KEY
Practice Problems: Nuclear Chemistry
CHEM 1B
1. Write the equation for the nuclear reaction described in each of the following processes:
a. Fluorine-18 (18F) undergoes positron emission (one of the radionuclides used in PET scans)
F 
O +
e
t ½ = 110. minutes
b. Technetium-99m (99mTc) undergoes gamma decay to form 99Tc (a diagnostic radioactive tracer
used to locate tumors, the “m” indicates a metastable excited nuclear state)
Tc 

Tc +
t ½ = 6.01 hours
c. Chromium-51 (51Cr) undergoes electron capture (a diagnostic radioactive tracer used to study
blood)
Cr +
e 
–
V
t ½ = 27.7 days
2. Would you expect the three radionuclides described in Problem 1 to have short half-lives (minutes
to days) or long half-lives (years)? Explain why.
Radionuclides used in medicine for imaging and as diagnostic tracers have short half-lives.
Decaying quickly (a short burst of radiation) allows less radioactive material to be used, while
still emitting enough particles to be detectable from outside of the body. This also allows the
patient to remain radioactive for a shorter period of time.
3. The decay series of Americium-241 (used in smoke detectors) eventually forms bismuth-209.
Determine how many alpha decays and how many beta decays occur in the series, and write the
overall nuclear reaction. Explain your reasoning.
𝟐𝟒𝟏
Am
𝟗𝟓

𝟐𝟎𝟗
Bi
𝟖𝟑
+ 8 𝟒𝟐 He + 4 –𝟎𝟏 e
All of the difference between the mass numbers of americium-241 and bismuth-209 must come
from alpha decay, with each alpha particle emitted decreasing the mass number by 4 nucleons
(beta decay does not change the mass number).
Number of nucleons (mass number):
241 nucleons – 209 nucleons = 32 nucleons (lost)
32 nucleons
1-particle
4 nucleons
= 8 -particles (emitted)
The remaining difference in the charge numbers must be from beta decay, with each beta particle
1
emitted balancing the charge of 1 proton created in the nucleus ( n 
0
Number of protons (charge number):
95 protons = 83 protons + 8 -particles
X = 4 -particles (emitted)
2 protons
1-particle
4 nucleons
+X
1
p
1
+
–1 proton
1-particle
4 nucleons
0
–1
e ).
4. Rocks can be dated using the beta decay of 40K into 40Ar. Assuming that there was no gaseous argon
in the molten rock when it formed, and that all of the argon produced since solidification has
remained trapped in the rock until it was crushed for analysis, use the following information to
calculate the age of the rock in years BP (before present, 1950): Year analyzed: 2011, 40Ar/40K ratio
in the rock = 1.15, t ½ of 40K = 1.28 x 109 years.
Finding the original amount of 40K:
40
Arp
= 1.15
40
Kr
40
40
Subscript Key:
o = original
p = produced
r = remaining
t = at time t
40
Arp = 1.15 Kr
Ko = 40Arp + 40Kr = (1.15 40Kr) + 40Kr = 2.15 40Kr
From the 1st order integrated rate law:
t½ =
ln (2)
k
ln
[A]t
[A]o
ln
[A]t
[A]o
t =
–k
k=
ln (2)
ln (2)
=
= 5.41521235 x 10–10 years–1
t½
1.28 x 109 years
= – kt
40
ln
=
40
Kr
40
Ko
–k
ln
=
Kr
2.15 40Kr
– 5.41521235 x 10–10 years–1
= 1,413,550,925 years
Converting to years BP:
1,413,550,925 years (before analysis in 2011)
– 61 years (between 2011 and 1950)
1,413,550,864 years (before 1950)
1.41 x 109 years BP
Note: Since Arp > Kr , t > 1 t ½ , so the
answer is reasonable.
Also note: Simply multiplying the t ½
by the ratio does NOT give the
correct answer!
5. The energy yield of nuclear weapons is measured in Mt (megatonnes of TNT). If the explosion of
1 tonne of TNT yields 4.2 GJ (gigajoules) of energy, how much mass (in kg) of matter was
converted to energy in the 57 Mt blast of the largest weapon ever tested (USSR, 1961)?
c = 2.998 x 108 m/s
106 tonne
4.2 GJ
109 J
57 Mt
= 2.394 x 1017 J
1 Mt
1 tonne
1 GJ
E = mc2
m =
2.394 x 1017 J
E
=
= 2.663550216 kg = 2.7 kg
(2.998 x 108 m/s)2
c2
Note: Though a combination
of elements was used, this
mass corresponds to a cube
of uranium only 5.2 cm per
side! (dU = 19.1 g/cm3)
6. What is the binding energy of 12C (in kJ/mol and MeV/nucleon)?
Mass nucleons:
6 p (1.007276 u / p) + 6 n (1.008665 u / n) = 12.095646 u
Particle
Relative Mass (u)
12
C atom
electron
proton
neutron
12 (exactly)
5.485798 x 10–4
1.007276
1.008665
u = 1.660539 x 10–27 kg
c = 2.998 x 108 m/s
1 eV = 1.602176 x 10–19 J
Mass 12C nucleus:
12 u – 6 e– (5.485798 x 10–4 u / e–) = 11.99670852 u
Mass defect (per nucleus):
1.660539 x 10–27 kg
12.095646 u – 11.99670852 u = 0.0989374788 u
= 1.64289542 x 10–28 kg
1u
Binding energy (per nucleus):
E = mc2 = (1.64289542 x 10–28 kg) (2.998 x 108 m/s)2 = 1.47663506 x 10–11 J
Conversions:
1.47663506 x 10–11 J
1 12C nucleus
6.022 x 1023 12C nuclei
1 mol 12C
1.47663506 x 10–11 J
1 12C nucleus
1 12C nucleus
12 nucleons
1 kJ
103 J
9
= 8,892,296,341 kJ/mol = 8.892 x 10 kJ/mol
1 eV
1.602176 x 10–19 J
1 MeV
106 eV
= 7.680362322 MeV/nucleon
= 7.680 MeV/nucleon
7. How much energy (in kJ/mol) is released in the following fusion
reaction? Hint: The number of electrons is the same on both sides
of the reaction, therefore their masses do not need to be subtracted.
1
H + He 
He + 11 H
Mass products:
4.0026 u + 1.0078 u = 5.0104 u
Particle
Relative Mass (u)
1
H atom
H atom
3
He atom
4
He atom
2
1.0078
2.0141
3.0160
4.0026
u = 1.660539 x 10–27 kg
c = 2.998 x 108 m/s
Mass reactants:
2.0141 u + 3.0160 u = 5.0301 u
Mass change (per fusion event):
1.660539 x 10–27 kg
5.0104 u – 5.0301 u = – 0.0197 u
= – 3.2712618 x 10–29 kg
1u
Energy released (per fusion event):
E = mc2 = (3.2712618 x 10–29 kg) (2.998 x 108 m/s)2 = 2.94021144 x 10–12 J
Conversion:
2.94021144 x 10–12 J
1 fusion event
6.022 x 1023 events
1 mol (any species)
1 kJ
103 J
9
= 1,770,595,330 kJ/mol = 1.77 x 10 kJ/mol