CHAPTER 20
Nuclear Chemistry
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
20.1. The nuclide symbol for potassium-40 is 40
19 K . Similarly, the nuclide symbol for calcium-40 is
20 Ca . The equation for beta emission is
40
40
19 K 40
20 Ca + 0
1 e
20.2. Plutonium-239 has the nuclide symbol is 239
94 Pu . An alpha particle has the symbol 42 He . The
nuclear equation is
239
94 Pu A
Z X + 42 He
From the superscripts you can write
239 = A + 4, or A = 235
Similarly, from the subscripts you can write
94 = Z + 2, or Z = 92
Hence, A = 235 and Z = 92, so the product is 235
92 X . Because element 92 is uranium, symbol U,
you write the product nucleus as 235
92 U .
20.3. a. 118
50 Sn has atomic number 50. It has 50 protons and 68 neutrons. Because its atomic number
is less than 83 and it has an even number of protons and neutrons, it is expected to be
stable.
b. 76
33 As has atomic number 33. It has 33 protons and 43 neutrons. Because stable odd-odd
nuclei are rare, you would expect 76
33 As to be one of the radioactive isotopes.
c. 227
89 Ac has atomic number 89. Because its atomic number is greater than 83, 227
89 Ac is
radioactive.
20.4. a. 13
7 N has seven protons and six neutrons (fewer neutrons than protons). Its N/Z ratio is
smaller than that in the stable 14N nucleus, so it is expected to decay by electron capture or
positron emission (more likely, because this is a light isotope).
b. 26
11 Na has 11 protons and 15 neutrons. It is expected to decay by beta emission.
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, 818 Chapter 20: Nuclear Chemistry
20.5. a. The abbreviated notation is 40
20 Ca (d,p) 41
20 Ca .
b. The nuclear equation is 126 C + 21 H 136 C + 11 H .
20.6. You can write the nuclear equation as follows:
A
Z X + 01 n 14
6 C + 11 H
To balance this equation in charge (subscripts) and mass number (superscripts), write the
equations
A + 1 = 14 + 1 (from superscripts)
Z+0=6+1 (from subscripts)
Hence, A = 14 and Z = 7. Therefore, the nucleus that produces carbon-14 by this reaction is 14
7 N.
20.7. Because an activity of 1.0 Ci is 3.7 1010 nuclei/s, the rate of decay in this sample is
3.7 1010 nuclei/s
Rate = 13 Ci = 4.81 1011 nuclei/s
1.0 Ci
The number of nuclei in this 2.5-µg (2.5 10−6-g) sample of 99m
43 Tc is
1 mol Tc-99m 6.02 1023 Tc-99m nuclei
2.5 10−6 g Tc-99m
99 g Tc-99m 1 mol Tc-99m
= 1.52 1016 Tc-99m nuclei
The decay constant is
rate 4.81 1011 nuclei/s
k= = = 3.16 10−5 = 3.2 10−5/s
Nt 1.52 1016 nuclei
20.8. First, substitute the value of k into the equation relating half-life to the decay constant:
0.693 0.693
t½ = = = 1.658 108 s
k 4.18 109 / s
Convert the half-life from seconds to years:
1 min 1h 1d 1y
1.658 108 s = 5.257 = 5.26 y
60 s 60 min 24 h 365 d
20.9. The conversion of the half-life to seconds gives
365 d 24 h 60 min 60 s
28.1 y = 8.861 108 s
1y 1d 1h 1 min
Because t1/2 = 0.693/k, solve this for k, and substitute the half-life in seconds.
0.693 0.693
k= = = 7.8208 10−10 /s
t1/2 8.861 108 s
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, Chapter 20: Nuclear Chemistry 819
Before substituting into the rate equation, you need to know the number of nuclei in a sample
containing 5.2 10−9 g of strontium-90.
1 mol Sr-90 6.02 1023 Sr-90 nuclei
5.2 10−9 g Sr-90 = 3.478 1013 Sr-90 nuclei
90 g Sr-90 1 mol Sr-90
Now, substitute into the rate equation:
Rate = kNt = (7.8208 10−10/s) (3.478 1013 nuclei) = 2.72 104 nuclei/s
Calculate the activity by dividing the rate (disintegrations of nuclei per second) by 3.70 1010
disintegrations of nuclei per second per curie.
1.0 Ci
Activity = 2.72 104 nuclei/s = 7.351 10−7 = 7.4 10−7 Ci
3.7 1010 nuclei/s
20.10. The decay constant, k, is 0.693/t1/2. If you substitute this into the equation
Nt 0.693 t
ln = −kt = −
No t1/2
Substitute the values t = 25.0 y and t1/2 = 10.76 y to get
Nt 0.693 25.0 y
ln = = −1.6101
No 10.76 y
The fraction of krypton-85 remaining after 25.0 y is Nt/No.
Nt
= e−1.6101 = 0.1999 = 0.20
No
20.11. The equation applied in this problem is
Nt
ln = −kt
No
The ratio of rates of disintegration is
No 15.3
= = 3.400
Nt 4.5
Therefore, substituting this value of No/Nt and t1/2 = 5730 y into the previous equation gives
t1/2 N 5730 y
t= ln o = ln (3.400) = 1.01 104 = 1.0 104 y
0.693 N t 0.693
20.12. a. First, determine the nuclear masses.
Nuclear mass of 234
90 Th = 234.03660 amu − (90 0.000549 amu) = 233.987190 amu
Nuclear mass of 234
91 Pa = 234.04330 amu − (91 0.000549 amu) = 233.993341 amu
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 820 Chapter 20: Nuclear Chemistry
Write the appropriate nuclear mass below each nuclide symbol. Then calculate m.
234 234 0
90 Th Pa + e
91 -1
233.987190 233.993341 0.000549 amu
Hence,
m = (233.993341 + 0.000549 − 233.987190) amu = 0.006700 amu
The mass change for molar amounts in this reaction is 0.006700 g, or 6.700 10−6 kg. The
energy change is
E = (m)c2 = (6.700 10−6 kg)(3.00 108 m/s)2 = 6.030 1011 J/mol
For 1.00 g Th-234, the energy change is
1 mol Th-234 6.030 1011 J
1.00 g Th-234 = 2.576 109 = 2.58 109 J
234 g Th-234 1 mol Th-234
b. Convert the mass change for the reaction from amu to grams.
1g
m = 0.006700 amu = 1.112 10−26 g = 1.112 10−29 kg
1 amu 6.02 1023
E = (m)c2 = (1.112 10−29 kg)(3.00 108 m/s)2 = 1.001 10−12 J
Convert this to MeV:
1 MeV
E = 1.001 10−12 J = 6.252 = 6.25 MeV
1.602 1013 J
■ ANSWERS TO CONCEPT CHECKS
20.1. a. Yes. Isotopes have similar chemical properties.
b. No, since the 13 H 2 O molecule is more massive than 11 H 2 O .
c. 3
1 H 2 O should be radioactive.
20.2. For the same radiation dosage (10 rads), the form of radiation with the highest RBE will cause the
greatest biological damage. Therefore, the particle will cause the most damage since it has the
highest RBE (10).
20.3. After 50,000 years, enough half-lives have passed (about 10) so there would be almost no carbon-
14 present to detect and measure (about 0.1% would be left).
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Nuclear Chemistry
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
20.1. The nuclide symbol for potassium-40 is 40
19 K . Similarly, the nuclide symbol for calcium-40 is
20 Ca . The equation for beta emission is
40
40
19 K 40
20 Ca + 0
1 e
20.2. Plutonium-239 has the nuclide symbol is 239
94 Pu . An alpha particle has the symbol 42 He . The
nuclear equation is
239
94 Pu A
Z X + 42 He
From the superscripts you can write
239 = A + 4, or A = 235
Similarly, from the subscripts you can write
94 = Z + 2, or Z = 92
Hence, A = 235 and Z = 92, so the product is 235
92 X . Because element 92 is uranium, symbol U,
you write the product nucleus as 235
92 U .
20.3. a. 118
50 Sn has atomic number 50. It has 50 protons and 68 neutrons. Because its atomic number
is less than 83 and it has an even number of protons and neutrons, it is expected to be
stable.
b. 76
33 As has atomic number 33. It has 33 protons and 43 neutrons. Because stable odd-odd
nuclei are rare, you would expect 76
33 As to be one of the radioactive isotopes.
c. 227
89 Ac has atomic number 89. Because its atomic number is greater than 83, 227
89 Ac is
radioactive.
20.4. a. 13
7 N has seven protons and six neutrons (fewer neutrons than protons). Its N/Z ratio is
smaller than that in the stable 14N nucleus, so it is expected to decay by electron capture or
positron emission (more likely, because this is a light isotope).
b. 26
11 Na has 11 protons and 15 neutrons. It is expected to decay by beta emission.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 818 Chapter 20: Nuclear Chemistry
20.5. a. The abbreviated notation is 40
20 Ca (d,p) 41
20 Ca .
b. The nuclear equation is 126 C + 21 H 136 C + 11 H .
20.6. You can write the nuclear equation as follows:
A
Z X + 01 n 14
6 C + 11 H
To balance this equation in charge (subscripts) and mass number (superscripts), write the
equations
A + 1 = 14 + 1 (from superscripts)
Z+0=6+1 (from subscripts)
Hence, A = 14 and Z = 7. Therefore, the nucleus that produces carbon-14 by this reaction is 14
7 N.
20.7. Because an activity of 1.0 Ci is 3.7 1010 nuclei/s, the rate of decay in this sample is
3.7 1010 nuclei/s
Rate = 13 Ci = 4.81 1011 nuclei/s
1.0 Ci
The number of nuclei in this 2.5-µg (2.5 10−6-g) sample of 99m
43 Tc is
1 mol Tc-99m 6.02 1023 Tc-99m nuclei
2.5 10−6 g Tc-99m
99 g Tc-99m 1 mol Tc-99m
= 1.52 1016 Tc-99m nuclei
The decay constant is
rate 4.81 1011 nuclei/s
k= = = 3.16 10−5 = 3.2 10−5/s
Nt 1.52 1016 nuclei
20.8. First, substitute the value of k into the equation relating half-life to the decay constant:
0.693 0.693
t½ = = = 1.658 108 s
k 4.18 109 / s
Convert the half-life from seconds to years:
1 min 1h 1d 1y
1.658 108 s = 5.257 = 5.26 y
60 s 60 min 24 h 365 d
20.9. The conversion of the half-life to seconds gives
365 d 24 h 60 min 60 s
28.1 y = 8.861 108 s
1y 1d 1h 1 min
Because t1/2 = 0.693/k, solve this for k, and substitute the half-life in seconds.
0.693 0.693
k= = = 7.8208 10−10 /s
t1/2 8.861 108 s
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, Chapter 20: Nuclear Chemistry 819
Before substituting into the rate equation, you need to know the number of nuclei in a sample
containing 5.2 10−9 g of strontium-90.
1 mol Sr-90 6.02 1023 Sr-90 nuclei
5.2 10−9 g Sr-90 = 3.478 1013 Sr-90 nuclei
90 g Sr-90 1 mol Sr-90
Now, substitute into the rate equation:
Rate = kNt = (7.8208 10−10/s) (3.478 1013 nuclei) = 2.72 104 nuclei/s
Calculate the activity by dividing the rate (disintegrations of nuclei per second) by 3.70 1010
disintegrations of nuclei per second per curie.
1.0 Ci
Activity = 2.72 104 nuclei/s = 7.351 10−7 = 7.4 10−7 Ci
3.7 1010 nuclei/s
20.10. The decay constant, k, is 0.693/t1/2. If you substitute this into the equation
Nt 0.693 t
ln = −kt = −
No t1/2
Substitute the values t = 25.0 y and t1/2 = 10.76 y to get
Nt 0.693 25.0 y
ln = = −1.6101
No 10.76 y
The fraction of krypton-85 remaining after 25.0 y is Nt/No.
Nt
= e−1.6101 = 0.1999 = 0.20
No
20.11. The equation applied in this problem is
Nt
ln = −kt
No
The ratio of rates of disintegration is
No 15.3
= = 3.400
Nt 4.5
Therefore, substituting this value of No/Nt and t1/2 = 5730 y into the previous equation gives
t1/2 N 5730 y
t= ln o = ln (3.400) = 1.01 104 = 1.0 104 y
0.693 N t 0.693
20.12. a. First, determine the nuclear masses.
Nuclear mass of 234
90 Th = 234.03660 amu − (90 0.000549 amu) = 233.987190 amu
Nuclear mass of 234
91 Pa = 234.04330 amu − (91 0.000549 amu) = 233.993341 amu
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 820 Chapter 20: Nuclear Chemistry
Write the appropriate nuclear mass below each nuclide symbol. Then calculate m.
234 234 0
90 Th Pa + e
91 -1
233.987190 233.993341 0.000549 amu
Hence,
m = (233.993341 + 0.000549 − 233.987190) amu = 0.006700 amu
The mass change for molar amounts in this reaction is 0.006700 g, or 6.700 10−6 kg. The
energy change is
E = (m)c2 = (6.700 10−6 kg)(3.00 108 m/s)2 = 6.030 1011 J/mol
For 1.00 g Th-234, the energy change is
1 mol Th-234 6.030 1011 J
1.00 g Th-234 = 2.576 109 = 2.58 109 J
234 g Th-234 1 mol Th-234
b. Convert the mass change for the reaction from amu to grams.
1g
m = 0.006700 amu = 1.112 10−26 g = 1.112 10−29 kg
1 amu 6.02 1023
E = (m)c2 = (1.112 10−29 kg)(3.00 108 m/s)2 = 1.001 10−12 J
Convert this to MeV:
1 MeV
E = 1.001 10−12 J = 6.252 = 6.25 MeV
1.602 1013 J
■ ANSWERS TO CONCEPT CHECKS
20.1. a. Yes. Isotopes have similar chemical properties.
b. No, since the 13 H 2 O molecule is more massive than 11 H 2 O .
c. 3
1 H 2 O should be radioactive.
20.2. For the same radiation dosage (10 rads), the form of radiation with the highest RBE will cause the
greatest biological damage. Therefore, the particle will cause the most damage since it has the
highest RBE (10).
20.3. After 50,000 years, enough half-lives have passed (about 10) so there would be almost no carbon-
14 present to detect and measure (about 0.1% would be left).
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
