Ch30 Giancoli7e Manual

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30-1

Responses to Questions

1. Different isotopes of a given element have the same number of protons and electrons. Because they

have the same number of electrons, they have almost identical chemical properties. Each isotope has

a different number of neutrons from other isotopes of the same element. Accordingly, they have

different atomic masses and mass numbers. Since the number of neutrons is different, they may

have different nuclear properties, such as whether they are radioactive or not.

2. Identify the element based on the atomic number.

(a) Uranium ( Z = 92)

(b) Nitrogen ( Z = 7)

(c) Hydrogen ( Z = 1)

(d ) Strontium ( Z = 38)

(e) Fermium ( Z = 100)

3. The number of protons is the same as the atomic number, and the number of neutrons is the mass

number minus the number of protons.

(a) Uranium: 92 protons, 140 neutrons

(b) Nitrogen: 7 protons, 11 neutrons

(c) Hydrogen: 1 proton, 0 neutrons

(d ) Strontium: 38 protons, 48 neutrons

(e) Fermium: 100 protons, 152 neutrons

4. With 87 nucleons and 50 neutrons, there must be 37 protons. This is the atomic number, so the element

is rubidium. The nuclear symbol is 8737 Rb.

5. The atomic mass of an element as shown in the Periodic Table is the average atomic mass of all

naturally occurring isotopes. For example, chlorine occurs as roughly 75% 3517 Cl and 25%3717 Cl, so its

atomic mass is about 35.5 ( 0.75 35 0.25 37).= × + × Other smaller effects would include the fact that

the masses of the nucleons are not exactly 1 atomic mass unit and that some small fraction of the mass

energy of the total set of nucleons is in the form of binding energy.

NUCLEAR PHYSICS AND R ADIOACTIVITY 30


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30-2 Chapter 30



6. The alpha particle is a very stable nucleus. It has less energy when bound together than when split

apart into separate nucleons.

7. The strong force and the electromagnetic (EM) force are two of the four fundamental forces in nature.

They are both involved in holding atoms together: the strong force binds quarks into nucleons and

binds nucleons together in the nucleus; the EM force is responsible for binding negatively charged

electrons to positively charged nuclei and for binding atoms into molecules. The strong force is the

strongest fundamental force; the EM force is about 100 times weaker at distances on the order of1710 m.

− The strong force operates at short range and is negligible for distances greater than about

the size of the nucleus. The EM force is a long-range force that decreases as the inverse square of the

distance between the two interacting charged particles. The EM force operates only between charged

particles. The strong force is always attractive; the EM force can be attractive or repulsive. Both of

these forces have mediating field particles associated with them—the gluon for the strong force and

the photon for the EM force.

8. Quoting from Section 30–3, “… radioactivity was found in every case to be unaffected by the strongest

physical and chemical treatments, including strong heating or cooling and the action of strong

chemical reagents.” Chemical reactions are a result of electron interactions, not nuclear processes.The absence of effects caused by chemical reactions is evidence that the radioactivity is not due to

electron interactions. Another piece of evidence is the fact that the α particle emitted in manyradioactive decays is much heavier than an electron and has a different charge than the electron, so it

can’t be an electron. Therefore, it must be from the nucleus. Finally, the energies of the electrons or

photons emitted from radioactivity are much higher than those corresponding to electron orbital

transitions. All of these observations support radioactivity being a nuclear process.

9. The resulting nuclide for gamma decay is the same isotope in a lower energy state.

64 6429 29Cu* Cu γ → +

The resulting nuclide for β − decay is an isotope of zinc, 64

30Zn.

64 6429 30Cu Zn e v

−→ + +

The resulting nuclide for β + decay is an isotope of nickel, 6428 Ni.

64 6429 28Cu Ni e v

+→ + +

10.23892U decays by alpha emission into

23490Th, which has 144 neutrons.

11. Alpha (α ) particles are helium nuclei. Each α particle consists of 2 protons and 2 neutrons andtherefore has a charge of +2e and an atomic mass value of 4 u. They are the most massive of the three.

Beta ( β ) particles are electrons ( ) β − or positrons ( ). β + Electrons have a charge of – e and positrons

have a charge of +e. In terms of mass, beta particles are much lighter than protons or neutrons, by a

factor of about 2000, so are lighter than alpha particles by a factor of about 8000. Their emission is

always accompanied by either an antineutrino ( ) β − or a neutrino ( ). β + Gamma (γ) particle are

photons. They have no mass and no charge.


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Nuclear Physics and Radioactivity 30-3



12. (a) 45 4520 21Ca Sc e v−

→ + + Scandium-45 is the missing nucleus.

(b) 58 5829 29Cu Cu γ → + Copper-58 is the missing nucleus.

(c) 46 4624 23Cr V e v+

→ + + The positron and the neutrino are the missing particles.

(d )234 23094 92Pu U α → + Uranium-230 is the missing nucleus.

(e) 239 23993 94 Np Pu e v−

→ + + The electron and the antineutrino are the missing particles.

13. The two extra electrons held by the newly formed thorium will be very loosely held, as the number of

protons in the nucleus will have been reduced from 92 to 90, reducing the nuclear charge. It will be

easy for these extra two electrons to escape from the thorium atom through a variety of mechanisms.

They are in essence “free” electrons. They do not gain kinetic energy from the decay. They might get

captured by the alpha nucleus, for example.

14. When a nucleus undergoes either β − or β + decay, it becomes a different element, since it has

converted either a neutron to a proton or a proton to a neutron. Thus its atomic number ( Z ) haschanged. The energy levels of the electrons are dependent on Z , so all of those energy levels change to

become the energy levels of the new element. Photons (with energies on the order of tens of eV) are

likely to be emitted from the atom as electrons change energies to occupy the new levels.

15. In alpha decay, assuming the energy of the parent nucleus is known, the unknowns after the decay are

the energies of the daughter nucleus and the alpha. These two values can be determined by energy and

momentum conservation. Since there are two unknowns and two conditions, the values are uniquely

determined. In beta decay, there are three unknown postdecay energies since there are three particles

present after the decay. The conditions of energy and momentum conservation are not sufficient to

exactly determine the energy of each particle, so a range of values is possible.

16. In electron capture, the nucleus will effectively have a proton change to a neutron. This isotope will

then lie to the left and above the original isotope. Since the process would only occur if it made thenucleus more stable, it must lie BELOW the line of stability in Fig. 30–2.

17. Neither hydrogen nor deuterium can emit an α particle. Hydrogen has only one nucleon (a proton) in

its nucleus, and deuterium has only two nucleons (one proton and one neutron) in its nucleus. Neither

one has the necessary four nucleons (two protons and two neutrons) to emit an α particle.

18. Many artificially produced radioactive isotopes are rare in nature because they have decayed away

over time. If the half-lives of these isotopes are relatively short in comparison with the age of Earth

(which is typical for these isotopes), then there won’t be any significant amount of these isotopes left

to be found in nature. Also, many of these isotopes have a very high energy of formation, which is

generally not available under natural circumstances.

19. After two months the sample will not have completely decayed. After one month, half of the samplewill remain, and after two months, one-fourth of the sample will remain. Each month, half of the

remaining atoms decay.

20. For Z > 92, the short range of the attractive strong nuclear force means that no number of neutrons is

able to overcome the long-range electrostatic repulsion of the large concentration of protons.


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30-4 Chapter 30



21. There are a total of 4 protons and 3 neutrons in the reactant. The α particle has 2 protons and 2

neutrons, so 2 protons and 1 neutron are in the other product particle. It must be 32 He.

6 1 4 33 1 2 2Li p Heα + → +

22. The technique of146C would not be used to measure the age of stone walls and tablets. Carbon-14

dating is only useful for measuring the age of objects that were living at some earlier time.

23. The decay series of Fig. 30–11 begins with a nucleus that has many more neutrons than protons and

lies far above the line of stability in Fig. 30–2. The alpha decays remove both 2 protons and 2

neutrons, but smaller stable nuclei have a smaller percentage of neutrons. In β + decay, a proton is

converted to a neutron, which would take the nuclei in this decay series even farther from the line of

stability. Thus, β + decay is not energetically preferred.

24. There are four alpha particles and four β − particles (electrons) emitted, no matter which decay path is

chosen. The nucleon number drops by 16 as 22286Rn decays into20682Pb, indicating that four alpha

decays occurred. The proton number only drops by four, from Z = 86 to Z = 82, but four alpha decays

would result in a decrease of eight protons. Four β − decays will convert four neutrons into protons,

making the decrease in the number of protons only four, as required. (See Fig. 30–11.)

25. (i) Since the momentum before the decay was 0, the total momentum after the decay will also be 0.

Since there are only two decay products, they must move in opposite directions with equal

magnitude of momentum. Thus, (c) is the correct choice: both the same.

(ii) Since the products have the same momentum, the one with the smallest mass will have the

greater velocity. Thus (b) is the correct choice: the alpha particle.

(iii) We assume that the products are moving slowly enough that classical mechanics can be used. In

that case, 2KE /2 . p m= Since the particles have the same momentum, the one with the smallest

mass will have the greater kinetic energy. Thus, (b) is the correct choice: the alpha particle.

Responses to MisConceptual Questions

1. (a) A common misconception is that the elements of the Periodic Table are distinguished by the

number of electrons in the atom. This misconception arises because the number of electrons in a

neutral atom is the same as the number of protons in its nucleus. Elements can be ionized by

adding or removing electrons, but this does not change what type of element it is. When an

element undergoes a nuclear reaction that changes the number of protons in the nucleus, the

element does transform into a different element.

2. (b) The role of energy in binding nuclei together is often misunderstood. As protons and neutrons

are added to the nucleus, they release some mass energy, which usually appears as radiation or

kinetic energy. This lack of energy is what binds the nucleus together. To break the nucleus

apart, the energy must be added back in. As a result, a nucleus will have less energy than the

protons and neutrons that made up the nucleus.


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3. (c) Large nuclei are typically unstable, so increasing the number of nuclei does not necessarily make

the nucleus more stable. Nuclei such as 148O have more protons than neutrons, yet14

8O is not as

stable as 168O. Therefore, having more protons than neutrons does not necessarily make a

nucleus more stable. Large unstable nuclei, such as

238

94Pu, have a much larger total bindingenergy than small stable nuclei such as

42 He, so the total binding energy is not a measure of

stability. However, stable nuclei typically have large binding energies per nucleon, which means

that each nucleon is more tightly bound to the others.

4. (e) The Coulomb repulsive force does act inside the nucleus, pushing the protons apart. Another

larger attractive force is thus necessary to keep the nucleus together. The force of gravity is far

too small to hold the nucleus together. Neutrons are not negatively charged. It is the attractive

strong nuclear force that overcomes the Coulomb force to hold the nucleus together.

5. (b) The exponential nature of radioactive decay is a concept that can be misunderstood. It is

sometimes thought that the decay is linear, such that the time for a substance to decay

completely is twice the time for half of the substance to decay. However, radioactive decay is notlinear but exponential. That is, during each half-life, half of the remaining substance decays. If

half the original decays in the first half-life, then half remains. During the second half-life, half

of what is left decays, which would be one-quarter of the initial substance. In each subsequent

half-life, half of the remaining substance decays, so it takes many half-lives for a substance to

effectively decay away. The decay constant is inversely related to the half-life.

6. (e) The half-life is the time it takes for half of the substance to decay away. The half-life is a

constant determined by the composition of the substance and not by the quantity of the initial

substance. As the substance decays away, the number of nuclei is decreasing and the activity

(number of decays per second) is decreasing, but the half-life remains constant.

7. (d ) A common misconception is that it would take twice the half-life, or 20 years, for the substance

to completely decay. This is incorrect because radioactive decay is an exponential process. Thatis, during each half-life, 1/2 of the remaining substance decays. After the first 10 years, 1/2

remains. After the second 10-year period, 1/4 remains. After each succeeding half-life another

half of the remaining substance decays, leaving 1/8, then 1/16, then 1/32, and so forth. The

decays stop when none of the substance remains, and this time cannot be exactly determined.

8. (e) During each half-life, 1/2 of the substance decays, leaving 1/2 remaining. After the first day, 1/2

remains. After the second day, 1/2 of 1/2, or 1/4, remains. After the third day, 1/2 of 1/4, or 1/8,

remains. Since 1/8 remains, 7/8 of the substance has decayed.

9. (c) A common misconception is that after the second half-life none of the substance remains.

However, during each half-life, 1/2 of the remaining substance decays. After one half-life, 1/2

remains. After two half-lives, 1/4 remains. After three half-lives, 1/8 remains.

10. (a) The decay constant is proportional to the probability of a particle decaying and is inversely

proportional to the half-life. Therefore, the substance with the larger half-life (Tc) will have the

smaller decay constant and smaller probability of decaying. The activity is proportional to the

amount of the substance (number of atoms) and the decay constant, so the activity of Tc will be

smaller than the activity of Sr.


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30-6 Chapter 30



11. (d ) The element with the largest decay constant will have the shortest half-life. Converting each of

the choices to decays per second yields the following: (a) 100/s, (b) 81.6 10 /s,× (c) 102.5 10 /s,×

(d ) 138.6 10 /s.× Answer (d ) has the largest decay rate, so it will have the smallest half-life.

12. (d ) If U-238 only decayed by beta decay, the number of nucleons would not change. Pb-206 has 32nucleons less than U-238, so all decays cannot be beta decays. In each alpha decay, the number

of nucleons decreases by four and the number of protons decreases by two. U-238 has 92 protons

and Pb-206 has 82 protons, so if only alpha decays occurred, then the number of nucleons would

need to decrease by 20. But Pb-206 has 32 fewer nucleons than U-238. Therefore, the sequence

is a combination of alpha and beta decays. Gamma decays are common, since some of the alpha

and beta decays leave the nuclei in excited states where energy needs to be released.

13. (d ) A common misconception is that carbon dating is useful for very long time periods. C-14 has a

half-life about 6000 years. After about 10 half-lives only about 1/1000 of the substance remains.

It is difficult to obtain accurate measurements for longer time periods. For example, after

600,000 years, or about 100 half-lives, only one part in 3010 remains.

14. (b) The half-life of radon remains constant and is not affected by temperature. Radon that existed

several billion years ago will have completely decayed away. Small isotopes, such as carbon-14,

can be created from cosmic rays, but radon is a heavy element. Lightning is not energetic enough

to affect the nucleus of the atoms. Heavy elements such as plutonium and uranium can decay

into radon and thus are the source of present-day radon gas.

15. (d ) The gravitational force between nucleons is much weaker (about 50 orders of magnitude weaker)

than the repulsive Coulomb force; therefore, gravity cannot hold the nucleus together. Neutrons

are electrically neutral and therefore cannot overcome the Coulomb force. Covalent bonds exist

between the electrons in molecules, not among the nucleons. The actual force that holds the

nucleus together is the strong nuclear force, but this was not one of the options.

16. (a) The nature of mass and energy in nuclear physics is often misunderstood. When the neutron and proton are close together, they bind together by releasing mass energy that is equivalent to the

binding energy. This energy comes from a reduction in their mass. Therefore, when the neutron

and proton are far from each other, their net mass is greater than their net mass when they are

bound together.

Solutions to Problems

1. Convert the units from2

MeV/c to atomic mass units.

2

2

1 u(139 MeV/ ) 0.149 u

931.5 Me V/m c

c

⎛ ⎞= =⎜ ⎟⎜ ⎟

⎝ ⎠

2. The α particle is a helium nucleus and has A = 4. Use Eq. 30–1.

1 13 315 15 15(1.2 10 m) (1.2 10 m)(4) 1.9 10 m 1.9 fmr A− − −= × = × = × =


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3. The radii of the two nuclei can be calculated with Eq. 30–1. Take the ratio of the two radii.

1/315 1/3238

15 1/3232

(1.2 10 m)(238) 2381.00855

232(1.2 10 m)(232)

r

r

−

−

× ⎛ ⎞= = =⎜ ⎟

⎝ ⎠×

So the radius of23892U is 0.855% larger than the radius of

23292U.

4. Use Eq. 30–1 for both parts of this problem.

(a) 15 1/3 15 1/3 15(1.2 10 m) (1.2 10 m)(112) 5.8 10 m 5.8 fmr A− − −= × = × = × =

(b)

33 1515 1/3

15 15

3.7 10 m(1.2 10 m) 29.3 29

1 2 10 m 1.2 10 m

r r A A

−−

− −

⎛ ⎞×⎛ ⎞= × → = = = ≈⎜ ⎟⎜ ⎟ ⎜ ⎟

. × ×⎝ ⎠ ⎝ ⎠

5. To find the mass of an α particle, subtract the mass of the two electrons from the mass of a helium atom:

He e

22 2

2

931 5 MeV/(4.002603 u) 2(0 511 MeV/ ) 3727 MeV/

1 u

m m m

cc c

α = −

⎛ ⎞.= − . =⎜ ⎟⎜ ⎟

⎝ ⎠

6. Each particle would exert a force on the other through the Coulomb electrostatic force (given by

Eq. 16–1). The distance between the particles is twice the radius of one of the particles.

( )

29 2 2 19

2 21/3 15

(8.988 10 N m /C ) (2)(1 60 10 C)63.41 N 63 N

(2 ) (2) 4 (1 2 10 m)

Q Q F k

r

α α

α

−

−

⎡ ⎤× ⋅ . ×⎣ ⎦

= = = ≈

⎡ ⎤. ×⎣ ⎦

The acceleration is found from Newton’s second law. We use the mass of a “bare” alpha particle as

calculated in Problem 5.

27 2

272

2

63.41 N9 5 10 m/s

1 6605 10 kg3727 MeV/

931 5 MeV/

F F ma a

mc

c

−= → = = = . ×

⎛ ⎞. ×⎜ ⎟⎜ ⎟

.⎝ ⎠

7. First, we calculate the density of nuclear matter. The mass of a nucleus with mass number A is

approximately ( A u) and its radius is 15 1/3(1.2 10 m) .r A−= × Calculate the density.

27 2717 3

3 15 34 43 3

(1.6605 10 kg/u) (1.6605 10 kg/u)2.294 10 kg/m

(1.2 10 m)

m A A

V r A ρ

π π

− −

−

× ×= = = = ×

×

We see that this is independent of A. The value has 2 significant figures.

(a) We set the density of the Earth equal to the density of nuclear matter.

EarthEarth nuclear 34

matter Earth3

1/31/3

24Earth

Earth 4 17 34nuclear 3 3 matter

5 98 10 kg183.9 m 180 m

(2 294 10 kg/m )

M

R

M R

ρ ρ π

πρ π

= = →

⎛ ⎞⎛ ⎞⎜ ⎟ . ×⎜ ⎟= = = ≈⎜ ⎟⎜ ⎟. ×⎜ ⎟ ⎝ ⎠

⎝ ⎠


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30-8 Chapter 30



(b) Set the density of Earth equal to the density of uranium, and solve for the radius of the uranium.

Then compare that with the actual radius of uranium, using Eq. 30–1, with A = 238.

Earth U Earth U

3 3 3 34 4Earth U Earth U3 3

1/31/3 276 10U

U Earth 24Earth

104

15 1/3

(238 u)(1.6605 10 kg /u)(6.38 10 m) 2.58 10 m

(5.98 10 kg)

2.58 10 m3.5 10

(1.2 10 )(238)

M m M m

R r R r

mr R

M

ρ π π

−−

−

−

= = → = →

⎡ ⎤⎛ ⎞ ×= = × = ×⎢ ⎥⎜ ⎟

×⎢ ⎥⎝ ⎠ ⎣ ⎦

×= ×

×

8. Use Eq. 30–1 to find the value for A. We use uranium-238 since it is the most common isotope.

15 1/33unknown

15 1/3U

(1.2 10 m)0.5 238(0.5) 29.75 30

(1.2 10 m)(238)

r A A

r

−

−

×= = → = = ≈

×

From Appendix B, a stable nucleus with 30 A ≈ is 3115 P .

9. Use conservation of energy. Assume that the centers of the two particles are located a distance from

each other equal to the sum of their radii, and use that distance to calculate the initial electrical PE.

Then we also assume, since the nucleus is much heavier than the alpha, that the alpha has all of the

final KE when the particles are far apart from each other (so they have no PE).

FmKE PE KE PE KEi i f f

0 Fm

19 29 2 2 7

KE1/3 1/3 15 19

7

10 0

4 ( )

(2)(100)(1.60 10 C)(8.99 10 N m /C ) 3.017 10 eV

(4 257 )(1.2 10 m)(1.60 10 J/eV)

3.0 10 eV 30 MeV

q q

r r

α α

α

α

πε

−

− −

+ = + → + = + →+

×= × ⋅ = ×

+ × ×

≈ × =

10. (a) The hydrogen atom is made of a proton and an electron. Use values from Appendix B.

p

H

1 007276 u0.9994553 99.95%

1 007825 u

m

m

.= = ≈

.

(b) Compare the volume of the nucleus to the volume of the atom. The nuclear radius is given by

Eq. 30–1. For the atomic radius we use the Bohr radius, given in Eq. 27–13.

3334 15nucleus 14nucleus 3 nucleus

3 104atom atomatom3

(1.2 10 m)1.2 10

(0.53 10 m)

r V r

V r r

π

π

−−

−

⎡ ⎤⎛ ⎞ ×= = = = ×⎢ ⎥⎜ ⎟

×⎢ ⎥⎝ ⎠ ⎣ ⎦

11. Electron mass is negligible compared to nucleon mass, and one nucleon weighs about 1.0 atomic mass

unit. Therefore, in a 1.0-kg object, we find the following:

2626(1.0 kg)(6.02 10 u/kg) 6.0 10 nucleons

1.0 u/nucleon N

×= ≈ ×

No, it does not matter what the element is, because the mass of one nucleon is essentially the same for

all elements.


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12. The initial kinetic energy of the alpha must be equal to the electrical potential energy when the alpha

just touches the uranium. The distance between the two particles is the sum of their radii.

UKE PE KE PE KEi i f f

U

19 29 2 2 7

KE1/3 1/3 15 19

0 0( )

(2)(92)(1 60 10 C)(8.99 10 N m /C ) 2 852 10 eV

(4 232 )(1.2 10 m)(1.60 10 J/eV)

29 MeV

Q Qk

r r

α α

α

α

−

− −

+ = + → + = + →+

. ×= × ⋅ = . ×

+ × ×

≈

13. The text states that the average binding energy per nucleon between 40 A ≈ and 80 A ≈ is about

8.7 MeV. Multiply this by the number of nucleons in the nucleus.

(63)(8.7 MeV) 548.1MeV 550 MeV= ≈

14. (a) From Fig. 30–1, we see that the average binding energy per nucleon at 238 is 7.5 MeV. A =

Multiply this by the 238 nucleons.

3(238)(7.5 MeV) 1.8 10 MeV= ×

(b) From Fig. 30–1, we see that the average binding energy per nucleon at 84 is 8 7 MeV A = . .

Multiply this by the 84 nucleons.

(84)(8.7 MeV) 730 MeV=

15. 157 N consists of seven protons and eight neutrons. We find the binding energy from the masses of the

components and the mass of the nucleus, from Appendix B.

( ) ( ) ( )1 1 15 21 0 72

2

Binding energy 7 H 8 n N

931.5 MeV/[7(1.007825 u) 8(1.008665 u) (15.000109 u)]

u

115.49 MeV

Binding energy per nucleon (115.49 MeV)/15 7.699 MeV

m m m c

cc

⎡ ⎤= + −⎣ ⎦

⎛ ⎞= + − ⎜ ⎟⎜ ⎟

⎝ ⎠

=

= =

16. Deuterium consists of one proton, one neutron, and one electron. Ordinary hydrogen consists of one

proton and one electron. We use the atomic masses from Appendix B, and the electron masses cancel.

( ) ( ) ( )1 1 2 21 0 12

2

Binding energy H n H

931.5 MeV/[(1 007825 u) (1 008665 u) (2 014102 u)]

u

2 224 MeV

m m m c

cc

⎡ ⎤= + −⎣ ⎦

⎛ ⎞= . + . − . ⎜ ⎟⎜ ⎟

⎝ ⎠

= .


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30-10 Chapter 30



17. We find the binding energy of the last neutron from the masses of the isotopes.

( ) ( ) ( )22 1 23 211 0 112 2

Binding energy Na n Na

[(21 994437 u) (1 008665 u) (22 989769 u)] (931.5 MeV/ )

12.42 MeV

m m m c

c c

⎡ ⎤= + −⎣ ⎦

= . + . − .

=

18. (a) 73 Li consists of three protons and three neutrons. We find the binding energy from the masses,

using hydrogen atoms in place of protons so that we account for the mass of the electrons.

( ) ( ) ( )1 1 7 21 0 32

2

Binding energy 3 H 4 n Li

931 5 MeV/[3(1 007825 u) 4(1 008665 u) (7 016003 u)]

u

39 25 MeV

Binding energy 39 25 MeV

5 607 MeV/nucleonnucleon 7 nucleons

m m m c

cc

⎡ ⎤= + −⎣ ⎦

⎛ ⎞.= . + . − . ⎜ ⎟⎜ ⎟

⎝ ⎠

= .

.= = .

(b)19578Pt consists of 78 protons and 117 neutrons. We find the binding energy as in part (a).

( ) ( ) ( )1 1 195 21 0 782

2

Binding energy 78 H 117 n Pt

931.5 MeV/[78(1.007825 u) 117(1.008665 u) (194.964792 u)]

u

1545.7 MeV 1546 MeV

Binding energy 1545.7 MeV7.927 MeV/nucleon

nucleon 195 nucleons

m m m c

cc

⎡ ⎤= + −⎣ ⎦

⎛ ⎞= + − ⎜ ⎟⎜ ⎟

⎝ ⎠

= ≈

= =

19. 2311 Na consists of 11 protons and 12 neutrons. We find the binding energy from the masses.

( ) ( ) ( )1 1 23 21 0 112

2

Binding energy 11 H 12 n Na

931.5 MeV/[11(1.007825 u) 12(1.008665 u) (22.989769 u)]

u

186.6 MeV


nucleon 23

m m m c

cc

⎡ ⎤= + −⎣ ⎦

⎛ ⎞= + − ⎜ ⎟⎜ ⎟

⎝ ⎠

=

= =

We do a similar calculation for 2411 Na, consisting of 11 protons and 13 neutrons.

( ) ( ) ( )1 1 24 21 0 112

2

Binding energy 11 H 13 n Na

931.5 MeV/[11(1.007825 u) 13(1.008665 u) (23.990963 u)]

u

193.5 MeV


nucleon 24

m m m c

cc

⎡ ⎤= + −⎣ ⎦

⎛ ⎞= + − ⎜ ⎟⎜ ⎟

⎝ ⎠

=

= =


11/30




By this measure, the nucleons in 2311 Na are more tightly bound than those in2411 Na. Thus we expect

2311 Na to be more stable than

2411 Na.

20. We find the required energy by calculating the difference in the masses.

(a) Removal of a proton creates an isotope of carbon. To balance electrons, the proton is included as

a hydrogen atom: 15 1 147 1 6 N H C.→ +

( ) ( ) ( )14 1 15 26 1 72

Energy needed C H N

931 5 MeV/[(14 003242 u) (1 007825 u) (15 000109 u)]

u

10.21MeV

m m m c

c

⎡ ⎤= + −⎣ ⎦

⎛ ⎞.= . + . − . ⎜ ⎟⎜ ⎟

⎝ ⎠=

(b) Removal of a neutron creates another isotope of nitrogen: 15 1 147 0 7 N n N.→ +

( ) ( ) ( )14 1 15 27 0 7

2Energy needed N n N

931.5 MeV/[(14 003074 u) (1.008665 u) (15.000109 u)]

u

10.83 MeV

m m m c

c⎡ ⎤= + −⎣ ⎦

⎛ ⎞= . + − ⎜ ⎟⎜ ⎟

⎝ ⎠=

The nucleons are held by the attractive strong nuclear force. It takes less energy to remove the

proton because there is also the repulsive electric force from the other protons.

21. (a) We find the binding energy from the masses.

( ) ( )4 8 22 42

2

Binding energy 2 He Be

931.5 MeV/[2(4.002603 u) (8.005305 u)]u

0.092 MeV

m m c

cc

⎡ ⎤= −⎣ ⎦

⎛ ⎞= − ⎜ ⎟⎜ ⎟

⎝ ⎠= −

Because the binding energy is negative, the nucleus is unstable. It will be in a lower energy state

as two alphas instead of a beryllium.

(b) We find the binding energy from the masses.

( ) ( )4 12 22 62

2

Binding energy 3 He C

931.5 MeV/[3(4 002603 u) (12 000000 u)] 7.3 MeV

u

m m c

cc

⎡ ⎤= −⎣ ⎦

⎛ ⎞= . − . = +⎜ ⎟⎜ ⎟

⎝ ⎠

Because the binding energy is positive, the nucleus is stable.

22. The wavelength is determined from the energy change between the states.

34 812

13

(6.63 10 J s)(3.00 10 m/s)2.6 10 m

(0 48 MeV)(1 60 10 J/MeV)

c hc E hf h

E λ

λ

−−

−

× ⋅ ×∆ = = → = = = ×

∆ . . ×


12/30

30-12 Chapter 30



23. For the decay 11 10 16 5 1C B p,→ + we find the difference of the initial and the final masses. We use

hydrogen so that the electrons are balanced.

( ) ( ) ( )11 10 16 5 1C B H(11 011434 u) (10 012937 u) (1 007825 u) 0.009328 u

m m m m∆ = − −

= . − . − . = −

Since the final masses are more than the original mass, energy would not be conserved.

24. The decay is 3 3 01 2 1H He e .v−→ + + When we add one electron to both sides to use atomic masses, we

see that the mass of the emitted β particle is included in the atomic mass of 32He. The energy released

is the difference in the masses.

( ) ( )3 3 21 22

2

Energy released H He

931.5 MeV/[(3 016049 u) (3 016029 u)] 0 019 MeV

u

m m c

cc

⎡ ⎤= −⎣ ⎦

⎛ ⎞= . − . = .⎜ ⎟⎜ ⎟

⎝ ⎠

25. The decay is 1 1 00 1 1n p e .v−→ + + The electron mass is accounted for if we use the atomic mass of11H

as a product. If we ignore the recoil of the proton and the neutrino, and any possible mass of the

neutrino, then we get the maximum kinetic energy.

( ) ( )2

1 1 2 2KEmax 0 1

931.5 MeV/n H [(1 008665 u) (1 007825 u)]

u

0.782 MeV

cm m c c

⎛ ⎞⎡ ⎤= − = . − . ⎜ ⎟⎜ ⎟⎣ ⎦

⎝ ⎠

=

26. For each decay, we find the difference of the final masses and the initial mass. If the final mass is more

than the initial mass, then the decay is not possible.

(a) ( ) ( ) ( )232 1 23292 0 92U n U 232 037156 u 1 008665 u 233 039636 u 0 006185 um m m m∆ = + − = . + . − . = .

Because an increase in mass is required, the decay is not possible.

(b) ( ) ( ) ( )13 1 137 0 7 N n N 13 005739 u 1.008665 u 14.003074 u 0 011330 um m m m∆ = + − = . + − = .


(c) ( ) ( ) ( )39 1 4019 0 19K n K 38.963706 u 1.008665 u 39.963998 u 0.008373 um m m m∆ = + − = + − =


27. (a) From Appendix B,2411 Na is a emitter . β

−

(b) The decay reaction is24 2411 12 Na Mg .v β

−→ + + We add 11 electrons to both sides in order to

use atomic masses. Then the mass of the beta is accounted for in the mass of the magnesium.

The maximum kinetic energy of the β − corresponds to the neutrino having no kinetic energy

(a limiting case). We also ignore the recoil of the magnesium.


13/30




( ) ( )24 24 2KE 11 122

2

Na Mg

931.5 MeV/[(23 990963 u) (23 985042 u)] 5.515 MeV

u

m m c

cc

β − ⎡ ⎤= −

⎣ ⎦

⎛ ⎞= . − . =⎜ ⎟⎜ ⎟

⎝ ⎠

28. (a) We find the final nucleus by balancing the mass and charge numbers.

( ) (U) (He) 92 2 90

( ) (U) (He) 238 4 234

Z X Z Z

A X A A

= − = − =

= − = − =

Thus the final nucleus is 23490Th .

(b) If we ignore the recoil of the thorium, then the kinetic energy of the α particle is equal to the

difference in the mass energies of the components of the reaction. The electrons are balanced.

( ) ( ) ( )

( ) ( ) ( )

238 234 4 2KE 92 90 2

KE234 238 490 92 2 2

2 2

U Th He

Th U He

4 20 MeV 1 u238 050788 u 4 002603 u

931 5 MeV/

234.043676 u

m m m c

m m mc

c c

⎡ ⎤= − − →⎣ ⎦

= − −

⎡ ⎤⎛ ⎞.= . − . −⎢ ⎥⎜ ⎟⎜ ⎟

.⎢ ⎥⎝ ⎠⎣ ⎦

=

This answer assumes that the 4.20-MeV value does not limit the significant figures of the

answer.

29. The reaction is 60 6027 28Co Ni .v β −

→ + + The kinetic energy of the β − will be maximum if the

(essentially) massless neutrino has no kinetic energy. We also ignore the recoil of the nickel.

( ) ( )60 60 2KE 27 282

2

Co Ni

931.5 MeV/[(59 933816 u) (59 930786 u)] 2.822 MeV

u

m m c

cc

β ⎡ ⎤= −

⎣ ⎦

⎛ ⎞= . − . =⎜ ⎟⎜ ⎟

⎝ ⎠

30. We add three electron masses to each side of the reaction7 0 74 1 3Be e Li .v−+ → + Then for the mass of

the product side, we may use the atomic mass of73 Li. For the reactant side, including the three

electron masses and the mass of the emitted electron, we may use the atomic mass of 74 Be. The energy

released is the Q-value.

( ) ( )7 7 24 32

2

Be Li

931.5 MeV/[(7 016929 u) (7 016003 u)] 0 863 MeV

u

Q m m c

cc

⎡ ⎤= −⎣ ⎦

⎛ ⎞= . − . = .⎜ ⎟⎜ ⎟

⎝ ⎠


14/30

30-14 Chapter 30



31. For alpha decay we have 218 214 484 82 2Po Pb He.→ + We find the Q-value, which is the energy released.

( ) ( ) ( )218 214 4 284 82 22

2

Po Pb He

931 5 MeV/[218 008973 u 213 999806 u 4 002603 u]u

6.114 MeV

Q m m m c

cc

⎡ ⎤= − −⎣ ⎦

⎛ ⎞.= . − . − . ⎜ ⎟⎜ ⎟

⎝ ⎠

=

For beta decay we have 218 218 084 82 1Po At e .v−→ + + We assume that the neutrino is massless, and find

the Q-value. The original 84 electrons plus the extra electron created in the beta decay means that there

are 85 total electrons on the right side of the reaction, so we can use the mass of the astatine atom and

“automatically” include the mass of the beta decay electron.

( ) ( )218 218 284 852

2

Po At

931 5 MeV/[218 008973 u 218 008695 u] 0 259 MeVu

Q m m c

cc

⎡ ⎤= −⎣ ⎦

⎛ ⎞.= . − . = .⎜ ⎟⎜ ⎟

⎝ ⎠

32. (a) We find the final nucleus by balancing the mass and charge numbers.

( ) (P) (e) 15 ( 1) 16

( ) (P) (e) 32 0 32

Z X Z Z

A X A A

= − = − − =

= − = − =

Thus the final nucleus is 3216 S .

(b) If we ignore the recoil of the sulfur and the energy of the neutrino, then the maximum kinetic

energy of the electron is the Q-value of the reaction. The reaction is 32 3215 16P S .v β −

→ + + We

add 15 electrons to each side of the reaction, and then we may use atomic masses. The mass of

the emitted beta is accounted for in the mass of the sulfur.

( ) ( )

( ) ( )

32 32 2KE 15 16

KE32 3216 15 2 2 2

P S

1.71 MeV 1 uS P (31.973908 u) 31.97207 u

931.5 MeV/

m m c

m mc c c

⎡ ⎤= − →⎣ ⎦

⎡ ⎤= − = − =⎢ ⎥

⎢ ⎥⎣ ⎦

33. We find the energy from the wavelength.

34 8

13 13

(6 63 10 J s)(3.00 10 m/s)10.8 MeV

(1.15 10 m)(1.602 10 J/MeV)

hc E

λ

−

− −

. × ⋅ ×= = =

× ×

This has to be a γ ray from the nucleus rather than a photon from the atom. Electron transitions do not

involve this much energy. Electron transitions involve energies on the order of a few eV.

34. The emitted photon and the recoiling nucleus have the same magnitude of momentum. We find the

recoil energy from the momentum. We assume that the energy is small enough that we can use

classical relationships.


15/30




KEK K K

2 25

KEK 2 2

2K

2

(1 46 MeV)2.86 10 MeV 28.6 eV

2 931.5 MeV/2(39.963998 u) u

E p p m

c

E

m c cc

γ γ

γ −

= = = →

.= = = × =

⎛ ⎞

⎜ ⎟⎜ ⎟⎝ ⎠

35. The kinetic energy of the β + particle will be its maximum if the (almost massless) neutrino has no

kinetic energy. We ignore the recoil of the boron. Note that if the mass of one electron is added to the

mass of the boron, then we may use atomic masses. We also must include the mass of the . β + (See

Problem 36 for details.)

( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

11 11 0 06 5 1 1

11 11 0 0 2 11 11 0 2KE 6 5 1 1 6 5 1

22

C B e

C B e C B 2 e

931 5 MeV/[(11 011434 u) (11 009305 u) 2(0 00054858 u)] 0 9612 MeV

u

v

m m m m c m m m c

cc

β

β

+−

+− −

→ + + +

⎡ ⎤ ⎡ ⎤= − − − = − −⎣ ⎦ ⎣ ⎦

⎛ ⎞.= . − . − . = .⎜ ⎟⎜ ⎟⎝ ⎠

If the β + has no kinetic energy, then the maximum kinetic energy of the neutrino is also

0.9612 MeV . The minimum energy of each is 0, when the other has the maximum.

36. For the positron emission process, 1 N N e . A A Z Z v

+−→ ′ + + We must add Z electrons to the nuclear

mass of N to be able to use the atomic mass, so we must also add Z electrons to the reactant side. On

the reactant side, we use 1 Z − electrons to be able to use the atomic mass of N .′ Thus we have 1

“extra” electron mass and the β particle mass, which means that we must include 2 electron masses

on the right-hand side. We find the Q-value given this constraint.

2 2P D e P D e[ ( 2 )] ( 2 )Q M M m c M M m c= − + = − −

37. We assume that the energies are low enough that we may use classical kinematics. In particular, we

will use KE2 . p m= The decay is238 234 492 90 2U Th He.→ + If the uranium nucleus is at rest when it

decays, then the magnitude of the momentum of the two daughter particles must be the same.

22Th

KE KETh ThTh Th Th Th

2 4 u; (4.20 MeV) 0.0718 MeV

2 2 2 234 u

p m K m p p p

m m m m

α α α α α α

⎛ ⎞= = = = = = =⎜ ⎟

⎝ ⎠

The Q-value is the total kinetic energy produced.

KE KETh 4.20 MeV 0.0718 MeV 4.27 MeVQ α = + = + =


16/30

30-16 Chapter 30



38. (a) The decay constant can be found from the half-life, using Eq. 30–6.

10 1 18 1

91/2

ln 2 ln 21.5 10 yr 4.9 10 s

4.5 10 yr T λ − − − −= = = × = ×

×

(b) The half-life can be found from the decay constant, using Eq. 30–6.

1/2 5 1

ln 2 ln 221, 660 s 6.0 h

3.2 10 sT

λ − −= = = =

×

39. We find the half-life from Eq. 30–5 and Eq. 30–6.

1/ 2

ln 2

0 0 1/2

0

ln 2 ln 2(3.6 h) 1.2 h

140ln ln

1120

t T t R R e R e T t

R

R

λ −= = → = − = − =

2

40. We use Eq. 30–4 and Eq. 30–6 to find the fraction remaining.

1/ 2

(ln 2)(2 5 yr)(12 mo/yr)ln 2

9 mo0

0

0.0992 0.1 10%t

T t t N N N e e e e N

λ λ

⎡ ⎤.−− ⎢ ⎥

− − ⎣ ⎦= → = = = = ≈ ≈

Only 1 significant figure was kept since the Problem said “about” 9 months.

41. The activity at a given time is given by Eq. 30–3b. We also use Eq. 30–6. The half-life is found in

Appendix B.

20 9

71/2

ln 2 ln 2(6.5 10 nuclei) 2.5 10 decays/s

(5730 yr)(3.16 10 s/yr)

dN N N

dt T λ = = = × = ×

×

42. For every half-life, the sample is multiplied by one-half.

( ) ( )5

1 12 2

0

0.03125 1/32 3.125%n N

N = = = = =

43. We need the decay constant and the initial number of nuclei. The half-life is found in Appendix B. Use

Eq. 30–6 to find the decay constant.

7 1

1/2

ln 2 ln 29.99668 10 s

(8.0252 days)(24 h/day)(3600 s/h)T λ − −= = = ×

623 18

0

(782 10 g)

(6.02 10 atoms/mol) 3.5962 10 nuclei(130.906 g/mol) N

−⎡ ⎤×

= × = ×⎢ ⎥⎢ ⎥⎣ ⎦

(a) We use Eq. 30–3b and Eq. 30–4 to evaluate the initial activity.

7 1 18 0 120

12

Activity (9.99668 10 s )(3.5962 10 ) 3.5950 10 decays/s

3.60 10 decays/s

t N N e eλ λ λ − − − −= = = × × = ×

≈ ×


17/30




(b) We evaluate Eq. 30–5 at 1.5 h.t =

7 1 3600s(9 99668 10 s )(1 5 h)12 h

0

12

(3.5950 10 decays/s)

3.58 10 decays/s

t R R e e

λ − − ⎛ ⎞

− . × . ⎜ ⎟− ⎝ ⎠= = ×

≈ ×

(c) We evaluate Eq. 30–5 at t = 3.0 months. We use a time of 1/4 year for the 3.0 months.

7 1 712 (9 99668 10 s )(0 25yr)(3 156 10 s/yr)0

9

(3.5950 10 decays/s)

1.3497 decays/s 1.34 10 decays/s

t R R e eλ − −− − . × . . ×

= = ×

= ≈ ×

44. We find the number of nuclei from the activity of the sample, using Eq. 30–3b and Eq. 30–6. The half-

life is found in Appendix B.

1/29 7

191/2

ln 2

(4.468 10 yr)(3.156 10 s/yr)(420 decays/s) 8.5 10 nuclei

ln 2 ln 2

N N N

t T

T N N

t

λ ∆

= = →∆

∆ × ×= = = ×

∆

45. Each α emission decreases the mass number by 4 and the atomic number by 2. The mass number

changes from 235 to 207, a change of 28. Thus there must be 7 particlesα emitted. With the 7 α

emissions, the atomic number would have changed from 92 to 78. Each β − emission increases the

atomic number by 1, so to have a final atomic number of 82, there must be 4 particles β − emitted.

46. We use the decay constant often, so we calculate it here:1

1/2

ln 2 ln 20.022505 s .

30.8 sT

λ −= = =

(a) We find the initial number of nuclei from an estimate of the atomic mass.

623 16 16

0

(8.7 10 g)(6.02 10 atoms/mol) 4.223 10 4.2 10 nuclei

(124 g/mol) N

−×

= × = × ≈ ×

(b) Evaluate Eq. 30–4 at 2.6 min.t =

116 (0 022505 s )(2 6 min)(60 s/ min) 15 15

0 (4.223 10 ) 1.262 10 1.3 10 nucleit

N N e eλ −− − . .

= = × = × ≈ ×

(c) The activity is found from using the absolute value of Eq. 30–3b.

1 15 13 13(0.022505 s )(1.262 10 ) 2.840 10 decays/s 2.8 10 decays/s N λ −= × = × ≈ ×

(d ) We find the time from Eq. 30–4.

0

1 160

1

1 decay/slnln

(0 022505s )(4 223 10 ) decays/s1532 s 26 min

0.022505 s

t N N e

N

N t

λ λ λ

λ

λ

λ

−

−

−

= →

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟

. . ×⎢ ⎥⎝ ⎠ ⎣ ⎦= − = − = ≈


18/30

30-18 Chapter 30



47. Find the initial number of nuclei from the initial decay rate (activity) and then the mass from the

number of nuclei.

50 0

1/2

9 75 1 5 1 221/20

220 23

ln 2initial decay rate 2.0 10 decays/s

(1.248 10 yr)(3.156 10 s/yr )(2.4 10 s ) (2.4 10 s ) 1.364 10 nucleiln 2 ln 2

(atomic weight) g/mol (39.(1.364 10 nuclei)

6.02 10 nuclei/mol

N N T

T N

m N

λ

− −

= × = = →

× ×= × = × = ×

= = ××

23

963998 g)0.91g

(6.02 10 )=

×

48. The number of nuclei is found from the mass and the atomic weight. The activity is then found from

the number of nuclei and the half-life.

623 17

17 10

61/2

(6.7 10 g)(6.02 10 atoms/mol) 1.261 10 nuclei

(31.9739 g/mol)

ln 2 ln 2

(1.261 10 ) 7.1 10 decays/s(1.23 10 s)

N

N N T λ

−⎡ ⎤×= × = ×⎢ ⎥

⎢ ⎥⎣ ⎦

⎡ ⎤= = × = ×

⎢ ⎥×⎢ ⎥⎣ ⎦

49. (a) The decay constant is found from Eq. 30–6.

13 1 13 1

5 71/2

ln 2 ln 21.381 10 s 1.38 10 s

(1.59 10 yr)(3.156 10 s /yr)T λ − − − −= = = × ≈ ×

× ×

(b) The activity is the decay constant times the number of nuclei.

13 1 18 5

7

60 s(1.381 10 s )(4.50 10 ) 6.215 10 decays/s

1min

3.73 10 decays/min

N λ − − ⎛ ⎞

= × × = × ⎜ ⎟⎝ ⎠

= ×

50. We use Eq. 30–5.

( )

( ) ( )

1 1 10 06 6 6

1/2

1/2 1 16 6

ln 2ln

ln 2 ln 2(9.4 min) 3.6 min

ln ln

t t R R R e e t t T

T t

λ λ λ − −= = → = → = − = →

= = − =

2

2

51. Because the fraction of atoms that are 146C is so small, we use the atomic weight of126C to find the

number of carbon atoms in the sample. The activity is found from Eq. 30–3b.

14 126 6

23 13

12 12

13

71/2

1.3 1.3 (345 g)(6.02 10 nuclei/mol) 2.250 10 nucleiC C (12 g/mol)10 10

ln 2 ln 2(2.250 10 ) 86 decays/s

(5730 yr)(3.156 10 s/yr)

N N

N N T

λ

⎡ ⎤⎛ ⎞ ⎛ ⎞= = × = ×⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎣ ⎦

⎡ ⎤= = × =⎢ ⎥

×⎢ ⎥⎣ ⎦


19/30




52. We first find the number of nuclei from the activity and then find the mass from the number of nuclei

and the atomic weight. The half-life is found in Appendix B.

2

1/2

9 7 2 19

ln 2Activity 4.20 10 decays/s

(4.468 10 yr)(3.156 10 s/yr)(4.20 10 decays /s)8.544 10 nuclei

ln 2

N N T

N

λ = = = × →

× × ×= = ×

192

23

(8.544 10 nuclei)(238.05 g/mol) 3.38 10 g 33.8 mg

(6.02 10 atoms/mol)m

−⎡ ⎤×

= = × =⎢ ⎥×⎢ ⎥⎣ ⎦

53. We assume that the elapsed time is much smaller than the half-life, so we can approximate the decay

rate as being constant. We also assume that the8738Sr is stable, and there was none present when the

rocks were formed. Thus every atom of 8737 Rb that decayed is now an atom of8738Sr.

109Sr 1/2

Sr Rb Rb

Rb

4.75 10 yr (0.0260) 1.78 10 yr

ln 2 ln 2

N T N N N t t

N λ

×= −∆ = ∆ → ∆ = = = ×

This is 4%≈ of the half-life, so our original assumption is valid.

54. We are not including the neutrinos that will be emitted during beta decay.

First sequence: 232 228 490 88 2Th Ra ;α → + 228 22888 89Ra Ac ; β

−→ + 228 22889 90Ac Th ; β

−→ +

228 224 490 88 2Th Ra ;α → +

224 220 488 86 2Ra Rn α → +

Second sequence:235 231 492 90 2U Th ;α → +

231 23190 91Th Pa ; β

−→ +

231 227 491 89 2Pa Ac ;α → +

227 22789 90Ac Th ; β

−→ + 227 223 490 88 2Th Ra α → +

55. Because the fraction of atoms that are14

6C is so small, we use the atomic weight of126C to find the

number of carbon atoms in 73 g. We then use the given ratio to find the number of 146C atoms present

when the club was made. Finally, we use the activity as given in Eq. 30–5 to find the age of the club.

( ) ( )

126

146

14 146 6

23 24

12 24 12

today 0

(73 g)(6.02 10 atoms/mol) 3.662 10 atomsC (12 g/mol)

(1.3 10 )(3.662 10 ) 4.761 10 nucleiC

C Ct

N

N

N N e λ λ λ

−

−

⎡ ⎤= × = ×⎢ ⎥

⎣ ⎦

= × × = ×

= →

( )

( )

( )14 146 6

141466

today today1/2

01/2 0

12

70

C C1ln ln

ln 2 ln 2C C

5730 yr (7.0 decays/s) ln 7921 yr 7900 yr

ln 2 ln 2(4.761 10 nuclei)

(5730 yr)(3.156 10 s/yr)

N N T

t N N

T

λ λ

λ λ = − = −

⎛ ⎞⎜ ⎟⎝ ⎠

= − = ≈⎡ ⎤

×⎢ ⎥×⎢ ⎥⎣ ⎦


20/30

30-20 Chapter 30



56. The decay rate is given by Eq. 30–3b, . N

N t

λ ∆

= −∆

We assume equal numbers of nuclei decaying by

α emission.

1/2 4214 7218 218 218 218

214 214 214 1/2218

214

(1.6 10 s)8.6 10

(3.1 min)(60 s/ min)

N

T t N

N N T

t

λ λ

λ λ

−−

∆⎛ ⎞

⎜ ⎟∆ − ×⎝ ⎠= = = = = ×

∆ −⎛ ⎞⎜ ⎟

∆⎝ ⎠

57. The activity is given by Eq. 30–5. The original activity is 0 , N λ so the activity 31.0 hours later is

00.945 . N λ

0 01/2

1/2

ln 20.945 ln 0.945

ln 2 1 d(31.0 h) 379.84 h 15.8 d

ln 0.945 24 h

t N N e t t T

T

λ λ λ λ −= → = − = − →

⎛ ⎞= − = =⎜ ⎟

⎝ ⎠

58. The activity is given by Eq. 30–5.

(a)21/2

00 0

1 53 d 25 decays/sln ln ln 201.79 d 2.0 10 d

ln 2 ln 2 350 decays/s

t T R R R R e t

R R

λ

λ − ⎛ ⎞

= → = − = − = − = ≈ ×⎜ ⎟⎝ ⎠

(b) We find the mass from the activity. Note that A N is used to represent Avogadro’s number, and

A is the atomic weight.

0 A0 0

1/2

140 1/2

23A

ln 2

(350 decays/s)(53 d)(86,400 s/d)(7.017 g/mole)2.7 10 g

ln 2 (6.02 10 nuclei/mole) ln 2

m N R N

T A

R T Am

N

λ

−

= = →

= = = ××

59. The number of radioactive nuclei decreases exponentially, and every radioactive nucleus that decays

becomes a daughter nucleus.

0 D 0 0; (1 )t t N N e N N N N e

λ λ − −= = − = −

60. The activity is given by Eq. 30–5, with 00 01050 . R R= .

00 1/2

1/2 0

ln / ln 2 ln 2 (4.00 h) ln 20 6085 h 36 5 min

ln / ln 0.01050

t R R t R R e T t T R R

λ λ −= → = − = → = − = − = . = .

From Appendix B we see that the isotope is

211

82Pb.

61. Because the carbon is being replenished in living trees, we assume that the amount of 146C is constant

until the wood is cut, and then it decays. Use Eq. 30–4 and Eq. 30–7 to find the age of the tool.

1/ 2

0 693

41/20 0 0

ln(0.045) (5730 yr) ln (0.045)0.045 2.6 10 yr

0.693 0.693

t

T t T N N e N e N t λ

.−

−= = = → = − = − = ×


21/30




62. (a) The mass number is found from the radius, using Eq. 30–1.

3 3

15 1/3 55 55

15 15

5000 m(1 2 10 m) 7.23 10 7 10

1.2 10 m 1.2 10 m

r r A A−

− −

⎛ ⎞ ⎛ ⎞= . × → = = = × ≈ ×⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

× ×⎝ ⎠ ⎝ ⎠

(b) The mass of the neutron star is the mass number times the atomic mass unit conversion in kg.

27 55 27 29 29(1 66 10 kg/u) (7.23 10 u)(1.66 10 kg/u) 1.20 10 kg 1 10 kgm A − −= . × = × × = × ≈ ×

Note that this is about 6% of the mass of the Sun.

(c) The acceleration of gravity on the surface of the neutron star is found from Eq. 5–5 applied to

the neutron star.

11 2 2 2911 2 11 2

2 2

(6.67 10 N m /kg )(1.20 10 kg)3.20 10 m/s 3 10 m/s

(5000 m)

Gm g

r

−× ⋅ ×

= = = × ≈ ×

63. Because the tritium in water is being replenished, we assume that the amount is constant until the wine

is made, and then it decays. We use Eq. 30–4.

0 1/20

1/2 0

lnln 2 (12.3 yr) ln 0.10

ln 41 yr ln 2 ln 2

t

N

N T N N N e t

t T N

λ λ −= → = − = → = − = − =

64. (a) We assume a mass of 70 kg of water and find the number of protons, given that there are

10 protons in a water molecule.

323

protons

28

(70 10 g water) molecules water 10 protons6 02 10

(18 g water/mol water) mol water water molecule

2.34 10 protons

N ⎡ ⎤× ⎛ ⎞⎛ ⎞

= . ×⎢ ⎥ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦

= ×

We assume that the time is much less than the half-life so that the rate of decay is constant.

1/2

3341/2

28

ln 2

1 proton 10 yr 6.165 10 yr 60,000 yr

ln 2 ln 22.34 10 protons

N N N

t T

T N t

N

λ ⎛ ⎞∆

= = →⎜ ⎟∆ ⎝ ⎠

⎛ ⎞⎛ ⎞∆∆ = = = × ≈⎜ ⎟⎜ ⎟ ⎜ ⎟

×⎝ ⎠ ⎝ ⎠

This is about 880 times a normal life expectancy.

(b) Instead of waiting 61,650 consecutive years for one person to experience a proton decay,

we could interpret that number as there being 880 people, each living 70 years, to make that

61,650 years (since 880 70 61, 650).× ≈ We would then expect one person out of every 880 to

experience a proton decay during their lifetime. Divide 7 billion by 880 to find out how many

people on Earth would experience proton decay during their lifetime.

967 10 7.95 10 8 million people

880

×= × ≈


22/30

30-22 Chapter 30



65. We assume that all of the kinetic energy of the alpha particle becomes electrostatic potential energy at

the distance of closest approach. Note that the distance found is the distance from the center of the

alpha to the center of the gold nucleus.

AuKE PE KE PE KE

i i f f

19 29 2 2 14Au

13KE

14

0 0

(2)(79)(1.60 10 C)(8.988 10 N m /C ) 2.951 10 m

(7.7 MeV)(1.60 10 J/MeV)

3.0 10 m

Q Qk

r

Q Qr k

α

α

α

α

−−

−

−

+ = + → + = + →

×= = × ⋅ = ×

×

≈ ×

We use Eq. 30–1 to compare to the size of the gold nucleus.

14approach

1/3 15Au

2.951 10 m4.2

197 (1.2 10 m)

r

r

−

−

×= =

×

So the distance of approach is about 4.2 × the radius of the gold nucleus.

66. Use Eq. 30–5 and Eq. 30–6.

1/ 2

ln 2

0 0 01/2

ln (0 0200)0.0200 5.64

ln 2

t

T t t R R e R e RT

λ −

− .= = = → = − =

It takes 5.64 half-lives for a sample to drop to 2.00% of its original activity.

67. The number of 4019K nuclei can be calculated from the activity, using Eq. 30–3b and Eq. 30–5. The

half-life is found in Appendix B. A subscript is used on the variable N to indicate the isotope.

7

9

181/240 40

decays 3.156 10 s42 (1.248 10 yr)s 1 yr

2.387 10 nucleiln 2 ln 2

RT R R N N λ

λ

⎛ ⎞×⎛ ⎞× ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= → = = = = ×

The mass of 4019K in the milk is found from the atomic mass and the number of nuclei.

18 27 740 (2.387 10 )(39.964 u)(1.66 10 kg/u) 1.583 10 kg 0.16 mgm

− −= × × = × ≈

From Appendix B, in a natural sample of potassium, 0.0117% is 4019K and 93.2581% is3919 K. Find the

number of 3919 K nuclei and then use the atomic mass to find the mass of3919 K in the milk.

40 total 39 total

18 2239 40

(0.0117%) ; (93 2581%)

(93.2581%) (93.2581%)(2.387 10 nuclei) 1.903 10 nuclei

(0.0117%) (0.0117%)

N N N N

N N

= = . →

⎡ ⎤ ⎡ ⎤= = × = ×⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

The mass of 3919 K is the number of nuclei times the atomic mass.

22 27 339 (1.903 10 )(38.964 u)(1.66 10 kg/u) 1.231 10 kg 1.2 gm

− −= × × = × ≈


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68. In the Periodic Table, Sr is in the same column as Ca. If Sr is ingested, then the body may treat it

chemically as if it were Ca, which means it might be stored by the body in bones. Use Eq. 30–4 to find

the time to reach a 1% level.

0 1/20

1/2 0

lnln 2 (29 yr) ln 0.01ln 192.67 yr 200 yr

ln 2 ln 2

t

N

N T N N N e t t T N

λ λ −= → = − = → = = − = ≈2

Assume that both the Sr and its daughter undergo beta decay, since the Sr has too many neutrons.

90 90 0 90 90 038 39 1 39 40 1Sr Y e ; Y Zr eν ν − −→ + + → + +

69. The number of nuclei is found from Eq. 30–5 and Eq. 30–3b. The mass is then found from the number

of nuclei and the atomic weight. The half-life is given.

1/2 1/2

1/2

4 27

14 11

ln 2; (atomic weight) (atomic weight)

ln 2 ln 2

24 h 3600 s(87.37 d)1 d 1 h

(4.28 10 decays/s)(34.969 u)(1.66 10 kg/u)ln 2

2.71 10 kg 2.71 10 g

T T R N N N R m N R

T

m

λ

−

− −

= = → = = =

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= × ×

= × = ×

70. (a) We find the daughter nucleus by balancing the mass and charge numbers.

( ) (Os) (e ) 76 ( 1) 77 Z X Z Z −

= − = − − =

( ) (Os) (e ) 191 0 191 A X A A −

= − = − =

The daughter nucleus is 19177 Ir .

(b) See the diagram.

(c) Because there is only one β energy, the β decay must be to the higher excited state.

71. (a) The number of nuclei is found from the mass of the sample and the atomic mass. The activity is

found from the half-life and the number of nuclei, using Eq. 30–3b and Eq. 30–5.

23 21

21 15

4

1.0 g(6.02 10 nuclei/mol) 4.599 10 nuclei

130.91 g/mol

0.693

(4.599 10 ) 4.6 10 decays/s(8.02 d)(8.64 10 s/d)

N

R N λ

⎛ ⎞= × = ×⎜ ⎟

⎝ ⎠

⎡ ⎤= = × = ×

⎢ ⎥×⎢ ⎥⎣ ⎦

(b) Follow the same procedure as in part (a).

23 21

21 4

9 7

1.0 g(6.02 10 nuclei/mol) 2.529 10 nuclei

238.05 g/mol

0.693(2.529 10 ) 1.2 10 decays/s

(4.47 10 yr)(3.16 10 s/yr)

N

R N λ

⎛ ⎞= × = ×⎜ ⎟

⎝ ⎠

⎡ ⎤= = × = ×⎢ ⎥

× ×⎢ ⎥⎣ ⎦

19176Os

191

77

Ir*γ (0.042 MeV)

γ (0.129 MeV)

β – (0.14 MeV)

19177 Ir

19177 Ir*


24/30

30-24 Chapter 30



72. From Fig. 30–1, the average binding energy per nucleon at 63 A = is ~8.6 MeV. Use the average

atomic weight as the average number of nucleons for the two stable isotopes of copper to find the total

binding energy for one copper atom.

(63 5 nucleons)(8.6 MeV/nucleon) 546.1MeV 550 MeV. = ≈

Convert the mass of the penny to a number of atoms and then use the above value to calculate the

energy needed.

23 6 19

12

Energy ( atoms)(energy/atom)

(3.0g)(6.02 10 atoms/mol)(546.1 10 eV/atom)(1.60 10 J/eV)

(63.5 g/mol)

2.5 10 J

−

= #

⎡ ⎤= × × ×⎢ ⎥

⎣ ⎦

= ×

73. (a)

( ) ( ) ( )

4 4 42 2 2

2 2

He He He 4.002603 u 4 0.002603 u

(0.002603 u)(931.5 MeV/u ) 2.425 MeV/

m A

c c

∆ = − = − =

= =

(b) ( ) ( ) ( )12 12 126 6 6C C C 12.000000 u 12 0m A∆ = − = − =

(c) ( ) ( ) ( )86 86 8638 38 382 2

Sr Sr Sr 85.909261 u 86 0.090739 u

( 0.090739 u)(931.5 MeV/u ) 84.52 MeV/

m A

c c

∆ = − = − = −

= − = −

(d ) ( ) ( ) ( )235 235 23592 92 922 2

U U U 235.043930 u 235 0.043930 u

(0.043930 u)(931 5 MeV/u ) 40.92 MeV/

m A

c c

∆ = − = − =

= . =

(e) From Appendix B, we see the following:

0 for 0 8 and 85;

0 for 9 84

Z Z

Z

∆ ≥ ≤ ≤ ≥

∆ < ≤ ≤

0 for 0 15 and 218;

0 for 16 218

A A

A

∆ ≥ ≤ ≤ ≥

∆ < ≤ <

74. The reaction is 1 1 21 0 1H n H.+ → If we assume that the initial kinetic energies are small, then the energy

of the gamma is the Q-value of the reaction.

( ) ( ) ( )1 1 2 21 0 12 2

H n H

[(1.007825 u) (1 008665 u) (2 014102 u)] (931.5 MeV/u ) 2.224 MeV

Q m m m c

c c

⎡ ⎤= + −⎣ ⎦

= + . − . =


25/30


26/30

30-26 Chapter 30



78. (a) From Section 30–11, the usual fraction of 146C is121.3 10 .

−× Because the fraction of atoms that

are 146C is so small, use the atomic weight of126C to find the number of carbon atoms in 72 g.

Use Eq. 30–4 to find the time.

23 2412

24 12 1214

51/20 12

0 0

72 g(6.02 10 atoms/mol) 3.612 10 atoms

12 g/mol

(3.612 10 atoms)(1.3 10 ) 4.6956 10 atoms

1 (5730 yr) 1ln ln ln 2.4 10 yr

ln 2 ln 2 4.6956 10

t

N

N

T N N N N e t

N N

λ

λ

−

−

⎛ ⎞= × = ×⎜ ⎟⎝ ⎠

= × × = ×

= → = − = − = − = ××

(b) Do a similar calculation for an initial mass of 340 g.

23 12 1314

51/2

0 130 0

340 g(6.02 10 atoms/mol)(1.3 10 ) 2.217 10 atoms

12 g/mol

1 (5730 yr) 1ln ln ln 2.5 10 yr

ln 2 ln 2 2.217 10

t

N

T N N N N e t

N N

λ

λ

−

−

⎛ ⎞= × × = ×⎜ ⎟

⎝ ⎠

= → = − = − = − = ×

×

For times on the order of 510 yr, the sample amount has fairly little effect on the age

determined. Thus, times of this magnitude are not accurately measured by carbon dating.

79. (a) This reaction would turn the protons and electrons in atoms into neutrons. This would eliminate

chemical reactions and thus eliminate life as we know it.

(b) We assume that there is no kinetic energy brought into the reaction and solve for the increase of

mass necessary to make the reaction energetically possible. For calculating energies, we write

the reaction as 1 11 0H n ,ν → + and we assume that the neutrino has no mass or kinetic energy.

( ) ( )1 1 2 2 2

1 0H n [(1 007825 u) (1 008665 u)] (931.5 MeV/u )0.782 MeV

Q m m c c c

⎡ ⎤= − = . − .⎣ ⎦

= −

This is the amount that the proton would have to increase in order to make this energetically

possible. We find the percentage change.

2

2

(0.782 MeV/ )(100) (100) 0.083%

(938 27 MeV/ )

m c

m c

⎡ ⎤∆⎛ ⎞= =⎢ ⎥⎜ ⎟

⎝ ⎠ .⎢ ⎥⎣ ⎦

80. We assume that the particles are not relativistic, so that KE2 . p m= The radius is given in

Example 20–6 as .m

r

qB

υ = Set the radii of the two particles equal. Note that the charge of the alpha

particle is twice that of the electron (in absolute value). We also use the “bare” alpha particle mass,

subtracting the two electrons from the helium atomic mass.

2 22

mmm m p p

eB eB

β β α α α α β β α β

υ υ υ υ = → = → =


27/30




22

KE 4

2 2KE

4

42 2 4(0 000549 u)5.48 10

4 002603 u 2(0 000549 u)

2 2

p p

mm m

m p p

m m

β α

β α α α

β α β β

β β

−.= = = = = ×

. − .

81. Natural samarium has an atomic mass of 150.36 grams per mole. We find the number of nuclei in the

natural sample and then take 15% of that to find the number of 14762Sm nuclei. We first find the number

of 147Sm nuclei from the mass and proportion information.

14762Sm

2320 147

natural 62

(0.15)(1.00 g)(6.02 10 nuclei/mol)(0.15) 6.006 10 nuclei of Sm

150.36 g/mol N N

×= = = ×

The activity level is used to calculate the half-life.

1/2

20 18 111/2 7

ln 2Activity

ln 2 ln 2 1 yr (6.006 10 ) 3.469 10 s 1.1 10 yr

120 decays/s 3.156 10 s

R N N

T

T N R

λ = = = →

⎛ ⎞= = × = × = ×⎜ ⎟⎜ ⎟

×⎝ ⎠

82. Since amounts are not specified, assume that “today” there is 0.720 g of 23592U in our sample, and

100.000 0.720 99.280 g− = of23892U. Use Eq. 30–4.

(a) Relate the amounts today to the amounts 91.0 10× years ago.

1/ 2

9

81/ 2

9

91/ 2

ln 2

0 0

(1 0 10 yr) ln 2ln 2(7 04 10 yr)

0 235 235

(1 0 10 )ln 2ln 2

(4 468 10 )0 238 238

( ) ( ) (0720 g) 1.927 g

( ) ( ) (99.280 g) 115.94 g

t

T t t

t T

t

T

N N e N Ne Ne

N N e e

N N e e

λ λ −

. ×

. ×

. ×

. ×

= → = =

= = =

= = =

The percentage of 23592U was1.927

100% 1 63% .1.927 115.94

× = .+

(b) Relate the amounts today to the amounts 6100 10× years from now.

6

81/ 2

6

91/ 2

(100 10 yr)ln 2ln 2

(7 04 10 yr)0 235 0 235

(100 10 yr)ln 2ln 2

(4 468 10 yr)238 0 238

( ) ( ) (0.720 g) 0.6525 g

( ) ( ) (99.280 g) 97.752 g

t

T t

t

T

N N e N N e e

N N e e

λ

×−−

− . ×

×−−

. ×

= → = = =

= = =

The percentage of 23592U will be0.6525

100% 0.663% .0.6525 97.752

× =+


28/30

30-28 Chapter 30



83. We determine the number of4019K nuclei in the sample and then use the half-life to determine the

activity.

40

19

3 2317

naturallyoccurring K

(420 10 g)(6.02 10 atoms/mol)(0.000117) (0.000117) 7.566 10

K 39.0983 g/mol

N N −

× ×= = = ×

17

9 71/2

ln 2 ln 2 1 yr (7.566 10 ) 13 14 decays/s 13 decays/s

1.265 10 yr 3.156 10 s R N N

T λ

⎛ ⎞= = = × = . ≈⎜ ⎟

× ×⎝ ⎠

84. The kinetic energy of the electron will be at its maximum if the (essentially) massless neutrino has no

kinetic energy. We also ignore the recoil energy of the sodium. The maximum kinetic energy of the

reaction is then the Q-value of the reaction. Note that the emitted electron mass is accounted for by

using atomic masses.

( ) ( )23 23 2 2 2KE 10 11 Ne Na [(22.9947 u) (22 9898 u)] (931.5 MeV/u )

4.6 MeV

Q m m c c c⎡ ⎤= = − = − .⎣ ⎦

=

If the neutrino were to have all of the kinetic energy, then the minimum kinetic energy of the electron

is 0 . The sum of the kinetic energy of the electron and the energy of the neutrino must be the Q-value,

so the neutrino energies are 0 and 4.6 MeV, respectively.

85. (a) If the initial nucleus is at rest when it decays, then momentum conservation says that the

magnitude of the momentum of the alpha particle will be equal to the magnitude of the

momentum of the daughter particle. We use that to calculate the (nonrelativistic) kinetic energy

of the daughter particle. The mass of each particle is essentially equal to its atomic mass number,

in atomic mass units. Note that classically, KE2 . p m=

22KE

DKE KE KE KED DD D D D D D

KEKEKED D

1KE KE DD D4KE KEKE KE

D

2 4; 2 2 2

4

1

14

4

p m m A p p p m m m m A A

A

A A

A

α α α α α α α α α

α α

α α α α α

= = = = = = =

= = =+ +⎡ ⎤ ⎡ ⎤

++⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

(b) We specifically consider the decay of 22688Ra. The daughter has D 222 A = .

KED

1 1KE KED D4 4

1 10.017699 1.8%

1 1 (222) Aα = = = ≈

+ + +

Thus, the alpha particle carries away 1 0.0177 0.9823 98.2% .− = =

86. The mass number changes only with α decay, and it changes by 4− . If the mass number is 4n, then

the new number is 4 4 4( 1) 4 .n n n− = − = ′ There is a similar result for each family, as shown here.

4 4 4 4( 1) 4n n n n→ − = − = ′

4 1 4 4 1 4( 1) 1 4 1n n n n+ → − + = − + = ′ +


29/30




4 2 4 4 2 4( 1) 2 4 2n n n n+ → − + = − + = ′ +

4 3 4 4 3 4( 1) 3 4 3n n n n+ → − + = − + = ′ +

Thus, the daughter nuclides are always in the same family.

Solutions to Search and Learn Problems

1. The nucleus of an atom consists of protons (which carry a positive electric charge) and neutrons

(which are electrically neutral). The electric force between protons is repulsive and much larger than

the force of gravity. If the electric and gravitational forces are the only two forces present in the

nucleus, then the nucleus would be unstable, as the electric force would push the protons away from

each other. Nuclei are stable; therefore, another force must be present in the nucleus to overcome the

electric force. This force is the strong nuclear force.

2. (a) An einsteinium nucleus has 99 protons and a fermium nucleus has 100 protons. If the fermium

undergoes either electron capture or β + decay, then a proton would in effect be converted into a

neutron. The nucleus would now have 99 protons and be an einsteinium nucleus.

(b) If the einsteinium undergoes β − decay, then a neutron would be converted into a proton. The

nucleus would now have 100 protons and be a fermium nucleus.

3. Take the momentum of the nucleon to be equal to the uncertainty in the momentum of the nucleon, as

given by the uncertainty principle. The uncertainty in position is estimated as the radius of the nucleus.

With that momentum, calculate the kinetic energy, using the classical formula. Iron has about 56

nucleons (depending on the isotope).

2 2 34 2

KE2 2

27 1/3 15 13(1.055 10 J s)

2 2 2(1.67 10 kg) (56 )(1.2 10 m) (1.60 10 J/MeV)

0.988 MeV 1 MeV

p x p p x r

pm mr

−

− − −

∆ ∆ ≈ → ≈ ∆ ≈ =∆

× ⋅= = =

⎡ ⎤× × ×⎣ ⎦

= ≈

4. (a) From Fig. 30–17, the initial force on the detected particle is down. Using the right-hand rule, the

force on a positive particle would be upward. Thus the particle must be negative, and the decay

is β − decay.

(b) The magnetic force is producing circular motion. Set the expression for the magnetic force equal

to the expression for centripetal force and solve for the velocity.

2 19 3

631

(1 60 10 C)(0.012 T)(4.7 10 m)9.9 10 m/s

9.11 10 kg

m qBr q B

r m

υ υ υ

− −

−

. × ×

= → = = = ××

5. In β decay, an electron is ejected from the nucleus of the atom, and a neutron is converted into a

proton. The atomic number of the nucleus increases by one, and the element now has different

chemical properties. In internal conversion, an orbital electron is ejected from the atom. This does not

change either the atomic number of the nucleus or its chemical properties.


30/30

30-30 Chapter 30

Also, in β decay, both a neutrino and an electron will be emitted from the nucleus. Because there are

three decay products (the neutrino, the β particle, and the nucleus), the momentum of the β particle

can have a range of values. In internal conversion, since there are only two decay products (the

electron and the nucleus), the electron will have a unique momentum and, therefore, a unique energy.

6. If the Earth had been bombarded with additional radiation several thousand years ago, then there

would have been a larger abundance of carbon-14 in the atmosphere at that time. Organisms that died

in that time period would have had a greater percentage of carbon-14 in them than organisms that die

today. Since we assumed the amount of starting carbon-14 was the same, we would have previously

underestimated the age of the organisms. With the new discovery, we would re-evaluate the organisms

as older than previously calculated.

7. (a) The decay constant is calculated from the half-life using Eq. 30–6.

12 1

71/2

ln 2 ln 23.83 10 s

(5730 yr)(3.156 10 s/yr)T λ − −= = = ×

×

(b) In a living organism, the abundance of 146C atoms is 121.3 10−× per carbon atom. Multiply this

abundance by Avogadro’s number and divide by the molar mass of carbon to find the number of

carbon-14 atoms per gram of carbon.

14 14 23 1212 126 6 6A

12 12 126 6 6

14 12 10 14 126 6 6 6

C atoms C atoms (6.02 10 atoms C/mol)1.3 10 1.3 10

C atoms C atoms 12.0107 g C/mol

6.516 atoms C/g C 6.5 10 atoms C/ g C

N N

M

− −⎛ ⎞ ⎛ ⎞ ×

=

Ch30 Giancoli7e Manual

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