CHAPTER 24 NUCLEAR REACTIONS AND

THEIR APPLICATIONS

FOLLOW–UP PROBLEMS

24.1A

Plan: Write a skeleton equation that shows fluorine-20 undergoing beta decay. Conserve mass and atomic number

by ensuring the superscripts and subscripts equal one another on both sides of the equation. Determine the identity

of the daughter nuclide by using the periodic table.

Solution:

Fluorine-20 has Z = 9. Its symbol is 209F. When it undergoes beta decay, a beta particle, 10β , and a daughter

nuclide, AZX, are produced.

20

A

0

9F ZX + 1β

To conserve atomic number, Z must equal 10. Element is neon.

To conserve mass number, A must equal 20.

The identity of

A

ZX

is 20

10Ne.

The balanced equation is:

24.1B

20

20

9F 10Ne

+

0

1β

Plan: Write a skeleton equation that shows an unknown nuclide,

A

ZX,

undergoing beta decay,

0

1β

, to form

133

55 Cs .

cesium–133,

Conserve mass and atomic number by ensuring the superscripts and subscripts equal one

another on both sides of the equation. Determine the identity of X by using the periodic table to identify the

element with atomic number equal to Z.

Solution:

The unknown nuclide yields cesium-133 and a particle:

A

133

Z X 55 Cs

0

+ 1β

To conserve atomic number, Z must equal 54. Element is xenon.

To conserve mass number, A must equal 133.

133

54 Xe .

133

equation is: 54 Xe

The identity of

A

ZX

is

133

0

The balanced

55 Cs + -1 β

Check: A = 133 = 133 + 0 and Z = 54 = 55 + (–1).

24.2A

Plan: Look at the N/Z ratio, the ratio of the number of neutrons to the number of protons. If the N/Z ratio falls in

the band of stability, the nuclide is predicted to be stable. For stable nuclides of elements with atomic number

greater than 20, the ratio of number of neutrons to number of protons (N/Z) is greater than one. In addition, the

ratio increases gradually as atomic number increases. Also check for exceptionally stable numbers of neutrons

and/or protons – the “magic” numbers of 2, 8, 20, 28, 50, 82, and (N = 126). Also, even numbers of protons and

or neutrons are related to stability whereas odd numbers are related to instability.

Solution:

a) 105B appears stable because its N/Z ratio (10 5)/5 = 1.00 is in the band of stability.

b) 58

23V appears unstable/radioactive because its N/Z ratio (58 23)/23 = 1.52 is too high and is above the band of

stability. Additionally, this nuclide has both odd N(35) and Z(23).

24.2B

Plan: Nuclear stability is found in nuclides with an N/Z ratio that falls within the band of stability. Nuclides with

an even N and Z, especially those nuclides that have magic numbers, are exceptionally stable. Examine the two

nuclides to see which of these criteria can explain the difference in stability.

Solution:

Phosphorus-31 has 16 neutrons and 15 protons, with an N/Z ratio of 1.07. Phosphorus-30 has 15 neutrons and 15

protons, with an N/Z ratio of 1.00. 31P has an even N while 30P has both an odd Z and an odd N. 31P has a slightly

higher N/Z ratio that is closer to the band of stability.

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24-1

24.3A

Plan: Examine the N/Z ratio and determine which mode of decay will yield a product nucleus that is closer to the

band. Nuclides whose N/Z ratios are too high generally decay by beta emission while nuclides whose N/Z ratios

are too low decay by positron emission or electron capture. Nuclides with Z > 83 decay by alpha particle

emission.

Solution:

a) Iron-61 has an N/Z ratio of (61 – 26)/26 = 1.35, which is too high for this region of the band. Iron-61 will

undergo – decay.

61

26 Fe

61

0

27 Co + 1β

new N/Z = (61 – 27)/27 = 1.26

Additionally, iron has an atomic mass of 55.85 amu. The A value of 61 is higher, suggesting beta decay.

b) Americium–241 has Z > 83, so it undergoes decay.

241

95 Am

237

93 Np

4

+ 2 He

24.3B

Plan: Examine the N/Z ratio and determine which mode of decay will yield a product nucleus that is closer to the

band. Nuclides whose N/Z ratios are too high generally decay by beta emission while nuclides whose N/Z ratios

are too low decay by positron emission or electron capture. Nuclides with Z > 83 decay by alpha particle

emission.

Solution:

a) Titanium-40 has an N/Z ratio of (40 – 22)/22 = 0.81, which is too low for this region of the band. Titanium-40

will undergo positron decay or electron capture. Additionally, titanium’s atomic mass is 47.87 amu, which is

much higher than the A value of 40, also suggesting positron decay or electron capture.

b) Cobalt-65 has an N/Z ratio of (65 – 27)/27 = 1.40, which is too high for this region of the band. Cobalt-65 will

undergo beta decay. Additionally, cobalt’s atomic mass is 58.93 amu, which is much lower than the A value of

65, also suggesting beta decay.

24.4A

Plan: Specific activity of a radioactive sample is its decay rate per gram. Find the mass of the sample. Calculate

the specific activity by dividing the number of particles emitted per second (disintegrations per second = dps) by

the mass of the sample. Convert disintegrations per second to Ci by using the conversion factor between the two

units: 1 Ci = 3.70x1010 dps. Convert Ci to Bq by using the conversion factor between the two units:

1 Ci = 3.70x1010 Bq.

Solution:

a) Mass (g) of As = (3.4x10–8 mol As)

Specific activity (Ci/g) =

2.6x10–6 g

1.6x106 Ci

g

3.70x1010 dps

3.70x1010 Bq

Ci

= 1.5904x106 = 1.6x106 Ci/g

= 5.9200x1016 = 5.9x1016 Bq/g

Plan: The rate constant, k, relates the number of radioactive nuclei to their decay rate through the equation

A = kN. The number of radioactive nuclei is calculated by converting moles to atoms using Avogadro’s number.

The decay rate is 9.97x1012 beta particles/h or more simply, 9.97x1012 nuclei/h.

Solution:

Decay rate = A = kN

k=

24.5A

= 2.5840x10–6 = 2.6x10–6 g As

1 mol As

1.53x1011 dps

1 Ci

b) Specific activity (Bq/g) =

24.4B

76 g As

A

N

=

9.97x1012 nuclei/h

6.50x10–2 mol (6.022x1023 nuclei/h)

= 2.5471x10–10 = 2.55x10–10 h–1

Plan: Use the half-life of 24Na to find k. Substitute the value of k, initial activity (A0), and time of decay (4 days)

into the integrated first-order rate equation to solve for activity at a later time (At).

Solution:

ln 2

ln 2

k=

=

= 0.0462098 h–1

t1/2

15 h

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24-2

ln At = ln A0 – kt

ln At = ln (2.5x109) – (0.0462098 h–1)(4.0 days)(24 h/day)

ln At = 17.203416

At = 2.960388x107 = 3.0x107 d/s

24.5B

Plan: Use the half-life of 59Fe to find k. Substitute the value of k and time of decay (17 days) into the integrated

first-order rate equation. Assume that A0 is 1.0 (100% of the original sample) and solve for At, the fraction of the

sample remaining after 17 days. Subtract the fraction remaining from the original sample (1.0) to calculate the

fraction that has decayed.

Solution:

ln 2

ln 2

k=

=

= 1.5576x10–2 days–1

44.5 days

t1/2

ln At = ln A0 – kt

ln At = ln (1.0) – (1.5576x10–2 days–1)(17 days)

ln At = –0.2648

At = 0.7674 = fraction of iron-59 remaining

Fraction of iron-59 decayed = 1.0 – 0.7674 = 0.2326 = 0.23 = fraction of iron-59 that has decayed

24.6A

Plan: The wood from the case came from a living organism, so A0 equals 15.3 d/min•g. Substitute the current

activity of the case (At), A0, and k into the first-order rate expression and solve for t. Find k from the half-life of

carbon (5730 yr).

Solution:

ln 2

ln 2

k=

=

= 1.209680943x10–4 yr–1

5730 yr

t1/2

ln At = ln A0 – kt

ln [9.41 d/min•g] = ln [15.3 d/min•g] – (1.209680943x10–4 yr–1)(t)

–0.486079875 = – (1.209681x10–4 yr–1)(t)

t = 4018.2 = 4.02x103 years

24.6B

Plan: The woolen tunic came from a living organism, so A0 equals 15.3 d/min•g. Substitute the current activity of

the tunic (At), A0, and k into the first-order rate expression and solve for t. Find k from the half-life of carbon

(5730 yr).

Solution:

ln 2

ln 2

k=

=

= 1.209680943x10–4 yr–1

5730 yr

t1/2

ln At = ln A0 – kt

ln [12.87 d/min•g] = ln [15.3 d/min•g] – (1.209680943x10–4 yr–1)(t)

–0.172953806 = – (1.209681x10–4 yr–1)(t)

t = 1429.7473 = 1430 years

24.7A

Plan: Nickel-58 has 28 protons and 30 neutrons in its nucleus. Calculate the change in mass (m) in one 58Ni

atom, convert to MeV and divide by 58 to obtain binding energy/nucleon.

Solution:

m = [(28 x mass H atom) + (30 x mass neutron)] – mass 58Ni atom

m = [(28 x 1.007825 amu) + (30 x 1.008665)] – 57.935346 amu

m = 0.543704 amu

931.5 MeV

Binding energy (MeV) = (0.543704 amu)

Binding Energy/nucleon =

56

506.460276 MeV

58 nucleons

1 amu

= 506.460276 MeV

= 8.7321 = 8.732 MeV/nucleon

The BE/nucleon of Fe is 8.790 MeV/nucleon. The energy per nucleon holding the 58Ni nucleus together is less

than that for 56Fe (8.732 < 8.790), so 58Ni is less stable than 56Fe.

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24-3

24.7B

Plan: Uranium-235 has 92 protons and 143 neutrons in its nucleus. Calculate the change in mass (m) in one 235U

atom, convert to MeV and divide by 235 to obtain binding energy/nucleon.

Solution:

m = [(92 x mass H atom) + (143 x mass neutron)] – mass 235U atom

m = [(92 x 1.007825 amu) + (143 x 1.008665)] – 235.043924 amu

m = 1.915071 amu

931.5 MeV

Binding energy (MeV) = 1.915071 amu

= 1783.888637 MeV

1 amu

1783.888637 MeV

= 7.591015 = 7.591 MeV/nucleon

235 nucleons

12

The BE/nucleon of C is 7.680 MeV/nucleon. The energy per nucleon holding the 235U nucleus together is less than that

for 12C (7.591 < 7.680), so 235U is less stable than 12C.

Binding Energy/nucleon =

CHEMICAL CONNECTIONS BOXED READING PROBLEMS

B24.1

In the s-process, a nucleus captures a neutron sometime over a long period of time. Then the nucleus emits a beta

particle to form another element. The stable isotopes of most heavy elements up to 209Bi form by the s-process.

The r-process very quickly forms less stable isotopes and those with A greater than 230 by multiple neutron

captures, followed by multiple beta decays.

B24.2

Plan: Find the change in mass of the reaction by subtracting the mass of the products from the

mass of the reactants and convert the change in mass to energy with the conversion factor

between amu and MeV. Convert the energy per atom to energy per mole by multiplying by

Avogadro’s number.

Solution:

m = mass of reactants – mass of products

= [(4)(1.007825)] – [4.00260 + (2)(5.48580x10–4)]

= 4.031300 – 4.003697 = 0.02760 amu /4He atom = 0.02760284 g/mol 4He

0.02760284 amu 4 He 931.5 MeV

Energy (MeV/atom) =

= 25.7120 = 25.71 MeV/atom

1 atom

1 amu

Convert atoms to moles using Avogadro’s number.

23

25.7120 MeV 6.022x10 atoms

25

25

Energy =

= 1.54838x10 = 1.548x10 MeV/mol

atom

1

mol

B24.3

The simultaneous fusion of three nuclei is a termolecular process. Termolecular processes have a very low

probability of occurring. The bimolecular fusion of 8Be with 4He is more likely.

B24.4

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the

right side must be equal.

Solution:

210

83 Bi

210

84 Po

206

82 Pb

209

82 Pb

210

0

84 Po + 1

206

4

82 Pb + 2

1

209

+ 3 0 n 82 Pb

210

0

83 Bi + 1

210

84 Po is Nuclide A

206

82 Pb is Nuclide B

209

82 Pb is Nuclide C

210

83 Bi is Nuclide D

END–OF–CHAPTER PROBLEMS

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24-4

24.1

a) Chemical reactions are accompanied by relatively small changes in energy while nuclear reactions are

accompanied by relatively large changes in energy.

b) Increasing temperature increases the rate of a chemical reaction but has no effect on a nuclear reaction.

c) Both chemical and nuclear reaction rates increase with higher reactant concentrations.

d) If the reactant is limiting in a chemical reaction, then more reactant produces more product and the yield

increases in a chemical reaction. The presence of more radioactive reagent results in more decay product, so a

higher reactant concentration increases the yield in a nuclear reaction.

24.2

a) The percentage of sulfur atoms that are sulfur-32 is 95.02%, the same as the relative abundance of 32S.

b) The atomic mass is larger than the isotopic mass of 32S. Sulfur-32 is the lightest isotope, as stated in the

problem, so the other 5% of sulfur atoms are heavier than 31.972070 amu. The average mass of all the sulfur

atoms will therefore be greater than the mass of a sulfur-32 atom.

24.3

a) She found that the intensity of emitted radiation is directly proportional to the concentration of the element in

the various samples, not to the nature of the compound in which the element occurs.

b) She found that certain uranium minerals were more radioactive than pure uranium, which implied that they

contained traces of one or more as yet unknown, highly radioactive elements. Pitchblende is the principal ore of

uranium.

24.4

Plan: Radioactive decay that produces a different element requires a change in atomic number (Z, number of

protons).

Solution:

A

ZX

A = mass number (protons + neutrons)

Z = number of protons (positive charge)

X = symbol for the particle

N = A – Z (number of neutrons)

a) Alpha decay produces an atom of a different element, i.e., a daughter with two less protons and two less

neutrons.

A

ZX

A 4

4

Z 2Y + 2 He

A

Z 1Y

2 fewer protons, 2 fewer neutrons

b) Beta decay produces an atom of a different element, i.e., a daughter with one more proton and one less neutron.

A neutron is converted to a proton and particle in this type of decay.

0

+ 1

1 more proton, 1 less neutron

c) Gamma decay does not produce an atom of a different element and Z and N remain unchanged.

A

ZX

A

Z X*

ZA X + 00

( ZA X * = energy rich state), no change in number of protons or neutrons.

d) Positron emission produces an atom of a different element, i.e., a daughter with one less proton and one more

neutron. A proton is converted into a neutron and positron in this type of decay.

0

+ 1

1 less proton, 1 more neutron

e) Electron capture produces an atom of a different element, i.e., a daughter with one less proton and one more

neutron. The net result of electron capture is the same as positron emission, but the two processes are different.

A

ZX

A

ZX

A

Z 1Y

0

A

+ 1e Z 1Y

1 less proton, 1 more neutron

A different element is produced in all cases except (c).

24.5

The key factor that determines the stability of a nuclide is the ratio of the number of neutrons to the number of

protons, the N/Z ratio. If the N/Z ratio is either too high or not high enough, the nuclide is unstable and decays.

3

2 He

2

2 He

24.6

N/Z = 1/2

N/Z = 0/2, thus it is more unstable.

A neutron-rich nuclide decays to convert neutrons to protons while a neutron-poor nuclide decays to convert

protons to neutrons. The conversion of neutrons to protons occurs by beta decay:

1

0n

1

1p +

0

1

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24-5

The conversion of protons to neutrons occurs by either positron decay:

1

1p

1

0

0 n + 1

or electron capture:

1

1p

0

1

+ 1e 0 n

Neutron-rich nuclides, with a high N/Z, undergo decay. Neutron-poor nuclides, with a low N/Z, undergo

positron decay or electron capture.

24.7

Both positron emission and electron capture increase the number of neutrons and decrease the number of protons.

The products of both processes are the same. Positron emission is more common than electron capture among

lighter nuclei; electron capture becomes increasingly common as nuclear charge increases. For Z < 20, +

emission is more common; for Z > 80, electron capture is more common.

24.8

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the

right side must be equal.

Solution:

a)

b)

c)

24.9

a)

b)

c)

24.10

234

4

230

92 U 2 He + 90Th

232

0

232

93 Np + 1e 92 U

12

0

12

7 N 1 + 6 C

26

11 Na

223

87 Fr

212

83 Bi

Mass: 234 = 4 + 230;

Charge: 92 = 2 + 90

Mass: 232 + 0 = 232;

Charge: 93 + (–1) = 92

Mass: 12 = 0 + 12;

Charge: 7 = 1 + 6

0

26

1 + 12 Mg

0

223

1 + 88 Ra

4

208

2 + 81Tl

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the

right side must be equal.

Solution:

a) The process converts a neutron to a proton, so the mass number is the same, but the atomic number increases by

one.

27

12 Mg

0

23

12 Mg

0

27

1 + 13 Al

Mass: 27 = 0 + 27;

Charge: 12 = –1 + 13

b) Positron emission decreases atomic number by one, but not mass number.

23

1 + 11 Na

Mass: 23 = 0 + 23;

Charge: 12 = 1 + 11

c) The electron captured by the nucleus combines with a proton to form a neutron, so mass number is constant,

but atomic number decreases by one.

103

46 Pd

24.11

a)

b)

c)

24.12

+

0

1e

103

45 Rh

Mass: 103 + 0 = 103;

Charge: 46 + (–1) = 45

32

0

32

14 Si 1 + 15 P

218

4

214

84 Po 2 + 82 Pb

110

0

110

49 In + 1e 48 Cd

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the

right side must be equal.

Solution:

a) In other words, an unknown nuclide decays to give Ti-48 and a positron.

48

23V

48

0

22Ti + 1

Mass: 48 = 48 + 0;

Charge: 23 = 22 + 1

b) In other words, an unknown nuclide captures an electron to form Ag-107.

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24-6

107

48 Cd

0

107

+ 1e 47 Ag Mass: 107 + 0 = 107;

Charge: 48 + (–1) = 47

c) In other words, an unknown nuclide decays to give Po-206 and an alpha particle.

210

86 Rn

24.13

a)

b)

c)

24.14

241

94 Pu

228

88 Ra

207

85 At

206

84 Po

4

+ 2 He Mass: 210 = 206 + 4;

Charge: 86 = 84 + 2

241

0

95 Am + 1

228

0

89 Ac + 1

203

4

83 Bi + 2

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and

the right side must be equal.

Solution:

a) In other words, an unknown nuclide captures an electron to form Ir-186.

186

78 Pt

0

186

+ 1e 77 Ir

Mass: 186 + 0 = 186;

Charge: 78 + (–1) = 77

b) In other words, an unknown nuclide decays to give Fr-221 and an alpha particle.

225

89 Ac

221

4

87 Fr + 2 He Mass: 225 = 221 + 4;

Charge: 89 = 87 + 2

c) In other words, an unknown nuclide decays to give I-129 and a beta particle.

129

52Te

24.15

a)

b)

c)

24.16

129

53 I

+

0

1

Mass: 129 = 129 + 0;

52

52

0

26 Fe 25 Mn + 1

219

215

4

86 Rn 84 Po + 2

81

0

81

37 Rb + 1e 36 Kr

Plan: Look at the N/Z ratio, the ratio of the number of neutrons to the number of protons. If the N/Z ratio falls in

the band of stability, the nuclide is predicted to be stable. For stable nuclides of elements with atomic number

greater than 20, the ratio of number of neutrons to number of protons (N/Z) is greater than one. In addition, the

ratio increases gradually as atomic number increases. Also check for exceptionally stable numbers of neutrons

and/or protons – the “magic” numbers of 2, 8, 20, 28, 50, 82, and (N = 126). Also, even numbers of protons and

or neutrons are related to stability whereas odd numbers are related to instability.

Solution:

a)

20

8O

appears stable because its Z (8) value is a magic number, but its N/Z ratio (20 8)/8 = 1.50 is too high and

this nuclide is above the band of stability;

59

b) 27 Co

9

c) 3 Li

a)

b)

c)

24.18

20

8O

is unstable.

might look unstable because its Z value is an odd number, but its N/Z ratio (59 27)/27 = 1.19 is in the

band of stability, so

24.17

Charge: 52 = 53 + (–1)

59

27 Co

appears stable.

appears unstable because its N/Z ratio (9 3)/3 = 2.00 is too high and is above the band of stability.

146

60 Nd

114

48 Cd

88

42 Mo

N/Z = 86/60 = 1.4

Stable, N/Z ok

N/Z = 66/48 = 1.4

Stable, N/Z ok

N/Z = 46/42 = 1.1

Unstable, N/Z too small for this region of the band

Plan: Look at the N/Z ratio, the ratio of the number of neutrons to the number of protons. If the N/Z ratio falls in

the band of stability, the nuclide is predicted to be stable. For stable nuclides of elements with atomic number

greater than 20, the ratio of number of neutrons to number of protons (N/Z) is greater than one. In addition, the

ratio increases gradually as atomic number increases. Also check for exceptionally stable numbers of neutrons

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24-7

and/or protons – the “magic” numbers of 2, 8, 20, 28, 50, 82, and (N = 126). Also, even numbers of protons and

or neutrons are related to stability whereas odd numbers are related to instability.

Solution:

a) For the element iodine Z = 53. For iodine-127, N = 127 53 = 74. The N/Z ratio for 127I is 74/53 = 1.4. Of the

examples of stable nuclides given in the book, 107Ag has the closest atomic number to iodine. The N/Z ratio for

107

Ag is 1.3. Thus, it is likely that iodine with six additional protons is stable with an N/Z ratio of 1.4.

b) Tin is element number 50 (Z = 50). The N/Z ratio for 106Sn is (106 50)/50 = 1.1. The nuclide 106Sn is unstable

with an N/Z ratio that is too low.

c) For 68As, Z = 33 and N = 68 33 = 35 and N/Z = 1.1. The ratio is within the range of stability, but the nuclide is

most likely unstable because there is an odd number of both protons and neutrons.

24.19

a)

b)

c)

24.20

48

19 K

79

35 Br

33

18 Ar

N/Z = 29/19 = 1.5

Unstable, N/Z too large for this region of the band

N/Z = 44/35 = 1.3

Stable, N/Z okay

N/Z = 14/18 = 0.78

Unstable, N/Z too small

Plan: Calculate the N/Z ratio for each nuclide. A neutron-rich nuclide decays to convert neutrons to protons while

a neutron-poor nuclide decays to convert protons to neutrons. Neutron-rich nuclides, with a high N/Z, undergo

decay. Neutron-poor nuclides, with a low N/Z, undergo positron decay or electron capture. For Z < 20, +

emission is more common; for Z > 80, e– capture is more common. Alpha decay is the most common means of

decay for a heavy, unstable nucleus (Z > 83).

Solution:

a)

238

92 U:

Nuclides with Z > 83 decay through decay.

48

b) The N/Z ratio for 24 Cr is (48 – 24)/24 = 1.00. This number is below the band of stability because N is too

low and Z is too high. To become more stable, the nucleus decays by converting a proton to a neutron, which is

positron decay. Alternatively, a nucleus can capture an electron and convert a proton into a neutron through

electron capture.

50

c) The N/Z ratio for 25 Mn is (50 – 25)/25 = 1.00. This number is below the band of stability, so the nuclide

undergoes positron decay or electron capture.

24.21

a)

b)

c)

111

47 Ag

41

17 Cl

110

44 Ru

beta decay N/Z = 1.4 which is too high

beta decay N/Z = 1.4 which is too high

beta decay N/Z = 1.5 which is too high

24.22

Plan: Calculate the N/Z ratio for each nuclide. A neutron-rich nuclide decays to convert neutrons to protons while

a neutron-poor nuclide decays to convert protons to neutrons. Neutron-rich nuclides, with a high N/Z, undergo

decay. Neutron-poor nuclides, with a low N/Z, undergo positron decay or electron capture. For Z < 20, +

emission is more common; for Z > 80, e– capture is more common. Alpha decay is the most common means of

decay for a heavy, unstable nucleus (Z > 83).

Solution:

a) For carbon-15, N/Z = 9/6 = 1.5, so the nuclide is neutron-rich. To decrease the number of neutrons and increase

the number of protons, carbon-15 decays by beta decay.

b) The N/Z ratio for 120Xe is 66/54 = 1.2. Around atomic number 50, the ratio for stable nuclides is larger than 1.2,

so 120Xe is proton-rich. To decrease the number of protons and increase the number of neutrons, the xenon-120

nucleus either undergoes positron emission or electron capture.

c) Thorium-224 has an N/Z ratio of 134/90 = 1.5. All nuclides of elements above atomic number 83 are unstable

and decay to decrease the number of both protons and neutrons. Alpha decay by thorium-224 is the most likely

mode of decay.

24.23

a)

106

49 In

positron decay or electron capture N/Z = 1.2

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24-8

b)

c)

24.24

141

63 Eu

241

95 Am

positron decay or electron capture N/Z = 1.2

alpha decay

N/Z = 1.5

Plan: Stability results from a favorable N/Z ratio, even numbers of N and/or Z, and the occurrence of magic

numbers.

Solution:

The N/Z ratio of

52

24 Cr

is (52 24)/24 = 1.17, which is within the band of stability. The fact that Z is even does not

account for the variation in stability because all isotopes of chromium have the same Z. However,

neutrons, so N is both an even number and a magic number for this isotope only.

52

24 Cr has

28

40

20 Ca

24.25

N/Z = 20/20 = 1.0

It lies in the band of stability, and N and Z are both even and magic.

24.26

237

4

233

93 Np 2 + 91 Pa

233

0

233

91 Pa 1 + 92 U

233

4

229

92 U 2 + 90Th

229

4

225

90Th 2 + 88 Ra

24.27

Alpha emission produces helium ions which readily pick up electrons to form stable helium atoms.

24.28

The equation for the nuclear reaction is 92 U 82 Pb + __ 1 + __ 2 He

To determine the coefficients, notice that the beta particles will not impact the mass number. Subtracting the mass

number for lead from the mass number for uranium will give the total mass number for the alpha particles

released, 235 207 = 28. Each alpha particle is a helium nucleus with mass number 4. The number of helium

atoms is determined by dividing the total mass number change by 4, 28/4 = 7 helium atoms or seven alpha

particles. The equation is now

235

235

92 U

207

0

207

0

4

4

82 Pb + __ 1 + 7 2 He

To find the number of beta particles released, examine the difference in number of protons (atomic number)

between the reactant and products. Uranium, the reactant, has 92 protons. The atomic number in the products, lead

atom and 7 helium nuclei, total 96. To balance the atomic numbers, four electrons (beta particles) must be emitted

to give the total atomic number for the products as 96 4 = 92, the same as the reactant. In summary, seven alpha

particles and four beta particles are emitted in the decay of uranium-235 to lead-207.

235

92 U

207

82 Pb

+

0

1

4

+ 7 2 He

24.29

a) In a scintillation counter, radioactive emissions are detected by their ability to excite atoms and cause them to

emit light.

b) In a Geiger-Müller counter, radioactive emissions produce ionization of a gas that conducts a current to a

recording device.

24.30

Since the decay rate depends only on the number of radioactive nuclei, radioactive decay is a first-order process.

24.31

No, it is not valid to conclude that t1/2 equals 1 min because the number of nuclei is so small (six nuclei). Decay

rate is an average rate and is only meaningful when the sample is macroscopic and contains a large number of

nuclei, as in the second case. Because the second sample contains 6x1012 nuclei, the conclusion that

t1/2 = 1 min is valid.

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24-9

24.32

High-energy neutrons in cosmic rays enter the upper atmosphere and keep the amount of 14C nearly constant

through bombardment of ordinary 14N atoms. This 14 C is absorbed by living organisms, so its proportion stays

relatively constant there also.

14

7N

24.33

1

+ 0n

14

6C

1

+ 1H

Plan: Specific activity of a radioactive sample is its decay rate per gram. Calculate the specific activity by dividing

the number of particles emitted per second (disintegrations per second = dps) by the mass of the sample. Convert

disintegrations per second to Ci by using the conversion factor between the two units.

Solution:

1 Ci = 3.70x1010 dps

1.56x106 dps 1 mg

1 Ci

= 2.55528x10–2 = 2.56x10–2 Ci/g

Specific activity (Ci/g) =

1.65 mg 103 g 3.70x1010 dps

24.34

24.35

24.36

24.37

4.13x108 d 1 h

3600 s

h

1 Ci

–6

–6

Specific activity (Ci/g) =

= 1.1925x10 = 1.2x10 Ci/g

10

2.6 g

3.70x10

dps

Plan: Specific activity of a radioactive sample is its decay rate per gram. Calculate the specific activity by dividing

the number of particles emitted per second (disintegrations per second = dps) by the mass of the sample. Convert

disintegrations per second to Bq by using the conversion factor between the two units.

Solution:

A becquerel is a disintegration per second (dps).

7.4x104 d 1 min

min 60 s 1 Bq

8

8

Specific activity (Bq/g) =

= 1.43745x10 = 1.4x10 Bq/g

106 g 1 dps

8.58 g

1 g

3.77x107 d 1 min

min

60 s 1 Bq

Specific activity (Bq/g) =

1 dps = 587.2274 = 587 Bq/g

103 g

1.07 kg

1

kg

Plan: The decay constant is the rate constant for the first-order reaction.

Solution:

N

Decay rate =

= kN

t

1 atom

= k(1x1012 atom)

day

k = 1x1012 d1

24.38

N

= kN

t

(2.8x1012 atom/1.0 yr) = k(1 atom)

Decay rate =

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24-10

k = 2.8x1012 yr1

24.39

Plan: The rate constant, k, relates the number of radioactive nuclei to their decay rate through the equation

A = kN. The number of radioactive nuclei is calculated by converting moles to atoms using Avogadro’s number.

The decay rate is 1.39x105 atoms/yr or more simply, 1.39x105 yr–1 (the disintegrations are assumed).

Solution:

N

Decay rate = A = –

= kN

t

1.00x1012 mol 6.022x1023 atoms

1.39x105 atoms

= k

1.00 yr

1 mol

1.39x105 atom/yr = k(6.022x1011 atom)

k = (1.39x105 atom/yr)/6.022x1011 atom

k = 2.30820x10–7 = 2.31x10–7 yr–1

24.40

24.41

N

= kN

t

– (–1.07x1015 atom/1.00 h) = k[(6.40x10–9 mol)(6.022x1023 atom/mol)]

(1.07x1015 atom/1.00 h) = k (3.85408x1015 atom)

k = [(1.07x1015 atom/1.00 h)]/(3.85408x1015 atom)

k = 0.2776 = 0.278 h–1

Decay rate = A = –

Plan: Radioactive decay is a first-order process, so the integrated rate law is ln Nt = ln N0 – kt

First find the value of k from the half-life and use the integrated rate law to find Nt. The time unit in

the time and the k value must agree.

Solution:

t1/2 = 1.01 yr

t = 3.75x103 h

ln 2

ln 2

t1/2 =

or k =

t1/2

k

ln 2

= 0.686284 yr–1

1.01 yr

ln Nt = ln N0 – kt

k=

1 d 1 yr

ln Nt = ln [2.00 mg] – (0.686284 yr–1) 3.75x103 h

24 h 365 d

ln Nt = 0.399361

Nt = e0.399361

Nt = 1.49087 = 1.49 mg

24.42

t1/2 = 1.60x103 yr

t=?h

ln 2

ln 2

k=

=

= = 0.000433217 yr–1

t1/2 1.60x103 yr

ln [0.185 g] = ln [2.50 g] – (0.000433217 yr–1)(t)

t = 6010.129 yr

365 d 24 h

7

7

t = 6010.129 yr

= 5.264873x10 = 5.26x10 h

1

yr

1

d

24.43

Plan: Lead-206 is a stable daughter of 238U. Since all of the 206Pb came from 238U, the starting amount of 238U was

(270 mol + 110 mol) = 380 mol = N0. The amount of 238U at time t (current) is 270 mol = Nt. Find k from the

first-order rate expression for half-life, and then substitute the values into the integrated rate law and solve

for t.

Solution:

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24-11

ln 2

ln 2

or k =

t1/2

k

ln 2

k=

= 1.540327x10–10 yr–1

4.5x109 yr

t1/2 =

ln Nt = ln N0 – kt

ln

or

ln

N0

= kt

Nt

380 mol

= (1.540327x10–10 yr–1)(t)

270 mol

0.3417492937 = (1.540327x10–10 yr–1)(t)

t = 2.21868x109 = 2.2x109 yr

24.44

The ratio (0.735) equals Nt/N0 so N0/Nt = 1.360544218

ln 2

ln 2

k=

=

= = 1.2096809x10–4 yr–1

t1/2

5730 yr

N0

= kt

Nt

ln 1.360544218 = (1.2096809x10–4 yr–1)(t)

0.30788478 = (1.2096809x10–4 yr–1)(t)

t = 2.54517x103 = 2.54x103 yr

ln

24.45

Plan: The specific activity of the potassium-40 is the decay rate per mL of milk. Use the conversion factor

1 Ci = 3.70x1010 disintegrations per second (dps) to find the disintegrations per mL per s; convert the time

unit to min and change the volume to 8 oz.

Solution:

6 x1011 mCi 103 Ci 3.70x1010 dps 60 s 1000 mL 1 qt 1 cup

Activity =

8 oz

mL

1 Ci

1 mCi

1 min 1.057 qt 4 cups 8 oz

= 31.50426 = 30 dpm

24.46

Plutonium-239 (t1/2 = 2.41x104 yr)

Time = 7(t1/2) = 7(2.41x104 yr) = 1.6870x105 = 1.69x105 yr

24.47

Plan: Both Nt and N0 are given: the number of nuclei present currently, Nt, is found from the moles of 232Th. Each

fission track represents one nucleus that disintegrated, so the number of nuclei disintegrated is added to the

number of nuclei currently present to determine the initial number of nuclei, N0. The rate constant, k, is calculated

from the half-life. All values are substituted into the first-order decay equation to find t.

Solution:

ln 2

ln 2

t1/2 =

or k =

t1/2

k

ln 2

k=

=4.95105129x10–11 yr–1

10

1.4x10 yr

6.022x1023 Th atoms

9

Nt = 3.1x10 15 mol Th

= 1.86682x10 atoms Th

1 mol Th

9

4

N0 =1.86682x10 atoms + 9.5x10 atoms = 1.866915x109 atoms

N

ln 0 = kt

or

ln Nt = ln N0 – kt

Nt

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24-12

1.866915x109 atoms

= (4.95105129x10–11 yr–1)(t)

1.86682x109 atoms

5.088738x10–5 = (4.95105129x10–11 yr–1)(t)

t = 1.027809x106 = 1.0x106 yr

ln

24.48

The mole relationship between 40K and 40Ar is 1:1. Thus, 1.14 mmol 40Ar = 1.14 mmol 40K decayed.

ln 2

ln 2

k=

=

= 5.5451774x10–10 yr–1

t1/2

1.25x109 yr

ln

ln

N0

= kt

Nt

1.38 1.14 mmol

= (5.5451774x10–10 yr–1)(t)

1.38 mmol

0.6021754 = (5.5451774x10–10 yr–1)(t)

t = 1.08594x109 = 1.09x109 yr

27

13 Al

4

30

1

24.49

+ 2 He 15 P + 0 n

They experimentally confirmed the existence of neutrons, and were the first to produce an artificial radioisotope.

24.50

Both gamma radiation and neutron beams have no charge, so neither is deflected by electric or magnetic fields.

Neutron beams differ from gamma radiation in that a neutron has mass approximately equal to that of a proton.

Researchers observed that a neutron beam could induce the emission of protons from a substance. Gamma rays do

not cause such emissions.

24.51

A proton, for example, exits the first tube just when it becomes positive and the next tube becomes negative.

Pushed by the first tube and pulled by the second, the proton accelerates across the gap between them.

24.52

Protons are repelled from the target nuclei due to the interaction of like (positive) charges. Higher energy is

required to overcome the repulsion.

24.53

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the

right side must be equal. In the shorthand notation, the nuclide to the left of the parentheses is the reactant while

the nuclide written to the right of the parentheses is the product. The first particle inside the parentheses is the

projectile particle while the second substance in the parentheses is the ejected particle.

Solution:

a) An alpha particle is a reactant with 10B and a neutron is one product. The mass number for the reactants is

10 + 4 = 14. So, the missing product must have a mass number of 14 – 1 = 13. The total atomic number for the

reactants is 5 + 2 = 7, so the atomic number for the missing product is 7.

10

5B

4

1

13

28

14 Si

2

242

96 Cm

4

1

+ 2 He 2 0 n +

+ 2 He 0 n + 7 N

b) A deuteron (2H) is a reactant with 28Si and 29P is one product. For the reactants, the mass number is 28 + 2 = 30

and the atomic number is 14 + 1 = 15. The given product has mass number 29 and atomic number 15, so the

missing product particle has mass number 1 and atomic number 0. The particle is thus a neutron.

1

29

+ 1H 0 n + 15 P

c) The products are two neutrons and 244Cf with a total mass number of 2 + 244 = 246, and an atomic number of

98. The given reactant particle is an alpha particle with mass number 4 and atomic number 2. The missing reactant

must have mass number of 246 – 4 = 242 and atomic number 98 – 2 = 96. Element 96 is Cm.

24.54

a)

31

15 P

1

1

+ 1H + 0 n +

31

P (, p, n) 29Si

244

98 Cf

29

14 Si

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24-13

24.55

24.56

b)

252

98 Cf

+ 5 B 5 0 n + 103 Lr

252

Cf (10B, 5n) 257Lr

10

c)

238

92 U

4

1

+ 2 He 3 0 n +

238

U (, 3n) 239Pu

a)

249

98 Cf

249

98 Cf

249

98 Cf

+

+

1

257

239

94 Pu

12

257

1

6 C 104 Rf + 4 0 n

15

260

1

7 N 105 Db + 4 0 n

18

263

1

8 O 106 Sg + 4 0 n

+

b) 249Cf (12C, 4n) 257Rf

249

Cf (15N, 4n) 260Db

249

Cf (18O, 4n) 263Sg

Gamma radiation has no mass or charge while alpha particles are massive and highly charged. These differences

account for the different effect on matter that these two types of radiation have. Alpha particles interact with

matter more strongly than gamma particles due to their mass and charge. Therefore alpha particles penetrate

matter very little. Gamma rays interact very little with matter due to the lack of mass and charge. Therefore

gamma rays penetrate matter more extensively.

24.57

In the process of ionization, collision of matter with radiation dislodges an electron. The free electron and the

positive ion that result are referred to as an ion-pair.

24.58

Ionizing radiation is more dangerous to children because their rapidly dividing cells are more susceptible to

radiation than an adult’s slowly dividing cells.

24.59

The hydroxyl free radical forms more free radicals which go on to attack and change surrounding biomolecules,

whose bonding and structure are delicately connected with their function. These changes are irreversible, as

opposed to the reversible changes produced by OH–.

24.60

Plan: The rad is the amount of radiation energy absorbed in J per body mass in kg: 1 rad = 0.01 J/kg. Change the

mass unit from pounds to kilograms. The conversion factor between rad and gray is 1 rad = 0.01 Gy.

Solution:

3.3x10 7 J 2.205 lb

1 rad

a) Dose (rad) =

= 5.39x10–7 = 5.4x10–7 rad

135 lb 1 kg 1x102 J /kg

0.01

gy

–9

–9

b) Gray (rad) = 5.39x107 rad

= 5.39x10 = 5.4x10 Gy

1 rad

24.61

1 rad

a) Dose (rad) = (8.92x10–4 Gy)

= 0.0892 rad

0.01 Gy

0.01 J/kg

–3

–3

b) Energy (J) = (0.0892 rad)

3.6 kg = 3.2112x10 = 3.2x10 J

1

rad

24.62

Plan: Multiply the number of particles by the energy of one particle to obtain the total energy absorbed.

Convert the energy to dose in grays with the conversion factor 1 rad = 0.01 J/kg = 0.01 Gy. To find the millirems,

convert grays to rads and multiply rads by RBE to find rems. Convert rems to mrems. Convert the dose to

sieverts with the conversion factor 1 rem = 0.01 Sv.

Solution:

a) Energy (J) absorbed = 6.0x105 8.74x1014 J/ = 5.244x10–8 J

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24-14

5.244x108 J 1 rad 0.01 Gy

–10

–10

Dose (Gy) =

= 7.4914x10 = 7.5x10 Gy

J

70. kg

0.01 kg 1 rad

1 mrem

1 rad

–5

–5

b) rems = rads x RBE = 7.4914x1010 Gy

= 7.4914x10 = 7.5x10 mrem

1.0 3

0.01

Gy

10 rem

10 3 rem 0.01 Sv

= 7.4914x10–10 = 7.5x10–10 Sv

Sv = 7.4914x10 5 mrem

1 mrem 1 rem

1.77x10 2.20x10

a) Dose =

24.63

24.64

J / 103 g 1 rad

= 1.46943 = 1.47 rad

265 g

1 kg 0.01 J kg

b) Dose = (1.46943 rad)(0.01 Gy/1 rad) = 1.46943x10–2 = 1.47x10–2 Gy

c) Dose = (1.46943 rad)(0.75 rem/rad)(0.01 Sv/rem) = 1.10207x10–2 = 1.10x10–2 Sv

2.50 pCi 1x1012 Ci 3.70x1010 dps

3600 s 8.25x1013 J 1 rad

Dose =

65

h

1 Ci

95 kg 1 pCi

1 h disint. 0.01 J kg

= 1.8796974x10–8 = 1.9x10–8 rad

Dose = (1.8796974x10–8 rad)(0.01 Gy/1 rad) = 1.8796974x10–10 = 1.9x10–10 Gy

10

13

24.65

Use the time and disintegrations per second (Bq) to find the number of 60Co atoms that disintegrate, which equals

the number of particles emitted. The dose in rads is calculated as energy absorbed per body mass.

475 Bq 103 g 1 dps 5.05x1014 J

60 s 1 rad

Dose =

24.0

min

1.858 g 1 kg 1 Bq 1 disint.

1 min 0.01 J kg

= 1.8591x10–3 = 1.86x10–3 rad

24.66

A healthy thyroid gland incorporates dietary I – into I-containing hormones at a known rate. To assess thyroid

function, the patient drinks a solution containing a trace amount of Na131I, and a scanning monitor follows the

uptake of 131I into the thyroid. Technetium-99 is often used for imaging the heart, lungs, and liver.

24.67

NAA does not destroy the sample while chemical analysis does. Neutrons bombard a non-radioactive sample,

“activating” or energizing individual atoms within the sample to create radioisotopes. The radioisotopes decay

back to their original state (thus, the sample is not destroyed) by emitting radiation that is different for each

isotope.

24.68

In positron-emission tomography (PET), the isotope emits positrons, each of which annihilates a nearby electron.

In the process, two photons are emitted simultaneously, 180° apart from each other. Detectors locate the sites

and the image is analyzed by computer.

24.69

The concentration of 59 Fe in the steel sample and the volume of oil would be needed.

24.70

The oxygen in formaldehyde comes from methanol because the oxygen isotope in the methanol reactant appears

in the formaldehyde product. The oxygen isotope in the chromic acid reactant appears in the water product, not

the formaldehyde product. The isotope traces the oxygen in methanol to the oxygen in formaldehyde.

24.71

The mass change in a chemical reaction was considered too minute to be significant and too small to measure with

even the most sophisticated equipment.

24.72

When a nucleus forms from its nucleons, there is a decrease in mass called the mass defect. This decrease in mass

is due to mass being converted to energy to hold the nucleus together. This energy is called the binding energy.

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24-15

24.73

Energy is released when a nuclide forms from nucleons. The nuclear binding energy is the amount of energy

holding the nucleus together. Energy is absorbed to break the nucleus into nucleons and is released when

nucleons “come together.”

24.74

The binding energy per nucleon is the average amount of energy per each component (proton and neutron) part of

the nuclide. The binding energies per nucleon are helpful in comparing the stabilities of different combinations

and to provide information on the potential processes a nuclide can undergo to become more stable. The binding

energy per nucleon varies considerably. The greater the binding energy per nucleon, the more strongly the

nucleons are held together and the more stable the nuclide.

24.75

Plan: The conversion factors are: 1 MeV = 106 eV and 1 eV = 1.602x1019 J.

Solution:

106 eV

a) Energy (eV) = 0.01861 MeV

= 1.861x104 eV

1 MeV

106 eV 1.602x10 19 J

15

15

b) Energy (J) = 0.01861 MeV

= 2.981322x10 = 2.981x10 J

1 MeV

1 eV

24.76

1 eV

1000 J

= 9.8002x106 = 9.80x106 eV

a) Energy (eV) = 1.57x10 15 kJ

19

1

kJ

1.602x10

J

1 MeV

6

b) Energy (MeV) = 9.8002x10 eV 6

9.8002 = 9.80 MeV

10 eV

24.77

Plan: Convert moles of 239Pu to atoms of 239Pu using Avogadro’s number. Multiply the number of atoms by the

energy per atom (nucleus) and convert the MeV to J using the conversion 1 eV = 1.602x10–19 J.

Solution:

6.022x1023 atoms

23

Number of atoms = 1.5 mol 239 Pu

= 9.033x10 atoms

mol

5.243 MeV 106 eV 1.602x10 19 J

11

11

Energy (J) = 9.033x1023 atoms

= 7.587075x10 = 7.6x10 J

1

atom

1

MeV

1

eV

24.78

24.79

103 J

1 MeV

8.11x105 kJ

1 eV

1 mol 49 Cr

Energy (MeV) =

3.2x10 3 mol 49 Cr 1 kJ 1.602x1019 J 106 eV 6.022x1023 nuclei

= 2.6270 = 2.6 MeV

Plan: Oxygen-16 has eight protons and eight neutrons. First find the Δm for the nucleus by subtracting the given

mass of one oxygen atom from the sum of the masses of eight 1H atoms and eight neutrons. Use the conversion

factor 1 amu = 931.5 MeV to convert Δm to binding energy in MeV and divide the binding energy by the total

number of nucleons (protons and neutrons) in the oxygen nuclide to obtain binding energy per nucleon. Convert

Δm of one oxygen atom to MeV using the conversion factor for binding energy/atom. To obtain binding energy

per mole of oxygen, use the relationship E = mc2. m must be converted to units of kg/mol.

Solution:

Mass of 8 1H atoms = 8 x 1.007825 = 8.062600 amu

Mass of 8 neutrons = 8 x 1.008665 = 8.069320 amu

Total mass =16.131920 amu

m = 16.131920 15.994915 = 0.137005 amu/16O = 0.137005 g/mol 16O

0.137005 amu 16 O 931.5 MeV

a) Binding energy (MeV/nucleon) =

= 7.976259844 = 7.976 MeV/nucleon

16 nucleons

1 amu

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24-16

0.137005 amu 16 O 931.5 MeV

b) Binding energy (MeV/atom) =

= 127.6201575 = 127.6 MeV/atom

1 atom

1 amu

c) E = mc2

2

0.137005 g 16 O 1 kg

1 J 1 kJ

8

Binding energy (kJ/mol) =

3 2.99792x10 m/ s

3

2

mol

kg•m 2 10 J

10 g

s

= 1.23133577x1010 = 1.23134x1010 kJ/mol

24.80

m is calculated from the mass of 82 protons (1H) and 124 neutrons vs. the mass of the lead nuclide.

Mass of 82 1H atoms = 82 x 1.007825 = 82.641650 amu

Mass of 124 neutrons = 124 x 1.008665 = 125.074460 amu

Total mass = 207.716110 amu

m = 207.716110 205.974440 = 1.741670 amu/206Pb = 1.741670 g/mol 206Pb

1.741670 amu 206 Pb 931.5 MeV

a) Binding energy (MeV/nucleon) =

= 7.8755612 = 7.876 MeV/nucleon

206 nucleons

1 amu

1.741670 amu 206 Pb 931.5 MeV

b) Binding energy (MeV/atom) =

= 1622.3656 = 1622 MeV/atom

1 atom

1 amu

1.741670 g 206 Pb 1 kg

2.99792 x108 m / s

c) Binding energy (kJ/mol) =

103 g

mol

= 1.5653301x1011 = 1.56533x1011 kJ/mol

24.81

1kJ

1J

3

2

kg•m 2 10 J

s

2

Plan: Cobalt-59 has 27 protons and 32 neutrons. First find the Δm for the nucleus by subtracting the given mass of

one cobalt atom from the sum of the masses of 27 1H atoms and 32 neutrons. Use the conversion factor

1 amu = 931.5 MeV to convert Δm to binding energy in MeV and divide the binding energy by the total number

of nucleons (protons and neutrons) in the cobalt nuclide to obtain binding energy per nucleon. Convert Δm of

one cobalt atom to MeV using the conversion factor for binding energy/atom. To obtain binding energy per

mole of cobalt, use the relationship E = mc2. m must be converted to units of kg/mol.

Solution:

Mass of 27 1H atoms = 27 x 1.007825 = 27.211275 amu

Mass of 32 neutrons = 32 x 1.008665 = 32.27728 amu

Total mass = 59.488555 amu

m = 59.488555 – 58.933198 = 0.555357 amu/59Co = 0.555357 g/mol 59Co

0.555357 amu 59 Co 931.5 MeV

a) Binding energy (MeV/nucleon) =

= 8.768051619 = 8.768 MeV/nucleon

59 nucleons

1 amu

0.555357 amu 59 Co 931.5 MeV

b) Binding energy (MeV/atom) =

= 517.3150 = 517.3 MeV/atom

1 atom

1 amu

c) Use E = mc2

2

0.555357 g 59 Co 1 kg

1 J 1 kJ

8

Binding energy (kJ/mol) =

3 2.99792x10 m/s

3

2

mol

kg•m 2 10 J

10 g

s

= 4.9912845x1010 = 4.99128x1010 kJ/mol

24.82

m is calculated from the mass of 53 protons (1H) and 78 neutrons vs. the mass of the iodine nuclide.

Mass of 53 1H atoms = 53 x 1.007825 = 53.414725 amu

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24-17

Mass of 78 neutrons = 78 x 1.008665 = 78.675870 amu

Total mass = 132.090595 amu

m = 132.090595 – 130.906114 = 1.184481 amu/131I = 1.184481 g/mol 131I

1.184481 amu 131 I 931.5 MeV

= 8.422473676 = 8.422 MeV/nucleon

a) Binding energy (MeV/nucleon) =

131 nucleons 1 amu

1.184481 amu 131 I 931.5 MeV

b) Binding energy (MeV/atom) =

= 1103.34405 = 1103 MeV/atom

1 atom

1 amu

c) E = mc2

2

1.184481 g 131 I 1 kg

1 J 1 kJ

8

Binding energy (kJ/mol) =

3 2.99792x10 m/s

3

2

mol

kg•m 2 10 J

10 g

s

= 1.06455518x1011 = 1.06456x1011 kJ/mol

24.83

a)

80

35 Br

80

35 Br

0

1e

0

1

+

80

36 Kr

(reaction 1)

80

34 Se

+

(reaction 2)

b) Reaction 1:

m = 79.918528 – 79.916380 = 0.002148 amu

Reaction 2:

m = 79.918528 – 79.916520 = 0.002008 amu

Since E = (m)c2, the greater mass change (reaction 1) will release more energy.

24.84

The minimum number of neutrons from each fission event that must be absorbed by the nuclei to sustain the chain

reaction is one. In reality, due to neutrons lost from the fissionable material, two to three neutrons are generally

needed to continue a self-sustaining chain reaction.

24.85

In both radioactive decay and fission, radioactive particles are emitted, but the process leading to the emission is

different. Radioactive decay is a spontaneous process in which unstable nuclei emit radioactive particles and

energy. Fission occurs as the result of high-energy bombardment of nuclei with small particles that cause the

nuclides to break into smaller nuclides, radioactive particles, and energy.

In a chain reaction, all fission events are not the same. The collision between the small particle emitted in the

fission and the large nucleus can lead to splitting of the large nuclei in a number of ways to produce several

different products.

24.86

Enriched fissionable fuel is needed in the fuel rods to ensure a sustained chain reaction. Naturally occurring

235

U is only present in a concentration of 0.7%. This is consistently extracted and separated until its concentration

is between 3-4%.

24.87

a) Control rods are movable rods of cadmium or boron which are efficient neutron absorbers. In doing so, they

regulate the flux of neutrons to keep the reaction chain self-sustaining which prevents the core from overheating.

b) The moderator is the substance flowing around the fuel and control rods that slows the neutrons, making them

better at causing fission.

c) The reflector is usually a beryllium alloy around the fuel-rod assembly that provides a surface for neutrons that

leave the assembly to collide with and therefore, return to the fuel rods.

24.88

The water serves to slow the neutrons so that they are better able to cause a fission reaction. Heavy water

2

1

( 1 H2O or D2O) is a better moderator because it does not absorb neutrons as well as light water ( 1 H2O ) does, so

more neutrons are available to initiate the fission process. However, D2O does not occur naturally in great

abundance, so production of D2O adds to the cost of a heavy water reactor. In addition, if heavy water does

3

absorb a neutron, it becomes tritiated, i.e., it contains the isotope tritium, 1H , which is radioactive.

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24-18

24.89

The advantages of fusion over fission are the simpler starting materials (deuterium and tritium), and no long-lived

toxic radionuclide by-products.

24.90

Virtually all the elements heavier than helium, up to and including iron, are produced by nuclear fusion reactions

in successively deeper and hotter layers of massive stars. Iron is the point at which fusion reactions cease to be

energy producers. Elements heavier than iron are produced by a variety of processes, primarily during a

supernova event, which distribute the Sun’s ash into the cosmos to form next generation suns and planets. Thus,

the high cosmic and Earth abundance of iron is consistent with it being the most stable of all nuclei.

24.91

Mass of reactants: 3.01605 + 2.0140 = 5.03005 amu

Mass of products: 4.00260 + 1.008665 = 5.011265 amu

m = mass of reactants – mass of products = 5.03005 – 5.011265 = 0.018785 amu = 0.018785 g/mol

E = mc2

0.018785 g 1 kg

8

Energy (kJ/mol) =

3 2.99792x10 m/s

mol

10 g

= 1.6883064x109 = 1.69x109 kJ/mol

24.92

24.93

243

4

239

95 Am 2 He + 93 Np

239

0

239

93 Np 1 + 94 Pu

239

4

235

94 Pu 2 He + 92 U

239

239

235

93 Np , 94 Pu , and 92 U

1 J 1 kJ

3

2

kg•m 2 10 J

s

2

were present as products in the decay of Am-243.

Plan: Use the masses given in the problem to calculate the mass change (reactant – products) for the reaction.

The conversion factor between amu and kg is 1 amu = 1.66054x10–27 kg. Use the relationship E = mc2 to

convert the mass change to energy.

Solution:

243

239

4

a) 96 Cm 94 Pu + 2 He

m (amu) = 243.0614 amu (4.0026 + 239.0522) amu = 0.0066 amu

1.661x10 24 g 1 kg

–29

–29

m (kg) = 0.0066 amu

3 = 1.09626x10 = 1.1x10 kg

1

amu

10 g

1

J

–13

–13

b) E = mc2 = 1.09626x10 kg 2.99792x10 m/s

= 9.85266x10 = 9.9x10 J

2

kg•m 2

s

13

23

9.85266x10

J 6.022x10 reactions 1 kJ

8

8

c) E released =

3 = 5.93317x10 = 5.9x10 kJ/mol

reaction

mol

10 J

This is approximately one million times larger than a typical heat of reaction.

24.94

29

8

2

a) First, determine the amount of activity released by the 239Pu for the duration spent in the body (16 h) using the

relationship A = kN. The rate constant is derived from the half-life and N is calculated using the molar mass and

Avogadro’s number.

ln 2

1 day 1 h

ln 2

1 yr

–13 –1

k=

=

= 9.113904x10 s

4

365.25

day

24

h

3600

s

t1/2

2.41x10 yr

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24-19

10 6 g 1 mol Pu 6.022x1023 atoms Pu

15

N = 1.00 g Pu

= 2.5196653x10 atoms Pu

1 g 239 g Pu

1

mol

Pu

1 disint.

3

A = kN = (9.113904 x 10–13 s–1)( 2.5196653x1015 atoms Pu)

= 2.2963988x10 d/s

1

atom

Pu

Each disintegration releases 5.15 MeV, so d/s can be converted to MeV. Convert MeV to J (using

1.602x10–13 J = 1 MeV) and J to rad (using 0.01 J/kg = 1 rad).

2.2963988x103 d/s 5.15 MeV 1.602x10 13 J 1 rad 3600 s

Energy =

16 h

J

85 kg

MeV

disint.

0.01 kg 1 h

= 1.2838686x10–4 = 1.28x10–4 rads

b) Since 0.01 Gy = 1 rad, the worker receives:

Dose = (1.2838686x10–4 rad)(0.01 Gy/rad) = 1.2838686x10–6 = 1.28x10–6 Gy

24.95

Plan: Determine k for 14C using the half-life (5730 yr). Determine the mass of carbon in 4.58 g of CaCO3.

Divide the given activity of the C in d/min by the mass of carbon to obtain the activity in d/min•g; this is At and is

compared to the activity of a living organism (A0 = 15.3 d/min•g) in the integrated rate law, solving for t.

Solution:

ln 2

ln 2

k=

=

= 1.2096809x10–4 yr–1

t1/2

5730 yr

1 mol CaCO3 1 mol C 12.01 g C

Mass (g) of C = 4.58 g CaCO3

= 0.5495634 g C

100.09 g CaCO3 1 mol CaCO3 1 mol C

3.2 d/min

At =

= 5.8228 d/min•g

0.5495634 g

Using the integrated rate law:

A

ln t = –kt

A0 = 15.3 d/min•g (the ratio of 12 C:14 C in living organisms)

A

0

5.8228 d/min•g

–4

–1

ln

= – (1.2096809x10 yr )(t)

15.3 d/min•g

t = 7986.17 = 8.0x103 yr

24.96

Find the rate constant from the rate of decay and the initial number of atoms. Use rate constant to calculate halflife.

Initial number of atoms:

106 g 1 mol RaCl 2 1 mol Ra 6.022x1023 Ra atoms

Ra atoms = 5.4 g RaCl 2

1 g 297 g RaCl 1 mol RaCl

1 mol Ra

2

2

= 1.09490909x1016 Ra atoms

A

A = kN or k =

N

1d s

1.5x105 Bq

= 1.36997675x10–11 s–1

k=

1.09490909x1016 Ra atoms Bq

ln 2

ln 2

t1/2 =

=

= 5.05955x1010 = 5.1x1010 s

k

1.36997675x1011 s1

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24-20

24.97

k=

ln 2

ln 2

=

= 1.2096809x10–4 yr–1

t1/2

5730 yr

1 mol 14 C 6.022x1023 atoms 14 C

14

14

C

= 4.3014x10 atoms C

14

14 g 14 C

1

mol

C

A = kN = [(1.2096809x10–4 yr–1)(4.3014x1014 atoms 14C)](1 disintegration/1 atom) = 5.2033x1010 dpyr

13

5.2033x1010 dpyr

0.156 MeV 1.602x10 J 1 rad

Dose =

= 2.001x10–3 = 10–3 rad

yr

1 MeV

0.01 J

65

kg

disint.

kg

Number of atoms = 10 8 g

24.98

14

Plan: Determine how many grams of AgCl are dissolved in 1 mL of solution. The activity of the radioactive Ag+

indicates how much AgCl dissolved, given a starting sample with a specific activity (175 nCi/g). Convert g/mL to

mol/L (molar solubility) using the molar mass of AgCl.

Solution:

1.25x102 Bq 1 dps

1 nCi 1 g AgCl

1 Ci

–6

Concentration =

9

= 1.93050x10 g AgCl/mL

10

mL

1

Bq

175

nCi

3.70

x10

dps

10

Ci

1.93050x10 6 g AgCl 1 mol AgCl 1 mL

–5

–5

Molarity =

3 = 1.34623x10 = 1.35x10 M AgCl

mL

143.4

g

AgCl

10

L

24.99

a) The process shown is fission in which a neutron bombards a large nucleus, splitting that nucleus

into two nuclei of intermediate mass.

1

b) 0 n +

235

92 U

144

55 Cs

+

90

37 Rb

1

+ 2 0n

144

c) 55 Cs , with 55 protons and 89 neutrons, has a n/p ratio of 1.6. This ratio places this isotope above the band of

stability and decay by beta particle emission is expected.

24.100 Plan: Determine the value of k from the half-life. Then determine the fraction from the integrated rate law.

Solution:

ln 2

ln 2

k=

=

= 9.90210x10–10 yr–1

t1/2

7.0x108 yr

ln

N0

= kt = (9.90210x10–10 yr–1)(2.8x109 yr) = 2.772588

Nt

N0

= 15.99998844

Nt

Nt

= 0.062500 = 6.2x10–2

N0

24.101 Determine the value of k from the half-life. Then determine the age from the integrated rate law.

ln 2

ln 2

k=

=

= 1.540327x10–10 yr–1

9

t1/2

4.5x10 yr

N0

= kt

Nt

6 9

ln

=(1.540327x10-10 yr-1 )(t)

9

0.5108256 = (1.540327x10–10 yr–1)(t)

t = 3.316345x109 = 3.3x109 yr

ln

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24-21

24.102 Plan: Find the rate constant, k, using any two data pairs (the greater the time between the data points, the

greater the reliability of the calculation). Calculate t1/2 using k. Once k is known, use the integrated rate law to

find the percentage lost after 2 h. The percentage of isotope remaining is the fraction remaining after 2.0 h

(Nt where t = 2.0 h) divided by the initial amount (N0), i.e., fraction remaining is Nt/N0. Solve the first-order rate

expression for Nt/N0, and then subtract from 100% to get fraction lost.

Solution:

N

a) ln t = –kt

N0

495 photons/s

ln

= –k(20 h)

5000 photons/s

–2.312635 = –k(20 h)

k = 0.11563 h–1

ln 2

ln 2

t1/2 =

=

= 5.9945 = 5.99 h (Assuming the times are exact, and the emissions have three

k

0.11563 h 1

significant figures.)

b) ln

ln

Nt

= –kt

N0

Nt

= – (0.11563 h–1)(2.0 h) = –0.23126

N0

Nt

Nt

= 0.793533

x 100% = 79.3533%

N0

N0

The fraction lost upon preparation is 100% – 79.3533% = 20.6467% = 21%.

ln 2

ln 2

=

= 5.5451774x10–10 yr–1

9

t1/2

1.25x10 yr

A = kN where A = dps N = number of atoms

Number of atoms = (1.0 mol 40K)(6.022x1023 atoms/mol) = 6.022x1023 atoms 40K

24.103 k =

5.5451774x10 10

1 day 1 h 1 disint.

1 yr

7

23

A=

6.022x10 atoms

= 1.05816x10 dps

yr

365.25

day

24

h

3600

s

1

atom

Dose (Ci) = (1.05816x107 dps)(1 Ci/3.70x1010 dps) = 2.85990x10–4 = 2.9x10–4 Ci

Dose = (1.05816x107 dps)(1 Bq/1 dps) = 1.05816x107 = 1.1x107 Bq

24.104 Plan: Use the given relationship for the fraction remaining after time t, where t = 10.0 yr, 10.0x103 yr, and

10.0x104 yr.

Solution:

a) Fraction remaining after 10.0 yr =

1

2

t

t1

10.0

2

1

2

4

1

2

c) Fraction remaining after 10.0x10 yr =

1

2

10.0x103

3

b) Fraction remaining after 10.0x10 yr =

=

10.0x104

5730

5730

5730

= 0.998791 = 0.999

= 0.298292 = 0.298

= 5.5772795x10–6 = 5.58x10–6

d) Radiocarbon dating is more reliable for b) because a significant quantity of 14C has decayed and a significant

quantity remains. Therefore, a change in the amount of 14C would be noticeable. For the fraction in a), very little

14

C has decayed and for c) very little 14C remains. In either case, it will be more difficult to measure the change so

the error will be relatively large.

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24-22

24.105

210

86 Rn

0

210

+ 1e 85 At

Mass = (2.368 MeV)(1 amu/931.5 MeV) = 0.0025421363 amu

Mass 210At = mass 210Rn + electron mass – mass equivalent of energy emitted.

= (209.989669 + 0.000549 – 0.0025421363) amu = 209.9876759 = 209.98768 amu

24.106 Plan: At one half-life, the fraction of sample is 0.500. Find n for which (0.900)n = 0.500.

Solution:

(0.900)n = 0.500

n ln (0.900) = ln (0.500)

n = (ln 0.500)/(ln 0.900) = 6.578813 = 6.58 h

24.107 a) decay by vanadium-52 produces chromium-52.

51

23V

51

23V

1

+ 0n

52

23V

52

24 Cr

+

0

1

52

(n,) 24 Cr

b) Positron emission by copper-64 produces nickel-64.

63

29 Cu

63

29 Cu

1

+ 0n

64

29 Cu

64

28 Ni

0

+ 1

64

(n,+) 28 Ni

c) decay by aluminum-28 produces silicon-28.

27

1

28

13 Al + 0 n 13 Al

27

28

13 Al (n,) 14 Si

28

14 Si

+

0

1

24.108 Determine k for 90Sr.

ln 2

ln 2

k=

=

= 0.023902 yr–1

t1/2

29 yr

a) ln

ln

Nt

= –kt

N0

Nt

= – (0.023902 yr–1)(10 yr) = – 0.23902

0.0500 g

Nt

= 0.787399

0.0500 g

(0.787399)(0.0500 g) = 0.03936995 = 0.039 g 90Sr

N

b) ln t = –kt

N0

100 99.9%

= – (0.023902 yr–1)(t)

100%

t = 289.003 = 3x102 yr (The calculation 100 – 99.9 limits the answer to one significant figure.)

ln

24.109 a)

12

4

6 C + 2 He

16

8O

13

931.5 MeV 1.602x10

b) 7.7x10 2 amu

1 MeV

1 amu

J 1 kJ

–14

–14

3 = 1.1490425x10 = 1.1x10 kJ

10

J

24.110 Plan: The production rate of radon gas (volume/hour) is also the decay rate of 226Ra. The decay rate, or activity, is

proportional to the number of radioactive nuclei decaying, or the number of atoms in 1.000 g of 226Ra, using the

relationship A = kN. Calculate the number of atoms in the sample, and find k from the half-life. Convert the

activity in units of nuclei/time (also disintegrations per unit time) to volume/time using the ideal gas law.

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24-23

Solution:

226

88 Ra

4

2 He +

222

86 Rn

ln 2

ln 2

=

= 4.33879178x10–4 yr–1(1 yr/8766 h) = 4.94510515x10–8 h–1

t1/2

1599 yr

The mass of 226Ra is 226.025402 amu/atom or 226.025402 g/mol.

k=

6.022x1023 Ra atoms

1 mol Ra

21

N = 1.000 g Ra

= 2.6643023x10 Ra atoms

226.025402

g

Ra

1

mol

Ra

–8 –1

21

A = kN = (4.94510515x10 h )(2.6643023x10 Ra atoms) = 1.3175255x1014 Ra atoms/h

This result means that 1.318x1014 226Ra nuclei are decaying into 222Rn nuclei every hour. Convert atoms of 222Rn

into volume of gas using the ideal gas law.

1.3175255x1014 Ra atoms 1 atom Rn

1 mol Rn

Moles of Rn/h =

23

h

1 atom Ra 6.022x10 Rn atoms

= 2.1878537x10–10 mol Rn/h

L•atm

2.1878537x1010 mol Rn/h 0.08206

273.15 K

nRT

mol•K

V=

=

1 atm

P

–9

–9

= 4.904006x10 = 4.904x10 L/h

Therefore, radon gas is produced at a rate of 4.904x10–9 L/h. Note: Activity could have been calculated as decay

in moles/time, removing Avogadro’s number as a multiplication and division factor in the calculation.

24.111 Determine k:

ln 2

ln 2

k=

=

= 0.0216608 s–1

t1/2

32 s

ln

Nt

= –kt

N0

90%

= – (0.0216608 s–1)(t)

100%

t = 4.86411 = 4.9 s

ln

24.112 a)

b)

c)

d)

133

55 Cs

79

35 Br

24

12 Mg

14

7N

The N/Z ratio for 140Cs is too high.

It has an even number of neutrons compared with 78Br.

The N/Z ratio equals 1.

The N/Z ratio equals 1.

24.113 Plan: Determine k from the half-life and then use the integrated rate law, solving for time.

Solution:

ln 2

ln 2

k=

=

= 0.0239016 yr–1

t1/2

29 yr

ln

Nt

= –kt

N0

1.0x104 particles

= – (0.0239016 yr–1)(t)

ln

7.0x104 particles

–1.945910 = – (0.0239016 yr–1)(t)

t = 81.413378 = 81 yr

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24-24

6

6

24.114 a) 3 Li + 3 Li

12

6 C (dilithium)

6

12

3 Li ) – mass 6 C =

12

b) m = 2(mass

2(6.015121 amu) – 12.000000 amu = 0.030242 amu/atom 6 C (dilithium)

m = (0.030242 amu/atom)(1.66054x10–27 kg/amu) =5.02180507x10–29 kg/atom

2

1

J

29

8

–12

E = mc2 = 5.02180507x10 kg/atom 2.99792x10 m/s

= 4.5133595x10 J/atom

2

kg•m

s2

4.5133595 x10 12 J

1 atom

1 amu

E=

27

atom

12.000000 amu 1.66054x10 kg

= 2.2650059x1014 = 2.2650x1014 J/kg dilithium

1

4

0

c) 4 1H 2 He + 2 1

2 positrons are released.

12

( 6 C ):

d) For dilithium

Mass = (5.02180507x10–29 kg/atom)(1 atom/12.000000 amu) (1 amu/1.66054x10–27 kg 12C)

= 2.5201667x10–3 = 2.5202x10–3 kg/kg 12C

4

For 2 He (The mass of a positron is the same as the mass of an electron.)

1

4

0

m = 4 (mass 1H ) – [mass 2 He + 2 mass 1e ]

= 4(1.007825 amu) – [4.00260 amu + 2(5.48580x10–4 amu)] = 0.02760 amu/atom

= (0.02760 amu/atom)(1.66054x10–27 kg/amu) = 4.58309x10–29 kg/atom

Mass = (4.58309x10–29 kg/atom)(1 atom/4.00260 amu)(1 amu/1.66054x10–27 kg 4He)

= 6.895517x10–3 = 6.896x10–3 kg/kg 4He

2

3

4

1

e) 1H + 1H 2 He + 0 n

m = [2.0140 amu + 3.01605 amu] – [4.00260 amu + 1.008665 amu]

= 5.0300 amu – 5.01126 amu = 0.0188 amu

= (0.0188 amu/atom)(1.66054x10–27 kg/amu) = 3.1218152x10–29 kg/atom

Mass = (3.1218152x10–29 kg/atom)(1 atom/4.00260 amu)(1 amu/1.66054x10–27 kg 4He)

= 4.696947x10–3 = 4.70x10–3 kg/kg 4He

6

1

4

3

f) 3 Li + 0 n 2 He + 1H

1 mol 3 H 3.01605 g 3 H 1 mol 6 Li 103 g 6 Li 1 kg 3 H

3

Mass 1H / kg 6Li =

1 mol 6 Li 1 mol 3 H 6.015121 g 6 Li 1 kg 6 Li 103 g 3 H

= 0.5014113598 = 0.501411 kg 3H/kg 6Li

103 g 1 mol 3 H 6.022x1023 atom 3 H 3.121852x1029 kg

Mass = (0.5014113598 kg 3H)

1 kg 3.01605 g 3 H

atom

mol 3 H

= 3.1254222x10–3 = 3.125x10–3 kg

Change in mass for dilithium reaction:

Mass = (5.02180507x10–29 kg/atom 12C)(1 atom 12C/2 atoms 6Li)(6.022x1023 atoms 6Li/mol)

(1 mol 6Li/6.015121 g 6Li)(103 g/kg 6Li) = 2.513774x10–3 = 2.514x10–3 kg

The change in mass for the dilithium reaction is slightly less than that for the fusion of tritium with

deuterium.

24.115 Plan: Convert pCi to Bq using the conversion factors 1 Ci = 3.70x1010 Bq and 1 pCi = 10–12 Ci. For part b),

use the first-order integrated rate law to find the activity at the later time (t = 9.5 days). You will first need to

calculate k from the half-life expression. For part c), solve for the time at which Nt = the EPA recommended

level.

Solution:

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24-25

THEIR APPLICATIONS

FOLLOW–UP PROBLEMS

24.1A

Plan: Write a skeleton equation that shows fluorine-20 undergoing beta decay. Conserve mass and atomic number

by ensuring the superscripts and subscripts equal one another on both sides of the equation. Determine the identity

of the daughter nuclide by using the periodic table.

Solution:

Fluorine-20 has Z = 9. Its symbol is 209F. When it undergoes beta decay, a beta particle, 10β , and a daughter

nuclide, AZX, are produced.

20

A

0

9F ZX + 1β

To conserve atomic number, Z must equal 10. Element is neon.

To conserve mass number, A must equal 20.

The identity of

A

ZX

is 20

10Ne.

The balanced equation is:

24.1B

20

20

9F 10Ne

+

0

1β

Plan: Write a skeleton equation that shows an unknown nuclide,

A

ZX,

undergoing beta decay,

0

1β

, to form

133

55 Cs .

cesium–133,

Conserve mass and atomic number by ensuring the superscripts and subscripts equal one

another on both sides of the equation. Determine the identity of X by using the periodic table to identify the

element with atomic number equal to Z.

Solution:

The unknown nuclide yields cesium-133 and a particle:

A

133

Z X 55 Cs

0

+ 1β

To conserve atomic number, Z must equal 54. Element is xenon.

To conserve mass number, A must equal 133.

133

54 Xe .

133

equation is: 54 Xe

The identity of

A

ZX

is

133

0

The balanced

55 Cs + -1 β

Check: A = 133 = 133 + 0 and Z = 54 = 55 + (–1).

24.2A

Plan: Look at the N/Z ratio, the ratio of the number of neutrons to the number of protons. If the N/Z ratio falls in

the band of stability, the nuclide is predicted to be stable. For stable nuclides of elements with atomic number

greater than 20, the ratio of number of neutrons to number of protons (N/Z) is greater than one. In addition, the

ratio increases gradually as atomic number increases. Also check for exceptionally stable numbers of neutrons

and/or protons – the “magic” numbers of 2, 8, 20, 28, 50, 82, and (N = 126). Also, even numbers of protons and

or neutrons are related to stability whereas odd numbers are related to instability.

Solution:

a) 105B appears stable because its N/Z ratio (10 5)/5 = 1.00 is in the band of stability.

b) 58

23V appears unstable/radioactive because its N/Z ratio (58 23)/23 = 1.52 is too high and is above the band of

stability. Additionally, this nuclide has both odd N(35) and Z(23).

24.2B

Plan: Nuclear stability is found in nuclides with an N/Z ratio that falls within the band of stability. Nuclides with

an even N and Z, especially those nuclides that have magic numbers, are exceptionally stable. Examine the two

nuclides to see which of these criteria can explain the difference in stability.

Solution:

Phosphorus-31 has 16 neutrons and 15 protons, with an N/Z ratio of 1.07. Phosphorus-30 has 15 neutrons and 15

protons, with an N/Z ratio of 1.00. 31P has an even N while 30P has both an odd Z and an odd N. 31P has a slightly

higher N/Z ratio that is closer to the band of stability.

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24-1

24.3A

Plan: Examine the N/Z ratio and determine which mode of decay will yield a product nucleus that is closer to the

band. Nuclides whose N/Z ratios are too high generally decay by beta emission while nuclides whose N/Z ratios

are too low decay by positron emission or electron capture. Nuclides with Z > 83 decay by alpha particle

emission.

Solution:

a) Iron-61 has an N/Z ratio of (61 – 26)/26 = 1.35, which is too high for this region of the band. Iron-61 will

undergo – decay.

61

26 Fe

61

0

27 Co + 1β

new N/Z = (61 – 27)/27 = 1.26

Additionally, iron has an atomic mass of 55.85 amu. The A value of 61 is higher, suggesting beta decay.

b) Americium–241 has Z > 83, so it undergoes decay.

241

95 Am

237

93 Np

4

+ 2 He

24.3B

Plan: Examine the N/Z ratio and determine which mode of decay will yield a product nucleus that is closer to the

band. Nuclides whose N/Z ratios are too high generally decay by beta emission while nuclides whose N/Z ratios

are too low decay by positron emission or electron capture. Nuclides with Z > 83 decay by alpha particle

emission.

Solution:

a) Titanium-40 has an N/Z ratio of (40 – 22)/22 = 0.81, which is too low for this region of the band. Titanium-40

will undergo positron decay or electron capture. Additionally, titanium’s atomic mass is 47.87 amu, which is

much higher than the A value of 40, also suggesting positron decay or electron capture.

b) Cobalt-65 has an N/Z ratio of (65 – 27)/27 = 1.40, which is too high for this region of the band. Cobalt-65 will

undergo beta decay. Additionally, cobalt’s atomic mass is 58.93 amu, which is much lower than the A value of

65, also suggesting beta decay.

24.4A

Plan: Specific activity of a radioactive sample is its decay rate per gram. Find the mass of the sample. Calculate

the specific activity by dividing the number of particles emitted per second (disintegrations per second = dps) by

the mass of the sample. Convert disintegrations per second to Ci by using the conversion factor between the two

units: 1 Ci = 3.70x1010 dps. Convert Ci to Bq by using the conversion factor between the two units:

1 Ci = 3.70x1010 Bq.

Solution:

a) Mass (g) of As = (3.4x10–8 mol As)

Specific activity (Ci/g) =

2.6x10–6 g

1.6x106 Ci

g

3.70x1010 dps

3.70x1010 Bq

Ci

= 1.5904x106 = 1.6x106 Ci/g

= 5.9200x1016 = 5.9x1016 Bq/g

Plan: The rate constant, k, relates the number of radioactive nuclei to their decay rate through the equation

A = kN. The number of radioactive nuclei is calculated by converting moles to atoms using Avogadro’s number.

The decay rate is 9.97x1012 beta particles/h or more simply, 9.97x1012 nuclei/h.

Solution:

Decay rate = A = kN

k=

24.5A

= 2.5840x10–6 = 2.6x10–6 g As

1 mol As

1.53x1011 dps

1 Ci

b) Specific activity (Bq/g) =

24.4B

76 g As

A

N

=

9.97x1012 nuclei/h

6.50x10–2 mol (6.022x1023 nuclei/h)

= 2.5471x10–10 = 2.55x10–10 h–1

Plan: Use the half-life of 24Na to find k. Substitute the value of k, initial activity (A0), and time of decay (4 days)

into the integrated first-order rate equation to solve for activity at a later time (At).

Solution:

ln 2

ln 2

k=

=

= 0.0462098 h–1

t1/2

15 h

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24-2

ln At = ln A0 – kt

ln At = ln (2.5x109) – (0.0462098 h–1)(4.0 days)(24 h/day)

ln At = 17.203416

At = 2.960388x107 = 3.0x107 d/s

24.5B

Plan: Use the half-life of 59Fe to find k. Substitute the value of k and time of decay (17 days) into the integrated

first-order rate equation. Assume that A0 is 1.0 (100% of the original sample) and solve for At, the fraction of the

sample remaining after 17 days. Subtract the fraction remaining from the original sample (1.0) to calculate the

fraction that has decayed.

Solution:

ln 2

ln 2

k=

=

= 1.5576x10–2 days–1

44.5 days

t1/2

ln At = ln A0 – kt

ln At = ln (1.0) – (1.5576x10–2 days–1)(17 days)

ln At = –0.2648

At = 0.7674 = fraction of iron-59 remaining

Fraction of iron-59 decayed = 1.0 – 0.7674 = 0.2326 = 0.23 = fraction of iron-59 that has decayed

24.6A

Plan: The wood from the case came from a living organism, so A0 equals 15.3 d/min•g. Substitute the current

activity of the case (At), A0, and k into the first-order rate expression and solve for t. Find k from the half-life of

carbon (5730 yr).

Solution:

ln 2

ln 2

k=

=

= 1.209680943x10–4 yr–1

5730 yr

t1/2

ln At = ln A0 – kt

ln [9.41 d/min•g] = ln [15.3 d/min•g] – (1.209680943x10–4 yr–1)(t)

–0.486079875 = – (1.209681x10–4 yr–1)(t)

t = 4018.2 = 4.02x103 years

24.6B

Plan: The woolen tunic came from a living organism, so A0 equals 15.3 d/min•g. Substitute the current activity of

the tunic (At), A0, and k into the first-order rate expression and solve for t. Find k from the half-life of carbon

(5730 yr).

Solution:

ln 2

ln 2

k=

=

= 1.209680943x10–4 yr–1

5730 yr

t1/2

ln At = ln A0 – kt

ln [12.87 d/min•g] = ln [15.3 d/min•g] – (1.209680943x10–4 yr–1)(t)

–0.172953806 = – (1.209681x10–4 yr–1)(t)

t = 1429.7473 = 1430 years

24.7A

Plan: Nickel-58 has 28 protons and 30 neutrons in its nucleus. Calculate the change in mass (m) in one 58Ni

atom, convert to MeV and divide by 58 to obtain binding energy/nucleon.

Solution:

m = [(28 x mass H atom) + (30 x mass neutron)] – mass 58Ni atom

m = [(28 x 1.007825 amu) + (30 x 1.008665)] – 57.935346 amu

m = 0.543704 amu

931.5 MeV

Binding energy (MeV) = (0.543704 amu)

Binding Energy/nucleon =

56

506.460276 MeV

58 nucleons

1 amu

= 506.460276 MeV

= 8.7321 = 8.732 MeV/nucleon

The BE/nucleon of Fe is 8.790 MeV/nucleon. The energy per nucleon holding the 58Ni nucleus together is less

than that for 56Fe (8.732 < 8.790), so 58Ni is less stable than 56Fe.

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24-3

24.7B

Plan: Uranium-235 has 92 protons and 143 neutrons in its nucleus. Calculate the change in mass (m) in one 235U

atom, convert to MeV and divide by 235 to obtain binding energy/nucleon.

Solution:

m = [(92 x mass H atom) + (143 x mass neutron)] – mass 235U atom

m = [(92 x 1.007825 amu) + (143 x 1.008665)] – 235.043924 amu

m = 1.915071 amu

931.5 MeV

Binding energy (MeV) = 1.915071 amu

= 1783.888637 MeV

1 amu

1783.888637 MeV

= 7.591015 = 7.591 MeV/nucleon

235 nucleons

12

The BE/nucleon of C is 7.680 MeV/nucleon. The energy per nucleon holding the 235U nucleus together is less than that

for 12C (7.591 < 7.680), so 235U is less stable than 12C.

Binding Energy/nucleon =

CHEMICAL CONNECTIONS BOXED READING PROBLEMS

B24.1

In the s-process, a nucleus captures a neutron sometime over a long period of time. Then the nucleus emits a beta

particle to form another element. The stable isotopes of most heavy elements up to 209Bi form by the s-process.

The r-process very quickly forms less stable isotopes and those with A greater than 230 by multiple neutron

captures, followed by multiple beta decays.

B24.2

Plan: Find the change in mass of the reaction by subtracting the mass of the products from the

mass of the reactants and convert the change in mass to energy with the conversion factor

between amu and MeV. Convert the energy per atom to energy per mole by multiplying by

Avogadro’s number.

Solution:

m = mass of reactants – mass of products

= [(4)(1.007825)] – [4.00260 + (2)(5.48580x10–4)]

= 4.031300 – 4.003697 = 0.02760 amu /4He atom = 0.02760284 g/mol 4He

0.02760284 amu 4 He 931.5 MeV

Energy (MeV/atom) =

= 25.7120 = 25.71 MeV/atom

1 atom

1 amu

Convert atoms to moles using Avogadro’s number.

23

25.7120 MeV 6.022x10 atoms

25

25

Energy =

= 1.54838x10 = 1.548x10 MeV/mol

atom

1

mol

B24.3

The simultaneous fusion of three nuclei is a termolecular process. Termolecular processes have a very low

probability of occurring. The bimolecular fusion of 8Be with 4He is more likely.

B24.4

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the

right side must be equal.

Solution:

210

83 Bi

210

84 Po

206

82 Pb

209

82 Pb

210

0

84 Po + 1

206

4

82 Pb + 2

1

209

+ 3 0 n 82 Pb

210

0

83 Bi + 1

210

84 Po is Nuclide A

206

82 Pb is Nuclide B

209

82 Pb is Nuclide C

210

83 Bi is Nuclide D

END–OF–CHAPTER PROBLEMS

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24-4

24.1

a) Chemical reactions are accompanied by relatively small changes in energy while nuclear reactions are

accompanied by relatively large changes in energy.

b) Increasing temperature increases the rate of a chemical reaction but has no effect on a nuclear reaction.

c) Both chemical and nuclear reaction rates increase with higher reactant concentrations.

d) If the reactant is limiting in a chemical reaction, then more reactant produces more product and the yield

increases in a chemical reaction. The presence of more radioactive reagent results in more decay product, so a

higher reactant concentration increases the yield in a nuclear reaction.

24.2

a) The percentage of sulfur atoms that are sulfur-32 is 95.02%, the same as the relative abundance of 32S.

b) The atomic mass is larger than the isotopic mass of 32S. Sulfur-32 is the lightest isotope, as stated in the

problem, so the other 5% of sulfur atoms are heavier than 31.972070 amu. The average mass of all the sulfur

atoms will therefore be greater than the mass of a sulfur-32 atom.

24.3

a) She found that the intensity of emitted radiation is directly proportional to the concentration of the element in

the various samples, not to the nature of the compound in which the element occurs.

b) She found that certain uranium minerals were more radioactive than pure uranium, which implied that they

contained traces of one or more as yet unknown, highly radioactive elements. Pitchblende is the principal ore of

uranium.

24.4

Plan: Radioactive decay that produces a different element requires a change in atomic number (Z, number of

protons).

Solution:

A

ZX

A = mass number (protons + neutrons)

Z = number of protons (positive charge)

X = symbol for the particle

N = A – Z (number of neutrons)

a) Alpha decay produces an atom of a different element, i.e., a daughter with two less protons and two less

neutrons.

A

ZX

A 4

4

Z 2Y + 2 He

A

Z 1Y

2 fewer protons, 2 fewer neutrons

b) Beta decay produces an atom of a different element, i.e., a daughter with one more proton and one less neutron.

A neutron is converted to a proton and particle in this type of decay.

0

+ 1

1 more proton, 1 less neutron

c) Gamma decay does not produce an atom of a different element and Z and N remain unchanged.

A

ZX

A

Z X*

ZA X + 00

( ZA X * = energy rich state), no change in number of protons or neutrons.

d) Positron emission produces an atom of a different element, i.e., a daughter with one less proton and one more

neutron. A proton is converted into a neutron and positron in this type of decay.

0

+ 1

1 less proton, 1 more neutron

e) Electron capture produces an atom of a different element, i.e., a daughter with one less proton and one more

neutron. The net result of electron capture is the same as positron emission, but the two processes are different.

A

ZX

A

ZX

A

Z 1Y

0

A

+ 1e Z 1Y

1 less proton, 1 more neutron

A different element is produced in all cases except (c).

24.5

The key factor that determines the stability of a nuclide is the ratio of the number of neutrons to the number of

protons, the N/Z ratio. If the N/Z ratio is either too high or not high enough, the nuclide is unstable and decays.

3

2 He

2

2 He

24.6

N/Z = 1/2

N/Z = 0/2, thus it is more unstable.

A neutron-rich nuclide decays to convert neutrons to protons while a neutron-poor nuclide decays to convert

protons to neutrons. The conversion of neutrons to protons occurs by beta decay:

1

0n

1

1p +

0

1

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24-5

The conversion of protons to neutrons occurs by either positron decay:

1

1p

1

0

0 n + 1

or electron capture:

1

1p

0

1

+ 1e 0 n

Neutron-rich nuclides, with a high N/Z, undergo decay. Neutron-poor nuclides, with a low N/Z, undergo

positron decay or electron capture.

24.7

Both positron emission and electron capture increase the number of neutrons and decrease the number of protons.

The products of both processes are the same. Positron emission is more common than electron capture among

lighter nuclei; electron capture becomes increasingly common as nuclear charge increases. For Z < 20, +

emission is more common; for Z > 80, electron capture is more common.

24.8

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the

right side must be equal.

Solution:

a)

b)

c)

24.9

a)

b)

c)

24.10

234

4

230

92 U 2 He + 90Th

232

0

232

93 Np + 1e 92 U

12

0

12

7 N 1 + 6 C

26

11 Na

223

87 Fr

212

83 Bi

Mass: 234 = 4 + 230;

Charge: 92 = 2 + 90

Mass: 232 + 0 = 232;

Charge: 93 + (–1) = 92

Mass: 12 = 0 + 12;

Charge: 7 = 1 + 6

0

26

1 + 12 Mg

0

223

1 + 88 Ra

4

208

2 + 81Tl

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the

right side must be equal.

Solution:

a) The process converts a neutron to a proton, so the mass number is the same, but the atomic number increases by

one.

27

12 Mg

0

23

12 Mg

0

27

1 + 13 Al

Mass: 27 = 0 + 27;

Charge: 12 = –1 + 13

b) Positron emission decreases atomic number by one, but not mass number.

23

1 + 11 Na

Mass: 23 = 0 + 23;

Charge: 12 = 1 + 11

c) The electron captured by the nucleus combines with a proton to form a neutron, so mass number is constant,

but atomic number decreases by one.

103

46 Pd

24.11

a)

b)

c)

24.12

+

0

1e

103

45 Rh

Mass: 103 + 0 = 103;

Charge: 46 + (–1) = 45

32

0

32

14 Si 1 + 15 P

218

4

214

84 Po 2 + 82 Pb

110

0

110

49 In + 1e 48 Cd

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the

right side must be equal.

Solution:

a) In other words, an unknown nuclide decays to give Ti-48 and a positron.

48

23V

48

0

22Ti + 1

Mass: 48 = 48 + 0;

Charge: 23 = 22 + 1

b) In other words, an unknown nuclide captures an electron to form Ag-107.

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24-6

107

48 Cd

0

107

+ 1e 47 Ag Mass: 107 + 0 = 107;

Charge: 48 + (–1) = 47

c) In other words, an unknown nuclide decays to give Po-206 and an alpha particle.

210

86 Rn

24.13

a)

b)

c)

24.14

241

94 Pu

228

88 Ra

207

85 At

206

84 Po

4

+ 2 He Mass: 210 = 206 + 4;

Charge: 86 = 84 + 2

241

0

95 Am + 1

228

0

89 Ac + 1

203

4

83 Bi + 2

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and

the right side must be equal.

Solution:

a) In other words, an unknown nuclide captures an electron to form Ir-186.

186

78 Pt

0

186

+ 1e 77 Ir

Mass: 186 + 0 = 186;

Charge: 78 + (–1) = 77

b) In other words, an unknown nuclide decays to give Fr-221 and an alpha particle.

225

89 Ac

221

4

87 Fr + 2 He Mass: 225 = 221 + 4;

Charge: 89 = 87 + 2

c) In other words, an unknown nuclide decays to give I-129 and a beta particle.

129

52Te

24.15

a)

b)

c)

24.16

129

53 I

+

0

1

Mass: 129 = 129 + 0;

52

52

0

26 Fe 25 Mn + 1

219

215

4

86 Rn 84 Po + 2

81

0

81

37 Rb + 1e 36 Kr

Plan: Look at the N/Z ratio, the ratio of the number of neutrons to the number of protons. If the N/Z ratio falls in

the band of stability, the nuclide is predicted to be stable. For stable nuclides of elements with atomic number

greater than 20, the ratio of number of neutrons to number of protons (N/Z) is greater than one. In addition, the

ratio increases gradually as atomic number increases. Also check for exceptionally stable numbers of neutrons

and/or protons – the “magic” numbers of 2, 8, 20, 28, 50, 82, and (N = 126). Also, even numbers of protons and

or neutrons are related to stability whereas odd numbers are related to instability.

Solution:

a)

20

8O

appears stable because its Z (8) value is a magic number, but its N/Z ratio (20 8)/8 = 1.50 is too high and

this nuclide is above the band of stability;

59

b) 27 Co

9

c) 3 Li

a)

b)

c)

24.18

20

8O

is unstable.

might look unstable because its Z value is an odd number, but its N/Z ratio (59 27)/27 = 1.19 is in the

band of stability, so

24.17

Charge: 52 = 53 + (–1)

59

27 Co

appears stable.

appears unstable because its N/Z ratio (9 3)/3 = 2.00 is too high and is above the band of stability.

146

60 Nd

114

48 Cd

88

42 Mo

N/Z = 86/60 = 1.4

Stable, N/Z ok

N/Z = 66/48 = 1.4

Stable, N/Z ok

N/Z = 46/42 = 1.1

Unstable, N/Z too small for this region of the band

Plan: Look at the N/Z ratio, the ratio of the number of neutrons to the number of protons. If the N/Z ratio falls in

the band of stability, the nuclide is predicted to be stable. For stable nuclides of elements with atomic number

greater than 20, the ratio of number of neutrons to number of protons (N/Z) is greater than one. In addition, the

ratio increases gradually as atomic number increases. Also check for exceptionally stable numbers of neutrons

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24-7

and/or protons – the “magic” numbers of 2, 8, 20, 28, 50, 82, and (N = 126). Also, even numbers of protons and

or neutrons are related to stability whereas odd numbers are related to instability.

Solution:

a) For the element iodine Z = 53. For iodine-127, N = 127 53 = 74. The N/Z ratio for 127I is 74/53 = 1.4. Of the

examples of stable nuclides given in the book, 107Ag has the closest atomic number to iodine. The N/Z ratio for

107

Ag is 1.3. Thus, it is likely that iodine with six additional protons is stable with an N/Z ratio of 1.4.

b) Tin is element number 50 (Z = 50). The N/Z ratio for 106Sn is (106 50)/50 = 1.1. The nuclide 106Sn is unstable

with an N/Z ratio that is too low.

c) For 68As, Z = 33 and N = 68 33 = 35 and N/Z = 1.1. The ratio is within the range of stability, but the nuclide is

most likely unstable because there is an odd number of both protons and neutrons.

24.19

a)

b)

c)

24.20

48

19 K

79

35 Br

33

18 Ar

N/Z = 29/19 = 1.5

Unstable, N/Z too large for this region of the band

N/Z = 44/35 = 1.3

Stable, N/Z okay

N/Z = 14/18 = 0.78

Unstable, N/Z too small

Plan: Calculate the N/Z ratio for each nuclide. A neutron-rich nuclide decays to convert neutrons to protons while

a neutron-poor nuclide decays to convert protons to neutrons. Neutron-rich nuclides, with a high N/Z, undergo

decay. Neutron-poor nuclides, with a low N/Z, undergo positron decay or electron capture. For Z < 20, +

emission is more common; for Z > 80, e– capture is more common. Alpha decay is the most common means of

decay for a heavy, unstable nucleus (Z > 83).

Solution:

a)

238

92 U:

Nuclides with Z > 83 decay through decay.

48

b) The N/Z ratio for 24 Cr is (48 – 24)/24 = 1.00. This number is below the band of stability because N is too

low and Z is too high. To become more stable, the nucleus decays by converting a proton to a neutron, which is

positron decay. Alternatively, a nucleus can capture an electron and convert a proton into a neutron through

electron capture.

50

c) The N/Z ratio for 25 Mn is (50 – 25)/25 = 1.00. This number is below the band of stability, so the nuclide

undergoes positron decay or electron capture.

24.21

a)

b)

c)

111

47 Ag

41

17 Cl

110

44 Ru

beta decay N/Z = 1.4 which is too high

beta decay N/Z = 1.4 which is too high

beta decay N/Z = 1.5 which is too high

24.22

Plan: Calculate the N/Z ratio for each nuclide. A neutron-rich nuclide decays to convert neutrons to protons while

a neutron-poor nuclide decays to convert protons to neutrons. Neutron-rich nuclides, with a high N/Z, undergo

decay. Neutron-poor nuclides, with a low N/Z, undergo positron decay or electron capture. For Z < 20, +

emission is more common; for Z > 80, e– capture is more common. Alpha decay is the most common means of

decay for a heavy, unstable nucleus (Z > 83).

Solution:

a) For carbon-15, N/Z = 9/6 = 1.5, so the nuclide is neutron-rich. To decrease the number of neutrons and increase

the number of protons, carbon-15 decays by beta decay.

b) The N/Z ratio for 120Xe is 66/54 = 1.2. Around atomic number 50, the ratio for stable nuclides is larger than 1.2,

so 120Xe is proton-rich. To decrease the number of protons and increase the number of neutrons, the xenon-120

nucleus either undergoes positron emission or electron capture.

c) Thorium-224 has an N/Z ratio of 134/90 = 1.5. All nuclides of elements above atomic number 83 are unstable

and decay to decrease the number of both protons and neutrons. Alpha decay by thorium-224 is the most likely

mode of decay.

24.23

a)

106

49 In

positron decay or electron capture N/Z = 1.2

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24-8

b)

c)

24.24

141

63 Eu

241

95 Am

positron decay or electron capture N/Z = 1.2

alpha decay

N/Z = 1.5

Plan: Stability results from a favorable N/Z ratio, even numbers of N and/or Z, and the occurrence of magic

numbers.

Solution:

The N/Z ratio of

52

24 Cr

is (52 24)/24 = 1.17, which is within the band of stability. The fact that Z is even does not

account for the variation in stability because all isotopes of chromium have the same Z. However,

neutrons, so N is both an even number and a magic number for this isotope only.

52

24 Cr has

28

40

20 Ca

24.25

N/Z = 20/20 = 1.0

It lies in the band of stability, and N and Z are both even and magic.

24.26

237

4

233

93 Np 2 + 91 Pa

233

0

233

91 Pa 1 + 92 U

233

4

229

92 U 2 + 90Th

229

4

225

90Th 2 + 88 Ra

24.27

Alpha emission produces helium ions which readily pick up electrons to form stable helium atoms.

24.28

The equation for the nuclear reaction is 92 U 82 Pb + __ 1 + __ 2 He

To determine the coefficients, notice that the beta particles will not impact the mass number. Subtracting the mass

number for lead from the mass number for uranium will give the total mass number for the alpha particles

released, 235 207 = 28. Each alpha particle is a helium nucleus with mass number 4. The number of helium

atoms is determined by dividing the total mass number change by 4, 28/4 = 7 helium atoms or seven alpha

particles. The equation is now

235

235

92 U

207

0

207

0

4

4

82 Pb + __ 1 + 7 2 He

To find the number of beta particles released, examine the difference in number of protons (atomic number)

between the reactant and products. Uranium, the reactant, has 92 protons. The atomic number in the products, lead

atom and 7 helium nuclei, total 96. To balance the atomic numbers, four electrons (beta particles) must be emitted

to give the total atomic number for the products as 96 4 = 92, the same as the reactant. In summary, seven alpha

particles and four beta particles are emitted in the decay of uranium-235 to lead-207.

235

92 U

207

82 Pb

+

0

1

4

+ 7 2 He

24.29

a) In a scintillation counter, radioactive emissions are detected by their ability to excite atoms and cause them to

emit light.

b) In a Geiger-Müller counter, radioactive emissions produce ionization of a gas that conducts a current to a

recording device.

24.30

Since the decay rate depends only on the number of radioactive nuclei, radioactive decay is a first-order process.

24.31

No, it is not valid to conclude that t1/2 equals 1 min because the number of nuclei is so small (six nuclei). Decay

rate is an average rate and is only meaningful when the sample is macroscopic and contains a large number of

nuclei, as in the second case. Because the second sample contains 6x1012 nuclei, the conclusion that

t1/2 = 1 min is valid.

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24-9

24.32

High-energy neutrons in cosmic rays enter the upper atmosphere and keep the amount of 14C nearly constant

through bombardment of ordinary 14N atoms. This 14 C is absorbed by living organisms, so its proportion stays

relatively constant there also.

14

7N

24.33

1

+ 0n

14

6C

1

+ 1H

Plan: Specific activity of a radioactive sample is its decay rate per gram. Calculate the specific activity by dividing

the number of particles emitted per second (disintegrations per second = dps) by the mass of the sample. Convert

disintegrations per second to Ci by using the conversion factor between the two units.

Solution:

1 Ci = 3.70x1010 dps

1.56x106 dps 1 mg

1 Ci

= 2.55528x10–2 = 2.56x10–2 Ci/g

Specific activity (Ci/g) =

1.65 mg 103 g 3.70x1010 dps

24.34

24.35

24.36

24.37

4.13x108 d 1 h

3600 s

h

1 Ci

–6

–6

Specific activity (Ci/g) =

= 1.1925x10 = 1.2x10 Ci/g

10

2.6 g

3.70x10

dps

Plan: Specific activity of a radioactive sample is its decay rate per gram. Calculate the specific activity by dividing

the number of particles emitted per second (disintegrations per second = dps) by the mass of the sample. Convert

disintegrations per second to Bq by using the conversion factor between the two units.

Solution:

A becquerel is a disintegration per second (dps).

7.4x104 d 1 min

min 60 s 1 Bq

8

8

Specific activity (Bq/g) =

= 1.43745x10 = 1.4x10 Bq/g

106 g 1 dps

8.58 g

1 g

3.77x107 d 1 min

min

60 s 1 Bq

Specific activity (Bq/g) =

1 dps = 587.2274 = 587 Bq/g

103 g

1.07 kg

1

kg

Plan: The decay constant is the rate constant for the first-order reaction.

Solution:

N

Decay rate =

= kN

t

1 atom

= k(1x1012 atom)

day

k = 1x1012 d1

24.38

N

= kN

t

(2.8x1012 atom/1.0 yr) = k(1 atom)

Decay rate =

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24-10

k = 2.8x1012 yr1

24.39

Plan: The rate constant, k, relates the number of radioactive nuclei to their decay rate through the equation

A = kN. The number of radioactive nuclei is calculated by converting moles to atoms using Avogadro’s number.

The decay rate is 1.39x105 atoms/yr or more simply, 1.39x105 yr–1 (the disintegrations are assumed).

Solution:

N

Decay rate = A = –

= kN

t

1.00x1012 mol 6.022x1023 atoms

1.39x105 atoms

= k

1.00 yr

1 mol

1.39x105 atom/yr = k(6.022x1011 atom)

k = (1.39x105 atom/yr)/6.022x1011 atom

k = 2.30820x10–7 = 2.31x10–7 yr–1

24.40

24.41

N

= kN

t

– (–1.07x1015 atom/1.00 h) = k[(6.40x10–9 mol)(6.022x1023 atom/mol)]

(1.07x1015 atom/1.00 h) = k (3.85408x1015 atom)

k = [(1.07x1015 atom/1.00 h)]/(3.85408x1015 atom)

k = 0.2776 = 0.278 h–1

Decay rate = A = –

Plan: Radioactive decay is a first-order process, so the integrated rate law is ln Nt = ln N0 – kt

First find the value of k from the half-life and use the integrated rate law to find Nt. The time unit in

the time and the k value must agree.

Solution:

t1/2 = 1.01 yr

t = 3.75x103 h

ln 2

ln 2

t1/2 =

or k =

t1/2

k

ln 2

= 0.686284 yr–1

1.01 yr

ln Nt = ln N0 – kt

k=

1 d 1 yr

ln Nt = ln [2.00 mg] – (0.686284 yr–1) 3.75x103 h

24 h 365 d

ln Nt = 0.399361

Nt = e0.399361

Nt = 1.49087 = 1.49 mg

24.42

t1/2 = 1.60x103 yr

t=?h

ln 2

ln 2

k=

=

= = 0.000433217 yr–1

t1/2 1.60x103 yr

ln [0.185 g] = ln [2.50 g] – (0.000433217 yr–1)(t)

t = 6010.129 yr

365 d 24 h

7

7

t = 6010.129 yr

= 5.264873x10 = 5.26x10 h

1

yr

1

d

24.43

Plan: Lead-206 is a stable daughter of 238U. Since all of the 206Pb came from 238U, the starting amount of 238U was

(270 mol + 110 mol) = 380 mol = N0. The amount of 238U at time t (current) is 270 mol = Nt. Find k from the

first-order rate expression for half-life, and then substitute the values into the integrated rate law and solve

for t.

Solution:

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24-11

ln 2

ln 2

or k =

t1/2

k

ln 2

k=

= 1.540327x10–10 yr–1

4.5x109 yr

t1/2 =

ln Nt = ln N0 – kt

ln

or

ln

N0

= kt

Nt

380 mol

= (1.540327x10–10 yr–1)(t)

270 mol

0.3417492937 = (1.540327x10–10 yr–1)(t)

t = 2.21868x109 = 2.2x109 yr

24.44

The ratio (0.735) equals Nt/N0 so N0/Nt = 1.360544218

ln 2

ln 2

k=

=

= = 1.2096809x10–4 yr–1

t1/2

5730 yr

N0

= kt

Nt

ln 1.360544218 = (1.2096809x10–4 yr–1)(t)

0.30788478 = (1.2096809x10–4 yr–1)(t)

t = 2.54517x103 = 2.54x103 yr

ln

24.45

Plan: The specific activity of the potassium-40 is the decay rate per mL of milk. Use the conversion factor

1 Ci = 3.70x1010 disintegrations per second (dps) to find the disintegrations per mL per s; convert the time

unit to min and change the volume to 8 oz.

Solution:

6 x1011 mCi 103 Ci 3.70x1010 dps 60 s 1000 mL 1 qt 1 cup

Activity =

8 oz

mL

1 Ci

1 mCi

1 min 1.057 qt 4 cups 8 oz

= 31.50426 = 30 dpm

24.46

Plutonium-239 (t1/2 = 2.41x104 yr)

Time = 7(t1/2) = 7(2.41x104 yr) = 1.6870x105 = 1.69x105 yr

24.47

Plan: Both Nt and N0 are given: the number of nuclei present currently, Nt, is found from the moles of 232Th. Each

fission track represents one nucleus that disintegrated, so the number of nuclei disintegrated is added to the

number of nuclei currently present to determine the initial number of nuclei, N0. The rate constant, k, is calculated

from the half-life. All values are substituted into the first-order decay equation to find t.

Solution:

ln 2

ln 2

t1/2 =

or k =

t1/2

k

ln 2

k=

=4.95105129x10–11 yr–1

10

1.4x10 yr

6.022x1023 Th atoms

9

Nt = 3.1x10 15 mol Th

= 1.86682x10 atoms Th

1 mol Th

9

4

N0 =1.86682x10 atoms + 9.5x10 atoms = 1.866915x109 atoms

N

ln 0 = kt

or

ln Nt = ln N0 – kt

Nt

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24-12

1.866915x109 atoms

= (4.95105129x10–11 yr–1)(t)

1.86682x109 atoms

5.088738x10–5 = (4.95105129x10–11 yr–1)(t)

t = 1.027809x106 = 1.0x106 yr

ln

24.48

The mole relationship between 40K and 40Ar is 1:1. Thus, 1.14 mmol 40Ar = 1.14 mmol 40K decayed.

ln 2

ln 2

k=

=

= 5.5451774x10–10 yr–1

t1/2

1.25x109 yr

ln

ln

N0

= kt

Nt

1.38 1.14 mmol

= (5.5451774x10–10 yr–1)(t)

1.38 mmol

0.6021754 = (5.5451774x10–10 yr–1)(t)

t = 1.08594x109 = 1.09x109 yr

27

13 Al

4

30

1

24.49

+ 2 He 15 P + 0 n

They experimentally confirmed the existence of neutrons, and were the first to produce an artificial radioisotope.

24.50

Both gamma radiation and neutron beams have no charge, so neither is deflected by electric or magnetic fields.

Neutron beams differ from gamma radiation in that a neutron has mass approximately equal to that of a proton.

Researchers observed that a neutron beam could induce the emission of protons from a substance. Gamma rays do

not cause such emissions.

24.51

A proton, for example, exits the first tube just when it becomes positive and the next tube becomes negative.

Pushed by the first tube and pulled by the second, the proton accelerates across the gap between them.

24.52

Protons are repelled from the target nuclei due to the interaction of like (positive) charges. Higher energy is

required to overcome the repulsion.

24.53

Plan: In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the

right side must be equal. In the shorthand notation, the nuclide to the left of the parentheses is the reactant while

the nuclide written to the right of the parentheses is the product. The first particle inside the parentheses is the

projectile particle while the second substance in the parentheses is the ejected particle.

Solution:

a) An alpha particle is a reactant with 10B and a neutron is one product. The mass number for the reactants is

10 + 4 = 14. So, the missing product must have a mass number of 14 – 1 = 13. The total atomic number for the

reactants is 5 + 2 = 7, so the atomic number for the missing product is 7.

10

5B

4

1

13

28

14 Si

2

242

96 Cm

4

1

+ 2 He 2 0 n +

+ 2 He 0 n + 7 N

b) A deuteron (2H) is a reactant with 28Si and 29P is one product. For the reactants, the mass number is 28 + 2 = 30

and the atomic number is 14 + 1 = 15. The given product has mass number 29 and atomic number 15, so the

missing product particle has mass number 1 and atomic number 0. The particle is thus a neutron.

1

29

+ 1H 0 n + 15 P

c) The products are two neutrons and 244Cf with a total mass number of 2 + 244 = 246, and an atomic number of

98. The given reactant particle is an alpha particle with mass number 4 and atomic number 2. The missing reactant

must have mass number of 246 – 4 = 242 and atomic number 98 – 2 = 96. Element 96 is Cm.

24.54

a)

31

15 P

1

1

+ 1H + 0 n +

31

P (, p, n) 29Si

244

98 Cf

29

14 Si

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24-13

24.55

24.56

b)

252

98 Cf

+ 5 B 5 0 n + 103 Lr

252

Cf (10B, 5n) 257Lr

10

c)

238

92 U

4

1

+ 2 He 3 0 n +

238

U (, 3n) 239Pu

a)

249

98 Cf

249

98 Cf

249

98 Cf

+

+

1

257

239

94 Pu

12

257

1

6 C 104 Rf + 4 0 n

15

260

1

7 N 105 Db + 4 0 n

18

263

1

8 O 106 Sg + 4 0 n

+

b) 249Cf (12C, 4n) 257Rf

249

Cf (15N, 4n) 260Db

249

Cf (18O, 4n) 263Sg

Gamma radiation has no mass or charge while alpha particles are massive and highly charged. These differences

account for the different effect on matter that these two types of radiation have. Alpha particles interact with

matter more strongly than gamma particles due to their mass and charge. Therefore alpha particles penetrate

matter very little. Gamma rays interact very little with matter due to the lack of mass and charge. Therefore

gamma rays penetrate matter more extensively.

24.57

In the process of ionization, collision of matter with radiation dislodges an electron. The free electron and the

positive ion that result are referred to as an ion-pair.

24.58

Ionizing radiation is more dangerous to children because their rapidly dividing cells are more susceptible to

radiation than an adult’s slowly dividing cells.

24.59

The hydroxyl free radical forms more free radicals which go on to attack and change surrounding biomolecules,

whose bonding and structure are delicately connected with their function. These changes are irreversible, as

opposed to the reversible changes produced by OH–.

24.60

Plan: The rad is the amount of radiation energy absorbed in J per body mass in kg: 1 rad = 0.01 J/kg. Change the

mass unit from pounds to kilograms. The conversion factor between rad and gray is 1 rad = 0.01 Gy.

Solution:

3.3x10 7 J 2.205 lb

1 rad

a) Dose (rad) =

= 5.39x10–7 = 5.4x10–7 rad

135 lb 1 kg 1x102 J /kg

0.01

gy

–9

–9

b) Gray (rad) = 5.39x107 rad

= 5.39x10 = 5.4x10 Gy

1 rad

24.61

1 rad

a) Dose (rad) = (8.92x10–4 Gy)

= 0.0892 rad

0.01 Gy

0.01 J/kg

–3

–3

b) Energy (J) = (0.0892 rad)

3.6 kg = 3.2112x10 = 3.2x10 J

1

rad

24.62

Plan: Multiply the number of particles by the energy of one particle to obtain the total energy absorbed.

Convert the energy to dose in grays with the conversion factor 1 rad = 0.01 J/kg = 0.01 Gy. To find the millirems,

convert grays to rads and multiply rads by RBE to find rems. Convert rems to mrems. Convert the dose to

sieverts with the conversion factor 1 rem = 0.01 Sv.

Solution:

a) Energy (J) absorbed = 6.0x105 8.74x1014 J/ = 5.244x10–8 J

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24-14

5.244x108 J 1 rad 0.01 Gy

–10

–10

Dose (Gy) =

= 7.4914x10 = 7.5x10 Gy

J

70. kg

0.01 kg 1 rad

1 mrem

1 rad

–5

–5

b) rems = rads x RBE = 7.4914x1010 Gy

= 7.4914x10 = 7.5x10 mrem

1.0 3

0.01

Gy

10 rem

10 3 rem 0.01 Sv

= 7.4914x10–10 = 7.5x10–10 Sv

Sv = 7.4914x10 5 mrem

1 mrem 1 rem

1.77x10 2.20x10

a) Dose =

24.63

24.64

J / 103 g 1 rad

= 1.46943 = 1.47 rad

265 g

1 kg 0.01 J kg

b) Dose = (1.46943 rad)(0.01 Gy/1 rad) = 1.46943x10–2 = 1.47x10–2 Gy

c) Dose = (1.46943 rad)(0.75 rem/rad)(0.01 Sv/rem) = 1.10207x10–2 = 1.10x10–2 Sv

2.50 pCi 1x1012 Ci 3.70x1010 dps

3600 s 8.25x1013 J 1 rad

Dose =

65

h

1 Ci

95 kg 1 pCi

1 h disint. 0.01 J kg

= 1.8796974x10–8 = 1.9x10–8 rad

Dose = (1.8796974x10–8 rad)(0.01 Gy/1 rad) = 1.8796974x10–10 = 1.9x10–10 Gy

10

13

24.65

Use the time and disintegrations per second (Bq) to find the number of 60Co atoms that disintegrate, which equals

the number of particles emitted. The dose in rads is calculated as energy absorbed per body mass.

475 Bq 103 g 1 dps 5.05x1014 J

60 s 1 rad

Dose =

24.0

min

1.858 g 1 kg 1 Bq 1 disint.

1 min 0.01 J kg

= 1.8591x10–3 = 1.86x10–3 rad

24.66

A healthy thyroid gland incorporates dietary I – into I-containing hormones at a known rate. To assess thyroid

function, the patient drinks a solution containing a trace amount of Na131I, and a scanning monitor follows the

uptake of 131I into the thyroid. Technetium-99 is often used for imaging the heart, lungs, and liver.

24.67

NAA does not destroy the sample while chemical analysis does. Neutrons bombard a non-radioactive sample,

“activating” or energizing individual atoms within the sample to create radioisotopes. The radioisotopes decay

back to their original state (thus, the sample is not destroyed) by emitting radiation that is different for each

isotope.

24.68

In positron-emission tomography (PET), the isotope emits positrons, each of which annihilates a nearby electron.

In the process, two photons are emitted simultaneously, 180° apart from each other. Detectors locate the sites

and the image is analyzed by computer.

24.69

The concentration of 59 Fe in the steel sample and the volume of oil would be needed.

24.70

The oxygen in formaldehyde comes from methanol because the oxygen isotope in the methanol reactant appears

in the formaldehyde product. The oxygen isotope in the chromic acid reactant appears in the water product, not

the formaldehyde product. The isotope traces the oxygen in methanol to the oxygen in formaldehyde.

24.71

The mass change in a chemical reaction was considered too minute to be significant and too small to measure with

even the most sophisticated equipment.

24.72

When a nucleus forms from its nucleons, there is a decrease in mass called the mass defect. This decrease in mass

is due to mass being converted to energy to hold the nucleus together. This energy is called the binding energy.

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24-15

24.73

Energy is released when a nuclide forms from nucleons. The nuclear binding energy is the amount of energy

holding the nucleus together. Energy is absorbed to break the nucleus into nucleons and is released when

nucleons “come together.”

24.74

The binding energy per nucleon is the average amount of energy per each component (proton and neutron) part of

the nuclide. The binding energies per nucleon are helpful in comparing the stabilities of different combinations

and to provide information on the potential processes a nuclide can undergo to become more stable. The binding

energy per nucleon varies considerably. The greater the binding energy per nucleon, the more strongly the

nucleons are held together and the more stable the nuclide.

24.75

Plan: The conversion factors are: 1 MeV = 106 eV and 1 eV = 1.602x1019 J.

Solution:

106 eV

a) Energy (eV) = 0.01861 MeV

= 1.861x104 eV

1 MeV

106 eV 1.602x10 19 J

15

15

b) Energy (J) = 0.01861 MeV

= 2.981322x10 = 2.981x10 J

1 MeV

1 eV

24.76

1 eV

1000 J

= 9.8002x106 = 9.80x106 eV

a) Energy (eV) = 1.57x10 15 kJ

19

1

kJ

1.602x10

J

1 MeV

6

b) Energy (MeV) = 9.8002x10 eV 6

9.8002 = 9.80 MeV

10 eV

24.77

Plan: Convert moles of 239Pu to atoms of 239Pu using Avogadro’s number. Multiply the number of atoms by the

energy per atom (nucleus) and convert the MeV to J using the conversion 1 eV = 1.602x10–19 J.

Solution:

6.022x1023 atoms

23

Number of atoms = 1.5 mol 239 Pu

= 9.033x10 atoms

mol

5.243 MeV 106 eV 1.602x10 19 J

11

11

Energy (J) = 9.033x1023 atoms

= 7.587075x10 = 7.6x10 J

1

atom

1

MeV

1

eV

24.78

24.79

103 J

1 MeV

8.11x105 kJ

1 eV

1 mol 49 Cr

Energy (MeV) =

3.2x10 3 mol 49 Cr 1 kJ 1.602x1019 J 106 eV 6.022x1023 nuclei

= 2.6270 = 2.6 MeV

Plan: Oxygen-16 has eight protons and eight neutrons. First find the Δm for the nucleus by subtracting the given

mass of one oxygen atom from the sum of the masses of eight 1H atoms and eight neutrons. Use the conversion

factor 1 amu = 931.5 MeV to convert Δm to binding energy in MeV and divide the binding energy by the total

number of nucleons (protons and neutrons) in the oxygen nuclide to obtain binding energy per nucleon. Convert

Δm of one oxygen atom to MeV using the conversion factor for binding energy/atom. To obtain binding energy

per mole of oxygen, use the relationship E = mc2. m must be converted to units of kg/mol.

Solution:

Mass of 8 1H atoms = 8 x 1.007825 = 8.062600 amu

Mass of 8 neutrons = 8 x 1.008665 = 8.069320 amu

Total mass =16.131920 amu

m = 16.131920 15.994915 = 0.137005 amu/16O = 0.137005 g/mol 16O

0.137005 amu 16 O 931.5 MeV

a) Binding energy (MeV/nucleon) =

= 7.976259844 = 7.976 MeV/nucleon

16 nucleons

1 amu

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24-16

0.137005 amu 16 O 931.5 MeV

b) Binding energy (MeV/atom) =

= 127.6201575 = 127.6 MeV/atom

1 atom

1 amu

c) E = mc2

2

0.137005 g 16 O 1 kg

1 J 1 kJ

8

Binding energy (kJ/mol) =

3 2.99792x10 m/ s

3

2

mol

kg•m 2 10 J

10 g

s

= 1.23133577x1010 = 1.23134x1010 kJ/mol

24.80

m is calculated from the mass of 82 protons (1H) and 124 neutrons vs. the mass of the lead nuclide.

Mass of 82 1H atoms = 82 x 1.007825 = 82.641650 amu

Mass of 124 neutrons = 124 x 1.008665 = 125.074460 amu

Total mass = 207.716110 amu

m = 207.716110 205.974440 = 1.741670 amu/206Pb = 1.741670 g/mol 206Pb

1.741670 amu 206 Pb 931.5 MeV

a) Binding energy (MeV/nucleon) =

= 7.8755612 = 7.876 MeV/nucleon

206 nucleons

1 amu

1.741670 amu 206 Pb 931.5 MeV

b) Binding energy (MeV/atom) =

= 1622.3656 = 1622 MeV/atom

1 atom

1 amu

1.741670 g 206 Pb 1 kg

2.99792 x108 m / s

c) Binding energy (kJ/mol) =

103 g

mol

= 1.5653301x1011 = 1.56533x1011 kJ/mol

24.81

1kJ

1J

3

2

kg•m 2 10 J

s

2

Plan: Cobalt-59 has 27 protons and 32 neutrons. First find the Δm for the nucleus by subtracting the given mass of

one cobalt atom from the sum of the masses of 27 1H atoms and 32 neutrons. Use the conversion factor

1 amu = 931.5 MeV to convert Δm to binding energy in MeV and divide the binding energy by the total number

of nucleons (protons and neutrons) in the cobalt nuclide to obtain binding energy per nucleon. Convert Δm of

one cobalt atom to MeV using the conversion factor for binding energy/atom. To obtain binding energy per

mole of cobalt, use the relationship E = mc2. m must be converted to units of kg/mol.

Solution:

Mass of 27 1H atoms = 27 x 1.007825 = 27.211275 amu

Mass of 32 neutrons = 32 x 1.008665 = 32.27728 amu

Total mass = 59.488555 amu

m = 59.488555 – 58.933198 = 0.555357 amu/59Co = 0.555357 g/mol 59Co

0.555357 amu 59 Co 931.5 MeV

a) Binding energy (MeV/nucleon) =

= 8.768051619 = 8.768 MeV/nucleon

59 nucleons

1 amu

0.555357 amu 59 Co 931.5 MeV

b) Binding energy (MeV/atom) =

= 517.3150 = 517.3 MeV/atom

1 atom

1 amu

c) Use E = mc2

2

0.555357 g 59 Co 1 kg

1 J 1 kJ

8

Binding energy (kJ/mol) =

3 2.99792x10 m/s

3

2

mol

kg•m 2 10 J

10 g

s

= 4.9912845x1010 = 4.99128x1010 kJ/mol

24.82

m is calculated from the mass of 53 protons (1H) and 78 neutrons vs. the mass of the iodine nuclide.

Mass of 53 1H atoms = 53 x 1.007825 = 53.414725 amu

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24-17

Mass of 78 neutrons = 78 x 1.008665 = 78.675870 amu

Total mass = 132.090595 amu

m = 132.090595 – 130.906114 = 1.184481 amu/131I = 1.184481 g/mol 131I

1.184481 amu 131 I 931.5 MeV

= 8.422473676 = 8.422 MeV/nucleon

a) Binding energy (MeV/nucleon) =

131 nucleons 1 amu

1.184481 amu 131 I 931.5 MeV

b) Binding energy (MeV/atom) =

= 1103.34405 = 1103 MeV/atom

1 atom

1 amu

c) E = mc2

2

1.184481 g 131 I 1 kg

1 J 1 kJ

8

Binding energy (kJ/mol) =

3 2.99792x10 m/s

3

2

mol

kg•m 2 10 J

10 g

s

= 1.06455518x1011 = 1.06456x1011 kJ/mol

24.83

a)

80

35 Br

80

35 Br

0

1e

0

1

+

80

36 Kr

(reaction 1)

80

34 Se

+

(reaction 2)

b) Reaction 1:

m = 79.918528 – 79.916380 = 0.002148 amu

Reaction 2:

m = 79.918528 – 79.916520 = 0.002008 amu

Since E = (m)c2, the greater mass change (reaction 1) will release more energy.

24.84

The minimum number of neutrons from each fission event that must be absorbed by the nuclei to sustain the chain

reaction is one. In reality, due to neutrons lost from the fissionable material, two to three neutrons are generally

needed to continue a self-sustaining chain reaction.

24.85

In both radioactive decay and fission, radioactive particles are emitted, but the process leading to the emission is

different. Radioactive decay is a spontaneous process in which unstable nuclei emit radioactive particles and

energy. Fission occurs as the result of high-energy bombardment of nuclei with small particles that cause the

nuclides to break into smaller nuclides, radioactive particles, and energy.

In a chain reaction, all fission events are not the same. The collision between the small particle emitted in the

fission and the large nucleus can lead to splitting of the large nuclei in a number of ways to produce several

different products.

24.86

Enriched fissionable fuel is needed in the fuel rods to ensure a sustained chain reaction. Naturally occurring

235

U is only present in a concentration of 0.7%. This is consistently extracted and separated until its concentration

is between 3-4%.

24.87

a) Control rods are movable rods of cadmium or boron which are efficient neutron absorbers. In doing so, they

regulate the flux of neutrons to keep the reaction chain self-sustaining which prevents the core from overheating.

b) The moderator is the substance flowing around the fuel and control rods that slows the neutrons, making them

better at causing fission.

c) The reflector is usually a beryllium alloy around the fuel-rod assembly that provides a surface for neutrons that

leave the assembly to collide with and therefore, return to the fuel rods.

24.88

The water serves to slow the neutrons so that they are better able to cause a fission reaction. Heavy water

2

1

( 1 H2O or D2O) is a better moderator because it does not absorb neutrons as well as light water ( 1 H2O ) does, so

more neutrons are available to initiate the fission process. However, D2O does not occur naturally in great

abundance, so production of D2O adds to the cost of a heavy water reactor. In addition, if heavy water does

3

absorb a neutron, it becomes tritiated, i.e., it contains the isotope tritium, 1H , which is radioactive.

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24-18

24.89

The advantages of fusion over fission are the simpler starting materials (deuterium and tritium), and no long-lived

toxic radionuclide by-products.

24.90

Virtually all the elements heavier than helium, up to and including iron, are produced by nuclear fusion reactions

in successively deeper and hotter layers of massive stars. Iron is the point at which fusion reactions cease to be

energy producers. Elements heavier than iron are produced by a variety of processes, primarily during a

supernova event, which distribute the Sun’s ash into the cosmos to form next generation suns and planets. Thus,

the high cosmic and Earth abundance of iron is consistent with it being the most stable of all nuclei.

24.91

Mass of reactants: 3.01605 + 2.0140 = 5.03005 amu

Mass of products: 4.00260 + 1.008665 = 5.011265 amu

m = mass of reactants – mass of products = 5.03005 – 5.011265 = 0.018785 amu = 0.018785 g/mol

E = mc2

0.018785 g 1 kg

8

Energy (kJ/mol) =

3 2.99792x10 m/s

mol

10 g

= 1.6883064x109 = 1.69x109 kJ/mol

24.92

24.93

243

4

239

95 Am 2 He + 93 Np

239

0

239

93 Np 1 + 94 Pu

239

4

235

94 Pu 2 He + 92 U

239

239

235

93 Np , 94 Pu , and 92 U

1 J 1 kJ

3

2

kg•m 2 10 J

s

2

were present as products in the decay of Am-243.

Plan: Use the masses given in the problem to calculate the mass change (reactant – products) for the reaction.

The conversion factor between amu and kg is 1 amu = 1.66054x10–27 kg. Use the relationship E = mc2 to

convert the mass change to energy.

Solution:

243

239

4

a) 96 Cm 94 Pu + 2 He

m (amu) = 243.0614 amu (4.0026 + 239.0522) amu = 0.0066 amu

1.661x10 24 g 1 kg

–29

–29

m (kg) = 0.0066 amu

3 = 1.09626x10 = 1.1x10 kg

1

amu

10 g

1

J

–13

–13

b) E = mc2 = 1.09626x10 kg 2.99792x10 m/s

= 9.85266x10 = 9.9x10 J

2

kg•m 2

s

13

23

9.85266x10

J 6.022x10 reactions 1 kJ

8

8

c) E released =

3 = 5.93317x10 = 5.9x10 kJ/mol

reaction

mol

10 J

This is approximately one million times larger than a typical heat of reaction.

24.94

29

8

2

a) First, determine the amount of activity released by the 239Pu for the duration spent in the body (16 h) using the

relationship A = kN. The rate constant is derived from the half-life and N is calculated using the molar mass and

Avogadro’s number.

ln 2

1 day 1 h

ln 2

1 yr

–13 –1

k=

=

= 9.113904x10 s

4

365.25

day

24

h

3600

s

t1/2

2.41x10 yr

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24-19

10 6 g 1 mol Pu 6.022x1023 atoms Pu

15

N = 1.00 g Pu

= 2.5196653x10 atoms Pu

1 g 239 g Pu

1

mol

Pu

1 disint.

3

A = kN = (9.113904 x 10–13 s–1)( 2.5196653x1015 atoms Pu)

= 2.2963988x10 d/s

1

atom

Pu

Each disintegration releases 5.15 MeV, so d/s can be converted to MeV. Convert MeV to J (using

1.602x10–13 J = 1 MeV) and J to rad (using 0.01 J/kg = 1 rad).

2.2963988x103 d/s 5.15 MeV 1.602x10 13 J 1 rad 3600 s

Energy =

16 h

J

85 kg

MeV

disint.

0.01 kg 1 h

= 1.2838686x10–4 = 1.28x10–4 rads

b) Since 0.01 Gy = 1 rad, the worker receives:

Dose = (1.2838686x10–4 rad)(0.01 Gy/rad) = 1.2838686x10–6 = 1.28x10–6 Gy

24.95

Plan: Determine k for 14C using the half-life (5730 yr). Determine the mass of carbon in 4.58 g of CaCO3.

Divide the given activity of the C in d/min by the mass of carbon to obtain the activity in d/min•g; this is At and is

compared to the activity of a living organism (A0 = 15.3 d/min•g) in the integrated rate law, solving for t.

Solution:

ln 2

ln 2

k=

=

= 1.2096809x10–4 yr–1

t1/2

5730 yr

1 mol CaCO3 1 mol C 12.01 g C

Mass (g) of C = 4.58 g CaCO3

= 0.5495634 g C

100.09 g CaCO3 1 mol CaCO3 1 mol C

3.2 d/min

At =

= 5.8228 d/min•g

0.5495634 g

Using the integrated rate law:

A

ln t = –kt

A0 = 15.3 d/min•g (the ratio of 12 C:14 C in living organisms)

A

0

5.8228 d/min•g

–4

–1

ln

= – (1.2096809x10 yr )(t)

15.3 d/min•g

t = 7986.17 = 8.0x103 yr

24.96

Find the rate constant from the rate of decay and the initial number of atoms. Use rate constant to calculate halflife.

Initial number of atoms:

106 g 1 mol RaCl 2 1 mol Ra 6.022x1023 Ra atoms

Ra atoms = 5.4 g RaCl 2

1 g 297 g RaCl 1 mol RaCl

1 mol Ra

2

2

= 1.09490909x1016 Ra atoms

A

A = kN or k =

N

1d s

1.5x105 Bq

= 1.36997675x10–11 s–1

k=

1.09490909x1016 Ra atoms Bq

ln 2

ln 2

t1/2 =

=

= 5.05955x1010 = 5.1x1010 s

k

1.36997675x1011 s1

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24-20

24.97

k=

ln 2

ln 2

=

= 1.2096809x10–4 yr–1

t1/2

5730 yr

1 mol 14 C 6.022x1023 atoms 14 C

14

14

C

= 4.3014x10 atoms C

14

14 g 14 C

1

mol

C

A = kN = [(1.2096809x10–4 yr–1)(4.3014x1014 atoms 14C)](1 disintegration/1 atom) = 5.2033x1010 dpyr

13

5.2033x1010 dpyr

0.156 MeV 1.602x10 J 1 rad

Dose =

= 2.001x10–3 = 10–3 rad

yr

1 MeV

0.01 J

65

kg

disint.

kg

Number of atoms = 10 8 g

24.98

14

Plan: Determine how many grams of AgCl are dissolved in 1 mL of solution. The activity of the radioactive Ag+

indicates how much AgCl dissolved, given a starting sample with a specific activity (175 nCi/g). Convert g/mL to

mol/L (molar solubility) using the molar mass of AgCl.

Solution:

1.25x102 Bq 1 dps

1 nCi 1 g AgCl

1 Ci

–6

Concentration =

9

= 1.93050x10 g AgCl/mL

10

mL

1

Bq

175

nCi

3.70

x10

dps

10

Ci

1.93050x10 6 g AgCl 1 mol AgCl 1 mL

–5

–5

Molarity =

3 = 1.34623x10 = 1.35x10 M AgCl

mL

143.4

g

AgCl

10

L

24.99

a) The process shown is fission in which a neutron bombards a large nucleus, splitting that nucleus

into two nuclei of intermediate mass.

1

b) 0 n +

235

92 U

144

55 Cs

+

90

37 Rb

1

+ 2 0n

144

c) 55 Cs , with 55 protons and 89 neutrons, has a n/p ratio of 1.6. This ratio places this isotope above the band of

stability and decay by beta particle emission is expected.

24.100 Plan: Determine the value of k from the half-life. Then determine the fraction from the integrated rate law.

Solution:

ln 2

ln 2

k=

=

= 9.90210x10–10 yr–1

t1/2

7.0x108 yr

ln

N0

= kt = (9.90210x10–10 yr–1)(2.8x109 yr) = 2.772588

Nt

N0

= 15.99998844

Nt

Nt

= 0.062500 = 6.2x10–2

N0

24.101 Determine the value of k from the half-life. Then determine the age from the integrated rate law.

ln 2

ln 2

k=

=

= 1.540327x10–10 yr–1

9

t1/2

4.5x10 yr

N0

= kt

Nt

6 9

ln

=(1.540327x10-10 yr-1 )(t)

9

0.5108256 = (1.540327x10–10 yr–1)(t)

t = 3.316345x109 = 3.3x109 yr

ln

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24-21

24.102 Plan: Find the rate constant, k, using any two data pairs (the greater the time between the data points, the

greater the reliability of the calculation). Calculate t1/2 using k. Once k is known, use the integrated rate law to

find the percentage lost after 2 h. The percentage of isotope remaining is the fraction remaining after 2.0 h

(Nt where t = 2.0 h) divided by the initial amount (N0), i.e., fraction remaining is Nt/N0. Solve the first-order rate

expression for Nt/N0, and then subtract from 100% to get fraction lost.

Solution:

N

a) ln t = –kt

N0

495 photons/s

ln

= –k(20 h)

5000 photons/s

–2.312635 = –k(20 h)

k = 0.11563 h–1

ln 2

ln 2

t1/2 =

=

= 5.9945 = 5.99 h (Assuming the times are exact, and the emissions have three

k

0.11563 h 1

significant figures.)

b) ln

ln

Nt

= –kt

N0

Nt

= – (0.11563 h–1)(2.0 h) = –0.23126

N0

Nt

Nt

= 0.793533

x 100% = 79.3533%

N0

N0

The fraction lost upon preparation is 100% – 79.3533% = 20.6467% = 21%.

ln 2

ln 2

=

= 5.5451774x10–10 yr–1

9

t1/2

1.25x10 yr

A = kN where A = dps N = number of atoms

Number of atoms = (1.0 mol 40K)(6.022x1023 atoms/mol) = 6.022x1023 atoms 40K

24.103 k =

5.5451774x10 10

1 day 1 h 1 disint.

1 yr

7

23

A=

6.022x10 atoms

= 1.05816x10 dps

yr

365.25

day

24

h

3600

s

1

atom

Dose (Ci) = (1.05816x107 dps)(1 Ci/3.70x1010 dps) = 2.85990x10–4 = 2.9x10–4 Ci

Dose = (1.05816x107 dps)(1 Bq/1 dps) = 1.05816x107 = 1.1x107 Bq

24.104 Plan: Use the given relationship for the fraction remaining after time t, where t = 10.0 yr, 10.0x103 yr, and

10.0x104 yr.

Solution:

a) Fraction remaining after 10.0 yr =

1

2

t

t1

10.0

2

1

2

4

1

2

c) Fraction remaining after 10.0x10 yr =

1

2

10.0x103

3

b) Fraction remaining after 10.0x10 yr =

=

10.0x104

5730

5730

5730

= 0.998791 = 0.999

= 0.298292 = 0.298

= 5.5772795x10–6 = 5.58x10–6

d) Radiocarbon dating is more reliable for b) because a significant quantity of 14C has decayed and a significant

quantity remains. Therefore, a change in the amount of 14C would be noticeable. For the fraction in a), very little

14

C has decayed and for c) very little 14C remains. In either case, it will be more difficult to measure the change so

the error will be relatively large.

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24-22

24.105

210

86 Rn

0

210

+ 1e 85 At

Mass = (2.368 MeV)(1 amu/931.5 MeV) = 0.0025421363 amu

Mass 210At = mass 210Rn + electron mass – mass equivalent of energy emitted.

= (209.989669 + 0.000549 – 0.0025421363) amu = 209.9876759 = 209.98768 amu

24.106 Plan: At one half-life, the fraction of sample is 0.500. Find n for which (0.900)n = 0.500.

Solution:

(0.900)n = 0.500

n ln (0.900) = ln (0.500)

n = (ln 0.500)/(ln 0.900) = 6.578813 = 6.58 h

24.107 a) decay by vanadium-52 produces chromium-52.

51

23V

51

23V

1

+ 0n

52

23V

52

24 Cr

+

0

1

52

(n,) 24 Cr

b) Positron emission by copper-64 produces nickel-64.

63

29 Cu

63

29 Cu

1

+ 0n

64

29 Cu

64

28 Ni

0

+ 1

64

(n,+) 28 Ni

c) decay by aluminum-28 produces silicon-28.

27

1

28

13 Al + 0 n 13 Al

27

28

13 Al (n,) 14 Si

28

14 Si

+

0

1

24.108 Determine k for 90Sr.

ln 2

ln 2

k=

=

= 0.023902 yr–1

t1/2

29 yr

a) ln

ln

Nt

= –kt

N0

Nt

= – (0.023902 yr–1)(10 yr) = – 0.23902

0.0500 g

Nt

= 0.787399

0.0500 g

(0.787399)(0.0500 g) = 0.03936995 = 0.039 g 90Sr

N

b) ln t = –kt

N0

100 99.9%

= – (0.023902 yr–1)(t)

100%

t = 289.003 = 3x102 yr (The calculation 100 – 99.9 limits the answer to one significant figure.)

ln

24.109 a)

12

4

6 C + 2 He

16

8O

13

931.5 MeV 1.602x10

b) 7.7x10 2 amu

1 MeV

1 amu

J 1 kJ

–14

–14

3 = 1.1490425x10 = 1.1x10 kJ

10

J

24.110 Plan: The production rate of radon gas (volume/hour) is also the decay rate of 226Ra. The decay rate, or activity, is

proportional to the number of radioactive nuclei decaying, or the number of atoms in 1.000 g of 226Ra, using the

relationship A = kN. Calculate the number of atoms in the sample, and find k from the half-life. Convert the

activity in units of nuclei/time (also disintegrations per unit time) to volume/time using the ideal gas law.

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24-23

Solution:

226

88 Ra

4

2 He +

222

86 Rn

ln 2

ln 2

=

= 4.33879178x10–4 yr–1(1 yr/8766 h) = 4.94510515x10–8 h–1

t1/2

1599 yr

The mass of 226Ra is 226.025402 amu/atom or 226.025402 g/mol.

k=

6.022x1023 Ra atoms

1 mol Ra

21

N = 1.000 g Ra

= 2.6643023x10 Ra atoms

226.025402

g

Ra

1

mol

Ra

–8 –1

21

A = kN = (4.94510515x10 h )(2.6643023x10 Ra atoms) = 1.3175255x1014 Ra atoms/h

This result means that 1.318x1014 226Ra nuclei are decaying into 222Rn nuclei every hour. Convert atoms of 222Rn

into volume of gas using the ideal gas law.

1.3175255x1014 Ra atoms 1 atom Rn

1 mol Rn

Moles of Rn/h =

23

h

1 atom Ra 6.022x10 Rn atoms

= 2.1878537x10–10 mol Rn/h

L•atm

2.1878537x1010 mol Rn/h 0.08206

273.15 K

nRT

mol•K

V=

=

1 atm

P

–9

–9

= 4.904006x10 = 4.904x10 L/h

Therefore, radon gas is produced at a rate of 4.904x10–9 L/h. Note: Activity could have been calculated as decay

in moles/time, removing Avogadro’s number as a multiplication and division factor in the calculation.

24.111 Determine k:

ln 2

ln 2

k=

=

= 0.0216608 s–1

t1/2

32 s

ln

Nt

= –kt

N0

90%

= – (0.0216608 s–1)(t)

100%

t = 4.86411 = 4.9 s

ln

24.112 a)

b)

c)

d)

133

55 Cs

79

35 Br

24

12 Mg

14

7N

The N/Z ratio for 140Cs is too high.

It has an even number of neutrons compared with 78Br.

The N/Z ratio equals 1.

The N/Z ratio equals 1.

24.113 Plan: Determine k from the half-life and then use the integrated rate law, solving for time.

Solution:

ln 2

ln 2

k=

=

= 0.0239016 yr–1

t1/2

29 yr

ln

Nt

= –kt

N0

1.0x104 particles

= – (0.0239016 yr–1)(t)

ln

7.0x104 particles

–1.945910 = – (0.0239016 yr–1)(t)

t = 81.413378 = 81 yr

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

24-24

6

6

24.114 a) 3 Li + 3 Li

12

6 C (dilithium)

6

12

3 Li ) – mass 6 C =

12

b) m = 2(mass

2(6.015121 amu) – 12.000000 amu = 0.030242 amu/atom 6 C (dilithium)

m = (0.030242 amu/atom)(1.66054x10–27 kg/amu) =5.02180507x10–29 kg/atom

2

1

J

29

8

–12

E = mc2 = 5.02180507x10 kg/atom 2.99792x10 m/s

= 4.5133595x10 J/atom

2

kg•m

s2

4.5133595 x10 12 J

1 atom

1 amu

E=

27

atom

12.000000 amu 1.66054x10 kg

= 2.2650059x1014 = 2.2650x1014 J/kg dilithium

1

4

0

c) 4 1H 2 He + 2 1

2 positrons are released.

12

( 6 C ):

d) For dilithium

Mass = (5.02180507x10–29 kg/atom)(1 atom/12.000000 amu) (1 amu/1.66054x10–27 kg 12C)

= 2.5201667x10–3 = 2.5202x10–3 kg/kg 12C

4

For 2 He (The mass of a positron is the same as the mass of an electron.)

1

4

0

m = 4 (mass 1H ) – [mass 2 He + 2 mass 1e ]

= 4(1.007825 amu) – [4.00260 amu + 2(5.48580x10–4 amu)] = 0.02760 amu/atom

= (0.02760 amu/atom)(1.66054x10–27 kg/amu) = 4.58309x10–29 kg/atom

Mass = (4.58309x10–29 kg/atom)(1 atom/4.00260 amu)(1 amu/1.66054x10–27 kg 4He)

= 6.895517x10–3 = 6.896x10–3 kg/kg 4He

2

3

4

1

e) 1H + 1H 2 He + 0 n

m = [2.0140 amu + 3.01605 amu] – [4.00260 amu + 1.008665 amu]

= 5.0300 amu – 5.01126 amu = 0.0188 amu

= (0.0188 amu/atom)(1.66054x10–27 kg/amu) = 3.1218152x10–29 kg/atom

Mass = (3.1218152x10–29 kg/atom)(1 atom/4.00260 amu)(1 amu/1.66054x10–27 kg 4He)

= 4.696947x10–3 = 4.70x10–3 kg/kg 4He

6

1

4

3

f) 3 Li + 0 n 2 He + 1H

1 mol 3 H 3.01605 g 3 H 1 mol 6 Li 103 g 6 Li 1 kg 3 H

3

Mass 1H / kg 6Li =

1 mol 6 Li 1 mol 3 H 6.015121 g 6 Li 1 kg 6 Li 103 g 3 H

= 0.5014113598 = 0.501411 kg 3H/kg 6Li

103 g 1 mol 3 H 6.022x1023 atom 3 H 3.121852x1029 kg

Mass = (0.5014113598 kg 3H)

1 kg 3.01605 g 3 H

atom

mol 3 H

= 3.1254222x10–3 = 3.125x10–3 kg

Change in mass for dilithium reaction:

Mass = (5.02180507x10–29 kg/atom 12C)(1 atom 12C/2 atoms 6Li)(6.022x1023 atoms 6Li/mol)

(1 mol 6Li/6.015121 g 6Li)(103 g/kg 6Li) = 2.513774x10–3 = 2.514x10–3 kg

The change in mass for the dilithium reaction is slightly less than that for the fusion of tritium with

deuterium.

24.115 Plan: Convert pCi to Bq using the conversion factors 1 Ci = 3.70x1010 Bq and 1 pCi = 10–12 Ci. For part b),

use the first-order integrated rate law to find the activity at the later time (t = 9.5 days). You will first need to

calculate k from the half-life expression. For part c), solve for the time at which Nt = the EPA recommended

level.

Solution:

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

24-25

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