective dose takes into account the type of radiation depositing the energy and is used to determine the biological damage done by the radiation. The effective dose is the absorbed dose multiplied by a relative biological effectiveness (RBE) factor. The effective dose is measured in rem, which are rad × RBE, or sieverts (Sv), which are gray × RBE. Sieverts is the SI unit, and 1 Sv = 100 rem. 1 Sv of any type of radiation does approximately the same amount of biological damage.

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22. Allow a radioactive tracer to be introduced into the liquid that flows through the pipe. Choose a tracer that emits particles that cannot penetrate the walls of the pipe. Then check the pipe with a Geiger counter. When tracer radiation is found on the outside of the pipe (radiation levels will be higher at that point), the leak will have been located.

Responses to MisConceptual Questions

1. (e) In nuclear reactions, energy is converted between mass, kinetic energy, and radiant energy, but the total energy is conserved. The momentum is conserved as particles are absorbed and released in the reactions. The net electric charge is conserved. That is, the total net charge before the reaction will always equal the net charge after the reaction. The new conserved quantity in this chapter is the conservation of nucleon number. In any reaction, the number of nucleons will remain unchanged.

2. (b) Large nuclei have more neutrons per proton than smaller nuclei. In a fission reaction, the daughter nuclei are neutron-rich and emit β− particles to convert excess neutrons into protons.

3. (c) The melting point, boiling point, and valence shells do not affect the critical mass for a chain reaction. The nuclear density is relatively constant for all substances. The critical mass is determined by the mass necessary for each reaction to produce, on average, one neutron that creates an additional reaction. If each fission can produce additional neutrons, then it is more likely that one of those neutrons will create an additional reaction and the critical mass can be smaller.

4. (b) Fusion occurs in small nuclei because the binding energy per nucleon of larger nuclei is greater than the binding energies of the small nuclei. Fission occurs in very large nuclei because the binding energy per nucleon of the smaller nuclei is greater than the binding energy per nucleon of the very large nuclei. If the binding energy per nucleon always increased, then fusion could occur through all elements, but fission could not occur.

5. (a) Nuclear reactions occur when uranium-235 absorbs a neutron. Thermal neutrons are more likely to be absorbed by uranium-235, so the moderator is used to slow down the neutrons.

6. (e) Fission and fusion are sometimes thought of as different names for the same physical process, but this is incorrect. Fission occurs when energy is released as large nuclei are broken apart. Fusion occurs when energy is released as small nuclei are put together.

7. (c) The primary fuel source for fusion is hydrogen, which is very plentiful. The by-product of nuclear fusion is helium, which is not radioactive. The problem with fusion is that the hydrogen nuclei repel each other; therefore, very high temperatures are necessary for them to get close enough to fuse.

8. (a) When the two hydrogen nuclei bind together to form helium, energy is released. This energy comes from the mass of the hydrogen nuclei, so the resulting helium nucleus has less mass than the mass of the two hydrogen nuclei.

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10. (a) A single fission reaction releases about 200 MeV of energy. A single fusion reaction produces less than 25 MeV. Therefore, the fission reaction releases more energy.

11. (a) The fuel for fusion is hydrogen, which is a component of water.

12. (a, b) Radiation is composed of high-energy particles that can disrupt cells. If a small amount of radiation hits the correct molecules of a cell (such as DNA), then it can alter the cell, causing cancer. Therefore, any amount of radiation can be harmful. Radiation is produced in natural nuclear decays and is therefore part of the normal environment. Gamma radiation can be very penetrating, but alpha and beta radiation do not penetrate very far into living tissue.

13. (e) Cell damage due to radiation occurs from extended exposure to high levels of radiation. The damage can be minimized by having the technician work farther away from the radiation, work for a shorter time, and be shielded from the radiation. A radiation badge can warn the worker when the allowed radiation exposure has been reached. Therefore, all of the answers can help reduce radiation damage.

14. (d) X-rays, gamma rays, and beta rays all produce about the same amount of radiation damage. Neutrons and alpha particles produce more cell damage per unit energy than X-rays, gamma rays, or beta rays due to their larger masses and slower speeds.

15. (b) The critical mass is the mass that must be present so that on average, a neutron from each fission is absorbed by another nucleus to instigate another fission. When more neutrons are released during each fission, another reaction is more likely with less mass present. Therefore, the plutonium has a smaller critical mass.

Solutions to Problems

1. When aluminum absorbs a neutron, the mass number increases by one and the atomic number is unchanged. The product nucleus is 2813Al . Since the nucleus now has an “extra” neutron, it will decay by β−, according to this reaction: 2813Al → 1428Si+β−+ve. Thus the product is 1428Si .

2. If the Q-value is positive, then no threshold energy is needed.

12 32

[

]

2

2 2 2 2

n

H He

MeV/

2 2(2 014102 u) 3 016029 u 1 008665 u 931 5

u 3.270 MeV

c

Q= m cm cm c = . − . − . ⎛⎜ . ⎞⎟c

⎝ ⎠

=

Thus, no threshold energy is required, since the Q-value is positive. The final mass is less than the initial mass, so energy is released in this process.

3. A “slow” neutron has negligible kinetic energy. If the Q-value is positive, then the reaction is possible.

[

]

238 239

92 92

2

2 2 2 2

n

U U

MeV/ 238.050788 u 1 008665 u 239 054294 u 931 5

u 4.806 MeV

c

Q m= c +m cm c = + . − . ⎛⎜ . ⎞⎟c

⎝ ⎠

=

(6)

4. (a) The product has 16 protons and 16 neutrons. Thus the reactants must have 16 protons and 16 neutrons. Thus, the missing nucleus has 15 protons and 16 neutrons, so it is 1531P.

(b) The Q-value tells us whether the reaction requires or releases energy. In order to balance the electrons, we use the mass of 11H for the proton.

[

]

31 32

15 16

2 2 2

p S S

2 2 MeV/

1.007825 u 30.973762 u 31.972071 u 931.5 8.864 MeV u

Q m c m c m c

c c

= + −

⎛ ⎞

= + − ⎜ =

⎝ ⎠

5. We assume that all of the particles are essentially at rest, so we ignore conservation of momentum.

To make just the fluorine nucleus, the Q-value plus the incoming kinetic energy should add to 0. In order to balance the electrons, we use the mass of 11H for the proton.

18 18

8 9

2 2 2 2

KE+ =Q KE+m cp +m Ocm cFm cn =0 →

[

]

18 18

9 8

18 9

2 2 2 2

KE p n

F O

2 2 4

4 2

F 2

MeV/ 2.438 MeV 1.007825 u 17.999160 u 1.008665 u 931.5

u

1.6767873 10 MeV

1 u

(1.6767873 10 MeV/ ) 18.000937 u

931.5 MeV/

m c m c m c m c

c c

m c

c

= + + − =

⎛ ⎞

= + + − ⎜

⎝ ⎠

= ×

⎛ ⎞

= × ⎜=

⎝ ⎠

6. (a) If the Q-value is positive, then no threshold energy is needed.

[

]

24 23

12 11

2 2 2 2

n Mg Na d

2 2 MeV/

1.008665 u 23.985042 u 22.989769 u 2.014102 u 931.5 9.468 MeV u

Q m c m c m c m c

c c

= + − −

⎛ ⎞

= + − − ⎜ = −

⎝ ⎠

Thus more energy is required if this reaction is to occur. The 18.00 MeV of kinetic energy is more than sufficient, so the reaction can occur.

(b) 18.00 MeV 9.468 MeV− = 8.53 MeV of energy is released

7. (a) If the Q-value is positive, then no threshold energy is needed. In order to balance the electrons, we use the mass of 11H for the proton and the mass of 42Hefor the alpha particle.

[

]

7 4

3 2

2 2 2 2

p Li He

2 2 MeV/

1.007825 u 7.016003 u 2(4.002603 u) 931.5 17.346 MeV u

Q m c m c m c m c

c c

α

= + − −

⎛ ⎞

= + − ⎜ =

⎝ ⎠

(7)

(b) The total kinetic energy of the products will be the Q-value plus the incoming kinetic energy.

KEtotal =KEreactants+ =Q 3.1 MeV 17.346 MeV+ = 20.4 MeV

8. (a) If the Q-value is positive, then no threshold energy is needed. In order to balance the electrons, we use the mass of 11H for the proton and the mass of 42He for the alpha particle.

[

]

14 17

7 8

2 2 2 2

p

N O

2 2 MeV/

4.002603 u 14.003074 u 16.999132 u 1.007825 u 931.5 1.192 MeV u

Q m c m c m c m c

c c

α

= + − −

⎛ ⎞

= + − − ⎜ = −

⎝ ⎠

Thus, more energy is required if this reaction is to occur. The 9.85 MeV of kinetic energy is more than sufficient, so the reaction can occur.

(b) The total kinetic energy of the products will be the Q-value plus the incoming kinetic energy.

KEtotal =KEreactants+ =Q 9.85 MeV 1.192 MeV− = 8.66 MeV

9. In order to balance the electrons, we use the mass of 42He for the alpha particle.

[

]

16 20

8 10

2 2

O Ne

2 2 MeV/

4.002603 u 15.994915 u 19.992440 u 931.5 4.730 MeV u

Q m c m c m

c c

α

= + −

⎛ ⎞

= + − ⎜ =

⎝ ⎠

10. The Q-value tells us whether the reaction requires or releases energy.

[

]

13 14

6 7

2 2 2 2

d C N n

2 2 MeV/

2.014102 u 13.003355 u 14.003074 u 1.008665 u 931.5 5.326 MeV u

Q m c m c m c m c

c c

= + − −

⎛ ⎞

= + − − ⎜ =

⎝ ⎠

The total kinetic energy of the products will be the Q-value plus the incoming kinetic energy.

KEtotal =KEreactants+ =Q 41.4 MeV 5.326 MeV+ = 46.7 MeV

11. The 147N absorbs a neutron, and 146C is a product. Thus the reaction is n+147N → 146C ?.+ The reactants have 7 protons and 15 nucleons, which means 8 neutrons. Thus the products also have 7 protons and 8 neutrons, so the unknown product must be a proton. The reaction is

14 14

7 6

n+ N → C p .+ In order to balance the electrons, we use the mass of 11H for the proton.

[

]

14 14

7 6

2 2 2 2

n N C p

2 2 MeV/

1.008665 u 14.003074 u 14.003242 u 1.007825 u 931.5 0.626 MeV u

Q m c m c m c m c

c c

= + − −

⎛ ⎞

= + − − ⎜ =

(8)

12. (a) The deuteron is 12H, so the reactants have 4 protons and 8 nucleons. Therefore, the reactants have 4 neutrons. Thus the products must have 4 protons and 4 neutrons. That means that X must have 3 protons and 4 neutrons, so X is 73Li .

(b) This is called a “stripping” reaction because the lithium nucleus has “stripped” a neutron from the deuteron.

(c) The Q-value tells us whether the reaction requires or releases energy. In order to balance the electrons, we use the mass of 11H for the proton

[

]

6 7

3 3

2 2 2 2

d Li Li p

2 2 MeV/

2.014102 u 6 015123 u 7.016003 u 1.007825 u 931.5 5.027 MeV u

Q m c m c m c m c

c c

= + − −

⎛ ⎞

= + . − − ⎜ =

⎝ ⎠

Since the Q-value is positive, the reaction is exothermic.

13. (a) This is called a “pickup” reaction because the helium has “picked up” a neutron from the carbon nucleus.

(b) The alpha is 42He. The reactants have 8 protons and 15 nucleons, so they have 7 neutrons. Thus, the products must also have 8 protons and 7 neutrons. The alpha has 2 protons and 2 neutrons, so X must have 6 protons and 5 neutrons. Thus, X is 116C .

(c) The Q-value tells us whether the reaction requires or releases energy. In order to balance the electrons, we use the mass of 11H for the proton and the mass of 42He for the alpha particle.

[

]

3 12 11

2 6 6

2 2 2 2

He C C

2 2 MeV/

3.016029 u 12.000000 u 11.011434 u 4.002603 u 931.5 1.856 MeV u

Q m c m c m c m c

c c

α

= + − −

⎛ ⎞

= + − − ⎜ =

⎝ ⎠

Since the Q-value is positive, the reaction is exothermic.

14. The Q-value tells us whether the reaction requires or releases energy. In order to balance the electron count, we use the mass of 11H for the proton and the mass of 42He for the alpha particle.

73 42

[

]

2

2 2 2 2 2

p Li He

MeV/ 1.007825 u 7 016003 u 2(4 002603) u 931.5

u 17.35 MeV

c

Q m c= +m cm cm cα = + . − . ⎛⎜ ⎞⎟c

⎝ ⎠

=

The reaction releases 17.35 MeV .

15. The Q-value tells us whether the reaction requires or releases energy. In order to balance the electron count, we use the mass of 42He for the alpha particle.

[

]

9 12

4 6

2 2 2 2

n

Be C

2 2 MeV/

4.002603 u 9.012183 u 12.000000 u 1.008665 u 931.5 5.702 MeV u

Q m c m c m c m c

c c

α

= + − −

⎛ ⎞

= + − − ⎜ =

⎝ ⎠

(9)

16. The Q-value gives the energy released in the reaction, assuming that the initial kinetic energy of the neutron is very small.

[

]

235 141 92

92 56 36

2 2 2 2 2

n U Ba Kr n

2 2 3

MeV/ 1.008665 u 235.043930 u 140.914411 u 91.926156 u 3(1.008665 u) 931.5

u

173.3 MeV

Q m c m c m c m c m c

c c

= + − − −

⎛ ⎞

= + − − − ⎜

⎝ ⎠

=

17. The Q-value gives the energy released in the reaction, assuming the initial kinetic energy of the neutron is very small.

[

]

235 88 136

92 38 54

2 2 2 2 2

n U Sr Xe n

2 2 12

MeV/ 1.008665 u 235.043930 u 87.905612 u 135.907214 u 12(1.008665 u) 931.5

u

126.5 MeV

Q m c m c m c m c m c

c c

= + − − −

⎛ ⎞

= + − − − ⎜

⎝ ⎠

=

18. The power released is the energy released per reaction times the number of reactions per second.

6

18

6 19

18 energy reactions reaction s

reactions 240 10 W

7.5 10 reactions/s energy

s (200 10 eV/reaction)(1.60 10 J/eV)

reaction

8 10 reactions/s P

P

#

= × →

# = = × = ×

× ×

≈ ×

19. Compare the energy per fission with the mass energy.

energy per fission2 200 MeV 2 2 9.1 10 4 1 (9 10 )%2 1100

mass energy mc (235 u)(931.5 MeV/ )c c

= = × 2 ≈ ≈ ×

20. Convert the 960 watts over a year’s time to a mass of uranium.

235 7

-4 235 92

92

13 23

235 92

0.235 kg U 960 J 3.156 10 s 1 MeV 1 fission

3.696 10 kg U 1 s 1 yr 1.60 10 J 200 MeV 6.02 10 atoms

0.4 g U

⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞

⎛ ⎞ ×

⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

⎜ ⎟ ⎟⎜ ⎟⎜

⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

× =

× ×

21. (a) The total number of nucleons for the reactants is 236, so the total number of nucleons for the products must also be 236. The two daughter nuclei have a total of 231 nucleons, so 5 neutrons must be produced in the reaction: 23592U n+ → 13351Sb+9841Nb 5n.+

(b)

[

]

235 133 98

92 51 41

2 2 2 2 2

n n

U Sb Nb

2 2 5

MeV/ 235.043930 u 1.008665 u 132.915250 u 97.910328 u 5(1.008665 u) 931.5

u

171.1 MeV

Q m c m c m c m c m c

c c

= + − − −

⎛ ⎞

= + − − − ⎜

⎝ ⎠

(10)

22. We assume as stated in Problems 18, 19, and 20 that an average of 200 MeV is released per fission of a uranium nucleus.

235 92

7 7

19 6 23

1 eV 1 MeV 1 nucleus 0.235 kg

(3 10 J) 3.7 10 kg

200 MeV

1.60 10 J 1 10 eV 6.02 10 nuclei U

− −

⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞

× ⎜ ⎟⎜⎟⎜⎜ ⎟⎜= ×

× × ⎝ ⎠ ×

⎝ ⎠⎝ ⎠ ⎝ ⎠

23. Since the reaction is 34% efficient, the fission needs to generate (950/0.34) MW of power. Convert the power rating to a mass of uranium using the factor-label method. We assume that 200 MeV is released per fission, as in other Problems.

6 7

235 92

6 19 23

235 92

950 10 J 1 atom 1 eV 0.235 kg U 3.156 10 s

1075 kg U 0.34 s 200 10 eV 1.60 10 J 6.02 10 atoms 1 yr

1100 kg U

× × × × × × =

× × ×

24. We find the number of collisions from the relationship

( )

1 0 2 ,

n n

E =E where n is the number of collisions.

( )

6 0

1

0 2 1 1

2 2

0.040 eV ln

ln

1.0 10 eV 24.58 25 collisions

ln ln

n n

n

E E

E =En= = × = ≈

25. If the uranium splits into two roughly equal fragments, then each will have an atomic mass number of half of 236, or 118. Each will have a nuclear charge of half of 92, or 46. Calculate the electrical potential energy using a relationship derived in Example 17–7. The distance between the nuclei will be twice the radius of a nucleus, and the radius is given in Eq. 30–1.

2 19 2

9 2 2

1 2 PE

15 1/3 13

0

1 (46) (1.60 10 C) 1 MeV

(8.99 10 N m /C ) 260 MeV

4 2(1.2 10 m)(118) 1.60 10 J

Q Q r πε

− −

⎛ ⎞

×

= = × ⋅ ⎜=

× ⎝ × ⎠

This is about 30% larger than the nuclear fission energy released.

26. The reaction rate is proportional to the number of neutrons causing the reactions. For each fission, the number of neutrons will increase by a factor of 1.0004, so in 1000 milliseconds, the number of neutrons will increase by a factor of (1.0004)1000=1.5 .

27. KE 3 3 23 7 16 16

2 2 19

1 eV (1.38 10 J/K)(2 10 K) 4 10 J 4 10 J

1.60 10 J

kT − − − ⎛

= = × × = × = × ⎜

×

⎝ ⎠

≈ 3000 eV

28. The Q-value gives the energy released in the reaction.

[

]

2 3 4

1 1 2

2 2 2 2

n

H H He

2 2 MeV/

2.014102 u 3.016049 u 4.002603 u 1.008665 u 931.5 17.59 MeV u

Q m c m c m c m c

c c

= + − −

⎛ ⎞

= + − − ⎜ =

(11)

29. Calculate the Q-value for the reaction 12H+21H → 32He n.+

[

]

2 3

1 2

2 2 2

n

H He

2 2 2

MeV/

2(2 014102 u) 3.016029 u 1.008665 u 931.5 3.27 MeV u

Q m c m c m c

c c

= − −

⎛ ⎞

= . − − ⎜ =

⎝ ⎠

30. For the reaction in Eq. 31–6a, if atomic masses are to be used, then one more electron needs to be added to the products side of the equation. Notice that charge is not balanced in the equation as written. The balanced reaction is 11H+11H → 21H e+ ++ +v e .−

[

]

1 2

1 1

2 2 2 2

H H e e

2 2 2

MeV/

2(1 007825 u) 2.014102 u 2(0 000549 u) 931.5 0.4192 MeV 0.42 MeV u

Q m c m c m c m c

c c

+ −

= − − −

⎛ ⎞

= . − − . ⎜ = ≈

⎝ ⎠

For the reaction in Eq. 31–6b, use atomic masses, since there would be two electrons on each side.

[

]

1 2 3

1 1 2

2 2 2

H H He

2 2 MeV/

1.007825 u 2.014102 u 3.016029 u 931.5 5.4940 MeV 5 49 MeV u

Q m c m c m c

c c

= + −

⎛ ⎞

= + − ⎜ = ≈ .

⎝ ⎠

For the reaction in Eq. 31–6c, use atomic masses, since there would be two electrons on each side.

[

]

3 4 1

2 2 1

2 2 2

He He H

2 2

2 2

MeV/

2(3.016029 u) 4.002603 2(1.007825 u) 931.5 12.8594 MeV 12.86 MeV u

Q m c m c m c

c c

= − −

⎛ ⎞

= − − ⎜ = ≈

⎝ ⎠

31. (a) Eq. 31–8a: 4.03 MeV 1 u 27 1 kg 6.03 1023MeV/g 2(2.014102 u)×1.66 10 × − kg×1000 g= ×

Eq. 31–8b: 3.27 MeV 1 u 27 1 kg 4.89 10 MeV/g23 2(2.014102 u)×1.66 10 × − kg×1000 g= ×

Eq. 31–8c: 17.59 MeV 1 u27 1 kg 2.11 1024MeV/g (2.014102 u 3.016049 u)+ ×1.66 10 × − kg×1000 g = ×

(b) Uranium fission (200 MeV per nucleus):

200 MeV 1 u27 1 kg 5.13 10 MeV/g23

(235 u) ×1.66 10 × − kg×1000 g= ×

Eq. 31–8a:

23 23

5.13 10 0 851 6.03 10

× = . ×

Eq. 31–8b:

23 23 5 13 10

1 05 4 89 10

. × = . . ×

Eq. 31–8c:

23 24

5 13 10 0 243 2 11 10

(12)

32. Calculate the Q-value for the reaction 23892U n+ → 23992U.

[

]

238 239

92 92

2 2 2

n

U U

2 2 MeV/

238.050788 u 1.008665 u 239 054294 u 931.5 4.806 MeV u

Q m c m c m c

c c

= + −

⎛ ⎞

= + − . ⎜ =

⎝ ⎠

33. The reaction of Eq. 31–8b consumes 2 deuterons and releases 3.23 MeV of energy. The amount of energy needed is the power times the elapsed time, and the energy can be related to the mass of deuterium by the reaction.

3 7

13 23

4

J s 1 MeV 2 d 2.014 10 kg

960 (1 yr) 3.156 10

s yr 1.60 10 J 3.23 MeV 6.02 10 d

3.923 10 kg 0.39 g

− −

⎛ ⎞

⎛ ⎞⎛ ⎞

⎛ ⎞ ×

⎛ ⎞ ×

⎟⎜ ⎟

⎜ ⎟

⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ × ⎠⎝ ⎠⎝ × ⎠

= × =

34. (a) The reactants have a total of 3 protons and 7 neutrons, so the products should have the same. After accounting for the helium, there are 3 neutrons and 1 proton in the other product, so it must be tritium, 31H . The reaction is 36Li+01n → 24He+31H.

(b) The Q-value gives the energy released.

[

]

6 1 4 3

3 0 2 1

2 2 2 2

Li n He H

2 2 MeV/

6.015123 u 1.008665 u 4.002603 u 3.016049 u 931.5 4.784 MeV u

Q m c m c m c m c

c c

= + − −

⎛ ⎞

= + − − ⎜ =

⎝ ⎠

35. Assume that the two reactions take place at equal rates, so they are both equally likely. Then from the reaction of 4 deuterons, there would be a total of 7.30 MeV of energy released, or an average of

1.825 MeV per deuteron. A total power of 1150 MW 3485 MW

0.33 = must be obtained from the fusion reactions to provide the required 1150-MW output, because of the 33% efficiency. We convert the power to a number of deuterons based on the energy released per reacting deuteron and then convert that to an amount of water using the natural abundance of deuterium.

6

13

2 2

23

J 3600 s 1 MeV 1 d 1 H atom

3485 10

s 1 h 1.60 10 J 1.825 MeV 0.000115 d's 3485 MW

1 H O molecule 0.018 kg H O 2 H atoms 6.02 10 molecules

5586 kg/h 5600 kg/h

⎡⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ ⎤

× ×

⎢⎜⎜ ⎟⎜⎜ ⎟⎜ ⎟ ⎥

×

⎢ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ ⎥

→ ⎢ ⎥

⎛ ⎞

⎛ ⎞

⎜ ⎟

⎠⎝×

⎣ ⎦

= ≈

36. We assume that the reactants are at rest when they react, so the total momentum of the system is 0. As a result, the momenta of the two products are equal in magnitude. The available energy of 17.57 MeV is much smaller than the masses involved, so we use the nonrelativistic relationship between

momentum and kinetic energy,

2

KE 2 KE.

2

p p m

m

(13)

(

)

4 4 4 4

2 2 2 2

4 4 4 4 4

2 2 2 2 2

4

2 4

2

n total n n n

He He He He

n n n total

He He He He He

n

total He

n He

KE KE KE KE KE

KE KE KE KE KE

KE KE

17.57 MeV 2 2

1.008665

17.57 MeV 3.536 MeV 3 4.002603 1.008665

p p m m

m m m m

m

m m

+ = = = → = →

= → = − →

= = = ≈

+ +

⎛ ⎞

⎜ ⎟

⎝ ⎠

4 2

n total He

KE KE KE

.5 MeV

17.57 MeV 3.54 MeV 14.03 MeV 14 MeV

= − = − = ≈

If the plasma temperature were significantly higher, then the approximation of 0 kinetic energy being brought into the reaction would not be reasonable. Thus the results would depend on plasma

temperature. A higher plasma temperature would result in higher values for the energies.

37. In Eq. 31–8a, 4.00 MeV of energy is released for every 2 deuterium atoms. The mass of water can be converted to a number of deuterium atoms.

23 4

21 2

2

2 2

6 19

21 9

6.02 10 H O 2 H 1.15 10 d

(1.00 kg H O) 7.692 10 d nuclei

0 018 kg H O 1 H O 1 H

4.00 10 eV 1.60 10 J

(7.692 10 d nuclei) 2.46 10 J

2 d atoms 1 eV

× ⎞⎛ ⎞⎛ ×

= × →

⎜ ⎟⎜ ⎟⎜⎜ ⎟

. ⎠⎝

⎝ ⎠

× ⎞⎛ ×

× ⎜ ⎟⎜⎟⎜= ×

⎝ ⎠⎝ ⎠

As compared to gasoline:

9 7 2.46 10 J

50 more than gasoline 5 10 J

× × ×

38. (a) We follow the method of Example 31–9. The reaction is 126C+21H → 137N+γ. We calculate the potential energy of the particles when they are separated by the sum of their radii. The radii are calculated from Eq. 30–1.

19 2

9 2 2

C H

KEtotal 15 1/3 1/3 13

0 C H

1 (8.99 10 N m /C ) (6)(1)(1 60 10 C) 1 MeV

4 (1.2 10 m)(1 12 ) 1.60 10 J

2.19 MeV Q Q

r r

πε

− −

⎛ ⎞

. ×

= = × ⋅ ⎜

+ × + ⎝ × ⎠

=

For the d–t reaction, Example 31–9 shows that KEtotal=0.45 MeV. Find the ratio of the two energies.

KEC

KEd t

2.19 MeV 4.9 0.45 MeV

− = =

The carbon reaction requires about 5 times more energy than the d–t reaction.

(b) The kinetic energy is proportional to the temperature by KE 3 2kT.

= Since the kinetic energy has

to increase by a factor of 5, so does the temperature. Thus we estimate T ≈1.5 10 K .× 9 39. Because the RBE of alpha particles is up to 20 and the RBE of X-rays is 1, it takes 20 times as many

(14)

40. Use Eq. 31–10b to relate Sv to Gy. From Table 31–1, the RBE of gamma rays is 1, so the number of Sv is equal to the number of Gy. Thus 4.0 Sv= 4.0 Gy .

41. The biological damage is measured by the effective dose, Eq. 31–10a, using Table 31–1.

(72 rads fast neutrons) 10 ( rads slow neutrons) 5 72 rads 10 144 rads slow neutrons

5

x

x

× = × →

×

= =

42. A gray is 1 joule per kilogram, according to Eq. 31–9.

(2.5 J/kg) 65 kg 162.5 J× = ≈160 J

43. (a) Since the RBE is 1, the effective dose (in rem) is the same as the absorbed dose (in rad). Thus the absorbed dose is 1.0 rad or 0.010 Gy.

(b) A Gy is 1 J/ kg.

(0.01 Gy) 1 J/kg (0.20 kg) 1 eV19 1 p6 1.0 10 p10

1 Gy 1.60 10− J 1.2 10 eV

⎛ ⎞⎛ ⎞

⎛ ⎞

= ×

⎜ ⎟⎜ ⎟

⎜ ⎟ ⎟⎜

× ×

⎝ ⎠ ⎝ ⎠⎝ ⎠

44. The counting rate will be 85% of 35% of the activity.

10

6 3.7 10 decays/s 1

(0.035 10 Ci) (0.35)(0.85) 385.3 counts/s 390 counts/s

1 Ci 1 decay

β

− ⎛ × ⎞⎛ ⎞

× ⎜ ⎟⎜⎟⎝ ⎟ = ≈

⎝ ⎠

45. The two definitions of roentgen are 1.6 10 ion pairs/g× 12 produced by the radiation and the newer definition of 0.878 10 × −2J/kg deposited by the radiation. Start with the current definition and relate them by the value of 35 eV per ion pair.

2 19

12

1 R (0.878 10 J/kg)(1 kg/1000 g)(1 eV/1.60 10 J)(1 ion pair/35 eV) 1.567 10 ion pairs/g

− −

= × ×

= ×

The two values of ion pairs per gram are within about 2% of each other.

46. We approximate the decay rate as constant and find the time to administer 32 Gy. If that calculated time is significantly shorter than the half-life of the isotope, then the approximation is reasonable. If 1.0 mCi delivers about 10 mGy/min, then 1.6 mCi would deliver 16 mGy/min.

dose rate time time dose 32 Gy3 1 day 1.39 days 1.4 days

rate 16 10− Gy/min 1440 min

⎛ ⎞

= × → = = ⎜ ⎟= ≈

× ⎝ ⎠

(15)

47. Since the half-life is long (5730 yr), we can consider the activity as constant over a short period of time. Use the definition of the curie from Section 31–5.

1 2 1

2

10

6 4

4 7 16

16

23

3.70 10 decays/s 0.693

(2.50 10 Ci) 9.25 10 decays/s

1 Ci

5730 yr

(9.25 10 decays/s) (3.156 10 s/yr) 2.414 10 nuclei

0.693 0.693

0.0140 kg

2.414 10 nuclei 5.

6.02 10 nuclei

N

N

t T

T N N

t

=

×Δ

× ⎜= × = = →

Δ

⎝ ⎠

Δ

= = × × = ×

Δ

⎛ ⎞

× ⎜

×

⎝ ⎠

10

61 10× − kg =0.561 gμ

48. Each decay releases one gamma ray of energy 122 keV. Half of that energy is deposited in the body. The activity tells at what rate the gamma rays are released into the body. We assume that the activity is constant.

6 10 16

7 7

/s 1

(1.55 10 Ci) 3.70 10 (86, 400 s/day)(0.50)(122 keV/ )(1.60 10 J/keV)

Ci 65 kg

J/kg Gy

7.44 10 7.4 10

day day

γ γ

− −

− −

⎛ ⎞

⎛ ⎞

× ⎜ × ⎟ × ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= × ≈ ×

49. Use the dose, the mass of the beef, and the energy per electron to find the number of electrons.

(4 5 10 Gy)3 1 J/kg (5 kg) 1 MeV13 1 e 8.78 10 e16 9 10 e16

1 Gy 1.60 10 J 1.6 MeV

− −

⎛ ⎞

⎛ ⎞

⎛ ⎞

. × ⎜ ⎟ ⎜⎟⎜⎜ ⎟= × ≈ ×

×

⎝ ⎠ ⎝ ⎠⎝ ⎠

50. (a) According to Appendix B, 13153I decays by beta decay. 13153I → 13154Xe+β−+v

(b) The number of nuclei present is given by Eq. 30–4, with N=0.05N0.

0 1/2 0

0

ln / ln / (8.0 d) ln 0.05 34.58 d 35 d

0 693 0.693

t N N T N N

N N e λ t

λ

= → = − = − = − = ≈

.

(c) The activity is given by Δ Δ =N t/ λN. This can be used to find the number of nuclei, and then the mass can be found.

3 10

1/2

13 13 12

23 /

( )( / )

/ (8.0 d)(86, 400 s/d)(1 10 Ci)(3.70 10 decays/s)

0 693 0 693

0.131 kg

3 69 10 nuclei; 3.69 10 nuclei 8 10 kg

6.02 10 nuclei

N t N

T N t

N t N

m λ

λ

Δ Δ = →

Δ Δ

Δ Δ × ×

= = =

. .

⎛ ⎞

= . × = × ⎜= ×

×

⎝ ⎠

(16)

12 10 6

13

6 5

Ci decays/s 3600 s L decays

2000 10 3.70 10 (12 h) 0.5 1.598 10

L 1 Ci h day day

decays days MeV 1 60 10 J

1.598 10 365 (0.10) 1.5 1.40 10

day yr decay MeV

⎛ ⎞ ⎛ ⎞

⎛ ⎞ ⎛ ⎞

× ⎜ × ⎟ ⎜ ⎟= ×

⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞

× ⎞⎛ ⎞ ⎛ ⎞ . × = ×

⎜ ⎟⎜ ⎟ ⎜ ⎟⎜

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝

J yr

(a) For the adult, use a mass of 60 kg.

5

5

7 2

7 2

J 1 1 Gy 1 Sv

Effective dose 1.40 10

yr 60 kg 1 J/kg 1 Gy

10 mrem

2.33 10 Sv/yr 2.33 10 mrem/yr

Sv

2 10 Sv/yr or 2 10 mrem/yr

− −

− −

⎛ ⎞⎛ ⎞⎛ ⎞

⎛ ⎞

=⎜ × ⎟⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

⎛ ⎞

= × ⎜= ×

⎝ ⎠

≈ × ×

2

4 mrem

2.33 10 year

Fraction of allowed dose 2 10 times the allowed dose mrem

100 year

×

= ≈ ×

(b) For the baby, the only difference is that the mass is 10 times smaller, so the effective dose is 10 times bigger. The results are as follows:

≈ × 2 10−6Sv/yr , 0.2 mrem/yr , and 2 10× −3times the allowed dose

52. (a) The reaction has Z = 86 and A = 222 for the parent nucleus. The alpha has Z = 2 and A = 4, so the daughter nucleus must have Z = 84 and A = 218. That makes the daughter nucleus 21884Po . (b) From Fig. 30–11, polonium-218 is radioactive. It decays via both alpha and beta decay. The

half-life for the alpha decay is 3.1 minutes.

(c) The daughter nucleus is not a noble gas, so it is chemically reacting. It is in the same group as oxygen, so it might react with many other elements chemically.

(d) The activity is given by Eq. 30–3b,

1/2 ln 2 .

R N N

T λ

= =

12

23 9

6 6 4

10

0 693 0.693 (1.4 10 g)6.02 10 nuclei

(3 8235 d)(86, 400 s/d) 222 g

1 Ci

7.964 10 decays/s 8.0 10 Bq 2.2 10 Ci

3.70 10 Bq

N N

t T

Δ = . = × ×

Δ .

= × ≈ × × = ×

×

To find the activity after 1 month, use Eq. 30–5.

1/ 2

0 693 0 693

(30 d)

6 (3 8235 d) 4

0

4 7

10

(7.964 10 decays/s) 3.465 10 decays/s

1 Ci

3 5 10 Bq 9.4 10 Ci

3 70 10 Bq t

T

N N

e e

t t

. .

− −

.

Δ =⎛Δ ⎞ = × = ×

⎜ ⎟

Δ ⎝ Δ ⎠

(17)

53. The frequency is given in Section 31–9 as 42.58 MHz. Use that to find the wavelength.

8 6 2.998 10 m/s

7.041 m 42.58 10 Hz

c

c f

f

λ λ ×

= → = = =

×

This lies in the radio wave portion of the spectrum.

54. (a) The mass of fuel can be found by converting the power to energy to number of nuclei to mass.

7 6

13 23

3.156 10 s 1 MeV 1 fission atom 0.235 kg (2100 10 J/s)(1 yr)

1 yr 1.60 10 J 200 MeV 6.02 10 atom

808.5 kg 810 kg

× ⎞⎛ ⎞⎛ ⎞⎛ ⎞

× ⎜ ⎟⎜⎜ ⎟⎜

× ⎝ ⎠ ×

⎝ ⎠ ⎝ ⎠

⎝ ⎠

= ≈

(b) The product of the first five factors above gives the number of U atoms that fission.

7

90 6

38 13

26 90 38

3.156 10 s 1 MeV 1 fission atom Sr nuclei 0 06(2100 10 J/s)(1 yr)

1 yr 1.60 10 J 200 MeV

(1.24 10 ) Sr nuclei

⎛ × ⎞⎛ ⎞⎛ ⎞

# = . × ⎜ ⎟⎜ ⎟⎜

× ⎝ ⎠

⎝ ⎠

⎝ ⎠

= ×

The activity is given by the absolute value of Eq. 30–3b.

26 16

7 1/2

16 6

10

0 693 0.693 (1.24 10 ) 9.389 10 decays

s (29 yr)(3.156 10 s/yr)

1 Ci

(9.389 10 decays/s) 2.5 10 Ci

3.70 10 decays/s

N N N

t λ T

Δ = = . = × = ×

Δ ×

⎛ ⎞

= × ⎜= ×

×

⎝ ⎠

55. (a) The reaction is 94Be+42He → +n ?. There are 6 protons and 13 nucleons in the reactants, so there must be 6 protons and 13 nucleons in the products. The neutron is 1 nucleon, so the other product must have 6 protons and 12 nucleons. Thus, it is 126C .

(b)

[

]

9 4 12

4 2 6

2 2 2 2

n

Be He C

2 2 MeV/

9.012183 u 4.002603 u 1.008665 u 12.000000 u 931.5 5.702 MeV u

Q m c m c m c m c

c c

= + − −

⎛ ⎞

= + − − ⎜ =

⎝ ⎠

56. If KE=kT, then to get from kelvins to keV, use the Boltzmann constant. It must be put in the proper units.

1.381 10 23 J 1 eV 19 1 keV 8.620 10 8keV/K

K 1.602 10 J 1000 eV

k= × − × × = × −

×

57. From Eq. 13–9, the average speed of a gas molecule (root-mean-square speed) is inversely

proportional to the square root of the mass of the molecule, if the temperature is constant. We assume that the two gases are in the same environment and therefore at the same temperature. We use UF6 molecules for the calculations.

23592 6 23892 6

238 235

92 6 92 6

UF UF

UF UF

238 05 6(19 00) 1.0043 :1 235 04 6(19 00)

m m υ

υ

. + .

= = =

(18)

58. (a) We assume that the energy produced by the fission was 200 MeV per fission, as in Problems 18, 19, and 20.

12

13 23

5 10 J 1 MeV 1 fission atom 0.235 kg (20 kilotons TNT)

1 kiloton 1.60 10 J 200 MeV 6.02 10 atoms

1.220 kg 1 kg

× ⎞⎛ ⎞⎛ ⎞⎛ ⎞

⎜ ⎟⎜⎜ ⎟⎜

⎜ ⎟ × ×

⎝ ⎠ ⎝ ⎠

⎝ ⎠

= ≈

(b) Use E mc= 2.

12

2 3

2 8 2

5 10 J (20 kilotons TNT)

1 kiloton

1.11 10 kg 1 g (3.0 10 m/s)

E

E mc m

c

− ⎛ ×

⎜ ⎟

⎜ ⎟

⎝ ⎠

= → = = = × ≈

×

59. The effective dose (in rem) is equal to the actual dose (in rad) times the RBE factor.

dose(rem) (32 mrad/yr X-ray, -ray)(1) (3.4 mrad/yr)(10)= γ + = 66 mrem/yr

60. Because the RBE factor for gamma rays is 1, the dose in rem is equal in number to the dose in rad. Since the intensity falls off as r2 (the square of the distance), the exposure rate times r2 will be constant.

3

3 2 2 2

2 2

3

1 m

5.0 rem 1 rad 1 yr 1 week rad

Allowed dose 2.747 10

yr 1 rem 52 weeks 35 h h

rad rad

2.747 10 4.8 10 (1 m)

h h

rad

4.8 10 ( )

h 4.180 m 4.2 m

rad 2.747 10

h

r

r

− −

=

=

= ≈

⎛ ⎞⎛ ⎞⎛ ⎞

= ×

⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

× ⎞ ⎛ ×

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

×

⎜ ⎟

⎝ ⎠ =

×

⎜ ⎟

⎝ ⎠

61. (a) The reaction is of the form ? → 42He+22286Rn. There are 88 protons and 226 nucleons as products, so there must be 88 protons and 226 nucleons as reactants. The parent nucleus is

226 88Ra.

22688Ra → 42He+22286Rn

(b) If we ignore the KE of the daughter nucleus, then the KE of the alpha particle is the Q-value of the reaction.

[

]

226 4 222

88 2 86

2 2 2

KE Ra He Rn

2 2 MeV/

226 025410 u 4 002603 u 222 017578 u 931.5 4.871 MeV u

m c m c m c

c c

α = − −

⎛ ⎞

= . − . − . ⎜ =

(19)

(c) From momentum conservation, the momentum of the alpha particle will be equal in magnitude to the momentum of the daughter particle. At the energy above, the alpha particle is not

relativistic, so

2

KE 2 KE .

2 a a a

p

p m

m

α α

α

= → =

2 KE 2(4.002603 u) 931.5 MeV/ 2 (4.871 MeV) 190.6 MeV/ 1 u

a a c

pα = m = ⎛⎜ ⎞⎟ = c

⎝ ⎠

The momentum of the daughter nucleus is the same as that of the alpha, 190.6 MeV/ .c

(d) Since pα = pdaughter,

2 2

daughter KEdaughter

daughter daughter .

2 2

p p

m m

α

= =

2 2

2

KEdaughter 2

daughter

(190.6 MeV/ )

8.78 10 MeV

2 931.5 MeV/

2(222.0 u)

1 u

p c

m c

α −

= = = ×

⎛ ⎞

⎜ ⎟

⎜ ⎟

⎝ ⎠

Thus we see that our original assumption of ignoring the kinetic energy of the daughter nucleus is valid. The kinetic energy of the daughter is less than 2% of the Q-value.

62. This “heat of combustion” is 26.2 MeV/4 hydrogen atoms.

13

14 27

26.2 MeV 1.60 10 J 1 H atom 1 u

6.26 10 J/kg 4 H atoms 1 MeV 1.007825 u 1.66 10 kg

× ⎞⎛ ⎞⎛ ⎞

= ×

⎜ ⎟⎜ ⎟⎜⎜ ⎟

⎜ ⎟ ⎠⎝ ×

⎝ ⎠

This is about 2 10 × 7 times the heat of combustion of coal.

63. (a) The energy is radiated uniformly over a sphere with a radius equal to the orbit radius of the Earth.

(1300 W/m )4 (1.496 10 m)2 π × 11 2=3.656 10× 26 W≈ 3.7 10× 26W

(b) After subtracting the energy of the neutrinos, the reaction of Eq. 31–7 releases 26.2 MeV for every 4 protons consumed.

26 38

13

38 J 4 protons 1 MeV

3.656 10 3.489 10 protons/s

s 26.2 MeV 1.6 10 J

3.5 10 protons/s

⎛ ⎞

⎛ ⎞

⎛ ⎞

× ⎜ ⎟⎜ ⎟= ×

⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ×

≈ ×

(c) Convert the Sun’s mass to a number of protons and then use the above result to estimate the Sun’s lifetime.

2.0 10 kg30 1 proton27 1 s38 1 yr 7 1.1 10 yr11

1.673 10− kg 3.489 10 protons 3.156 10 s

⎛ ⎞⎛ ⎞⎛ ⎞

× ⎜ ⎟⎜⎟⎜ ⎟⎜⎟⎜≈ ×

× × ×

⎝ ⎠⎝ ⎠⎝ ⎠

64. For the net proton cycle, Eq. 31–7, there are two neutrinos produced for every four protons consumed. Thus the net number of neutrinos generated per second from the Sun is just half the value of protons consumed per second. That proton consumption rate is calculated in Problem 63b.

38 38

(20)

Assume that the neutrino distribution is spherically symmetric, centered at the Sun. The fraction that would pass through the area of the ceiling can be found by a ratio of areas, assuming that the ceiling is perpendicular to the neutrino flux. But since the ceiling is not perpendicular, a cosine factor is included to account for the angle difference, as discussed in Eq. 21–1. Finally, we adjust for the one-hour duration, assuming that the relative angle is constant over that hour.

2

38 20

11 2 180 m

(1.712 10 /s) (cos 44 )(3600 s) 2.8 10

4 (1.496 10 m)

v v

π

× ° = ×

×

65. Use the common value of 200 MeV of energy released per fission, as used in Problems 18, 19, and 20. Multiply that by the number of fissions, which is 5.0% of the number of U-238 atoms.

235 13

238 92

92 235 238

92 92 23 238

92 238

92 12 12

U nuclei

200 MeV 1.602 10 J

0.05 2.0 kg U

1 MeV

1 nucleus of U U nuclei

Total energy

6.022 10 nuclei of U nuclei 0.238 kg U

8.107 10 J 8 10 J

( )

×

=

× ×

= ≈ ×

⎡ ⎛ ⎞⎛ ⎞ ⎤

⎜ ⎟

⎢ ⎜⎟⎜

⎢ ⎝ ⎠⎝ ⎠ ⎥

⎜ ⎟

⎣ ⎦

×

66. (a) The energy released is given by the Q-value.

[

]

12 24

6 12

2

2 2 2

C Mg

MeV/

2 2(12 000000 u) 23.985042 u 931.5 13.93 MeV

u c

Q= m cm c = . − ⎛⎜ ⎞⎟c =

⎝ ⎠

(b) The total KE of the two nuclei must equal their PE when separated by 6.0 fm.

1 2 KE

0

2 19

9 2 2 13

1 2

1 1

KE

2 2 15

0

13 19 6

1 2

4

(6)(1 60 10 C)

1 (9.0 10 N m /C ) 6.912 10 J

4 6 0 10 m

6 912 10 J(1 eV/1.60 10 J)(1 MeV/10 eV) 4.23 MeV 4.3 MeV Q Q

r

Q Q r πε

πε

− −

− −

= →

. ×

⎣ ⎦

= = × ⋅ = ×

. ×

= . × × = ≈

(c) The kinetic energy and temperature are related by Eq.13–8.

13

KE 10

3 2 2

KE 2 3 3

23 6.912 10 J

3.3 10 K 1.38 10 J/K

kT T

k

− −

×

= → = = = ×

×

67. (a) A Curie is 3.7 10 decays/s. × 10

(0.10 10 × −6Ci)(3.7 10 decays/s) × 10 = 3700 decays/s

(21)

13 7

4 4 4

1 (3700 decays/s)(1.4 MeV/decay)(1.60 10 J/MeV)(3.156 10 s/yr)

65 kg

4.024 10 J/kg/yr 4.024 10 Gy/yr 4.0 10 Sv/yr

− − −

⎛ ⎞

× × ⎜ ⎟

⎝ ⎠

= × = × = ×

This is about

4 3 4.024 10 Sv/yr

0 11 3.6 10 Sv/yr

− −

× = .

× or 11% of the background rate.

68. The surface area of a sphere is 4πr2.

7 7 10

3

2 2 6 2 2

Earth

Activity 2.0 10 Ci (2 0 10 Ci)(3.7 10 decays/s) 1.4 10 decays/s

m 4πr 4 (6.38 10 m)π m

× . × ×

= = = ×

×

69. 4 12

[

]

2 6

2

2 2 2

He C

MeV/

3 3(4 002603 u) 12 000000 u 931.5 7.274 MeV

u c

Q= m cm c = . − . ⎛⎜ ⎞⎟c =

⎝ ⎠

70. Since the half-life is 30 years, we assume that the activity does not change during the 1.4 hours of exposure. We calculate the total energy absorbed and then calculate the effective dose. The two energies can be added directly since the RBE factor for both gammas and betas is about 1.

6 10

5

3 19

5

5 5

decays s

(1.2 10 Ci) 3 7 10 (1 4 h) 3600

s 1 h

Energy 3.043 10 J

eV J

850 10 1.60 10

decay eV

3 043 10 J 100 rad

dose 4.909 10 rad 4.9 10 rem

62 kg 1 J/kg

− −

− −

⎡ ⎛ ⎞ ⎛ ⎞⎤

× . × .

⎢ ⎜ ⎜ ⎟⎥

⎝ ⎠

⎢ ⎥

= = ×

⎛ ⎞⎛ ⎞

× × ⎟⎜ ×

⎢ ⎝ ⎠⎝ ⎠ ⎥

⎣ ⎦

. ×

= × = × ≈ ×

71. The actual power created by the reaction of Eq. 31–8a must be 1000 MW/0.33 = 3030 MW. We convert that to grams of deuterium using Eq. 31–8a.

3 7

9

13 23

J 2 d nuclei 1 MeV 2 10 kg d 3.156 10 s

Mass d 3 03 10

s 4 03 MeV 1.60 10 J 6.02 10 nuclei 1 yr

985.4 kg d 990 kg d

− −

⎛ ⎞⎛ ⎞

⎛ ⎞

⎛ ⎞ × ×

⎛ ⎞

= . × ⎜ ⎟. ⎟⎜⎟⎜⎟⎜⎟⎜ ⎟

× ×

⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

= ≈

72. In the first scenario, slow neutrons have an RBE of 5, from Table 31–1. The maximum dose for a radiation worker is given in the text as 50 mSv/yr, so this worker has been exposed to 50 mSv. Use Eq. 31–10b to convert the Sv to Gy and then convert that to J by the definition of Gy.

Energy Gy mass (50 10 Sv)3 1 Gy 1 J/kg (65 kg) 0.65 J 5 Sv 1 Gy

− ⎛ ⎞⎛ ⎞

= × = × ⎜ ⎟⎜ ⎟ =

⎝ ⎠⎝ ⎠

In the second scenario, the RBE changes to a value of 2 for the protons.

Energy Gy mass (50 10 Sv)3 1 Gy 1 J/kg (65 kg) 1.6 J 2 Sv 1 Gy

− ⎛ ⎞⎛ ⎞

= × = × ⎜ ⎟⎜ ⎟ =

(22)

73. (a) There are 92 protons in the reactants, so X must have 92 – 38 = 54 protons. There are 236 total nucleons in the reactants, so X must have 236 – 92 – 3 = 141 total nucleons. Thus, X is 14154Xe . (b) If the reaction is “barely critical,” then every nucleus that fissions leads to another nucleus that fissions. Thus one of the produced neutrons causes another fission. The other two either escape the reactor vessel or are absorbed by some nucleus without causing a fission.

(c) The masses needed are 14154Xe: 140.92665 u and 9238Sr: 91.911038 u. Use those to calculate the Q-value.

[

]

235 1 92 141 1

92 0 38 54 0

2

n Sr Xe n

2 2

( U 3 )

235 043930 u 1 008665 u 91 911038 u 140 92665 u 3(1 008665 u)

MeV/

(0 188912 u) 931 5 176 0 MeV u

Q m m m m m c

c

c

= + − − −

= . + . − . − . − .

⎛ ⎞

= . ⎜ . ⎟= .

⎝ ⎠

74. The half-life of the strontium isotope is 28.90 years. Use that with Eq. 30–5 to find the time for the activity to be reduced to 15% of its initial value.

1/ 2 1/ 2 1/ 2

0 693 0 693 0 693

0 0 0

1/2 1/2

0 15 0 15

ln(0.15)

0.693 (28.90 yr)ln(0.15)

ln(0.15) 79 yr

0 693 0.693

t t t

T T T

N N N N

e e e

t t t t

T

t t

T

. . .

− − −

Δ =⎛Δ ⎞ . ⎛Δ ⎞ =⎛Δ ⎞ . =

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

Δ ⎝ Δ ⎠ ⎝ Δ ⎠ ⎝ Δ ⎠

= − → = − = − ≈

.

75. Source B is more dangerous than source A because of its higher energy. Since the sources have the same activity, they both emit the same number of gammas. Source B can deposit twice as much energy per gamma and therefore causes more biological damage.

Source C is more dangerous than source B because the alphas have an RBE factor up to 20 times larger than the gammas. Thus a number of alphas may have an effective dose up to 20 times higher than the effective dose of the same number of like-energy gammas.

So from most dangerous to least dangerous, the ranking of the sources is C > B > A.

We might say that source B is twice as dangerous as source A, and source C is 20 times more dangerous than source B.

76. The whole-body dose can be converted into a number of decays, which would be the maximum number of nuclei that could be in the Tc sample. The RBE factor of gammas is 1.

3 17

19

17 12

3

1 kg 1 eV

50 mrem 50 mrad (50 10 rad) (55 kg) 1.719 10 eV

100 rad 1.60 10 J

1 effective 2 decays 1 nucleus

(1.719 10 eV) 2.455 10 nuclei

1 effective 1 decay 140 10 eV

γ γ

γ γ

⎛ ⎞

⎛ ⎞

= → × ⎜ ⎟ ⎜= ×

×

⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞⎛ ⎞

× ⎜ ⎟⎜ ⎟⎜ ⎟= ×

× ⎝ ⎠⎝ ⎠

⎝ ⎠

This then is the total number of decays that will occur. The activity for this number of nuclei can be calculated from Eq. 30–3b.

12

10 1/2

3

0 693 0.693(2 455 10 decays) 1 Ci Activity

(6 h)(3600 s/h) 3.70 10 decays/s

2.129 10 Ci 2 mCi

N N

N

t λ T

⎛ ⎞

Δ . . ×

= = = = ⎜

Δ ⎝ × ⎠

(23)

Solutions to Search and Learn Problems

1. (a) Three problems that must be overcome to make a functioning fission nuclear reactor: (i) the neutrons must be moderated (slowed) so they will react with the fissionable nuclei; (ii) the neutrons must be prevented from being absorbed by the wrong nuclei;

(iii) the neutrons must be prevented from escaping the reaction vessel.

(b) Five environmental problems or dangers resulting from a nuclear fission reactor: (i) thermal pollution—the warming of the environment around the nuclear reactor; (ii) the need for safe storage of fuel before using it in the reactor;

(iii) the safe disposal of radioactive waste; (iv) radiation damage to the power plant itself;

(v) accidental release of radiation into the environment.

(c) Breeder reactors make fissionable material that can be used in fission bombs.

2. (a) For small nuclei, the binding energy per nucleon increases as the number of nucleons increases. When two small nuclei are fused together, the total energy of the resulting nucleus is smaller than the energy of the two initial nuclei, with the excess energy released in the fusion process. (b) The initial proton–proton reaction has a much smaller probability of occurring than the other

reactions. Since the other reactions can take place relatively quickly compared with the first reaction, it determines the time scale for the full fusion process.

(c) Iron (Fe).

(d) The force of gravity between all of the atoms in the Sun (or star) holds it together. (e) (i) A tokamak uses magnetic fields to confine the fusion material.

(ii) Lasers are used to heat the fusion material so quickly that the nuclei cannot move away before they fuse (inertial confinement).

3. The oceans cover about 70% of the Earth, to an average depth of approximately 4 km. The density of the water is approximately 1000 kg/m .3 The mass of water is found by multiplying the volume of water by its density.

Mass of water 0 7(surface area)(depth)(density)6 2 3 21

(0 7)4 (6.38 10 m) (4000 m)(1000 kg/m ) 1.43 10 kg waterπ = .

= . × = ×

Dividing the mass by the molar mass of water and multiplying by Avogadro’s number gives the number of water molecules. There are two hydrogen atoms per water molecule. Multiplying the number of hydrogen atoms by the percentage of deuterium gives the number of deuterium atoms.

If the reactions of Eqs. 31–8a and 31–8b are carried out at the same rate, then 4 deuterons would produce 7.30 MeV of energy. Use that relationship to convert the number of deuterons in the oceans to energy.

13

43 7.30 MeV 1.60 10 J 30 30

(1.10 10 d) 3.21 10 J 3 10 J

4 d 1 MeV

×

× × × = × ≈ ×

(24)

4. (a) No carbon is consumed in this cycle because one 126C nucleus is required in the first step of the cycle, and one 126C nucleus is produced in the last step of the cycle. The net effect of the cycle can be found by adding all the reactants and all the products together and canceling those that appear on both sides of the reaction.

12 1 13

6 1 7

13 13

7 6

13 1 14

6 1 7

14 1 15

7 1 8

15 15

8 7

15 1 12 4

7 1 6 2

12 1 13 13 1 14 1 15 15 1

6 1 7 6 1 7 1 8 7 1

13 13 14 15 15 12 4

7 6 7 8 7 6 2

1 4

1 2

C H N

N C e

C H N

N H O

O N e

N H C He

C H N C H N H O N H

N C e N O N e C He

4 H He

v

v

v v

γ

γ γ

γ γ γ

+

+

+ +

+ → +

→ + +

+ → +

+ → +

→ + +

+ → +

+ + + + + + + + +

→ + + + + + + + + + + + + +

→ +2 e++2v+3γ

Eq. 31–7 shows the proton–proton chain, which is only different from the carbon cycle by the emission of one additional gamma ray in the carbon cycle.

(b) To use the values from Appendix B, we must be sure that the number of electrons is balanced as well as the number of protons and neutrons. The above “net” equation does not consider the electrons that neutral nuclei would have, because it does not conserve charge. What the above reaction really represents (ignoring the gammas and neutrinos) is the following:

4 H11 → 42He 2 e+ + ⇒ 4 p11 → 2 p 2 n 2e11 + 10 + +

To use the values from Appendix B, we must add 4 electrons to each side of the reaction.

(4 p 4e )11 + − → (2 p 2e11 + −+2 n) 2e10 + ++2e− ⇒ 4 H11 → 24He 2e+ ++2e− The energy produced in the reaction is the Q-value.

[

]

1 4

1 2

2 2

e

H He

2 2

4 4

<

References

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