Chapter 16
What is radioactivity?
Radioactivity is the spontaneous disintegration of nuclei. The first radioactive
elements discovered were the heavy atoms thorium and uranium.
These heavy
atoms and others near to them in the Periodic Table emit alpha particles, beta
particles and gamma rays. It is now clear that other radioactive nuclei are found in
nature.
Some of these are the result of the impact of high energy radiation on the
nucleus of a non-radioactive atom.
Radioactive carbon-14, used in radiocarbon
dating, is formed in constantly replenished trace amounts by the interaction of high
energy cosmic rays with atoms in the upper atmosphere. Others, such as radon, a
radioactive gas associated with granite; result from the radioactive decay of natural
uranium ores. Large numbers of other artificial radioactive elements have now been
prepared.
These lighter elements tend to emit electrons and positrons rather than
alpha particles.
How does carbon dating work?
To date organic materials less than about 50,000 years old, the radioactive
decay of carbon-14 is preferred.
Radiocarbon dating rests upon two main facts.
The first is that the production of radioactive carbon
4
6C
in the upper atmosphere due
to the interaction of cosmic ray neutrons with nitrogen is constant and has been over
past ages. The production reaction is:
14
7N
+
1
0n

14
6C
+
1
1H
The second is that the rate of decay of this radioactive carbon is constant and has
not changed over time via the reaction:
14
6C

14
7N
+
0
-1e;
t½ = 5730 years.
The carbon-14 is distributed throughout the atmosphere in the form of carbon
dioxide, CO2, molecules and due to atmospheric diffusion a fairly constant proportion
of all CO2 is radioactive due to the presence of carbon-14. Living plants absorb CO2
and so incorporate carbon-14 into their tissues. The relative quantity of carbon-14 in
an organism remains constant until it dies. At this point the carbon-14 begins to
decay at its normal rate.
A measurement of the radioactivity of the once-living
samples can then be used to determine the age of the sample itself.
What produces energy in a nuclear power station?
Energy is produced in a nuclear power station by nuclear fission of U-235.
Nuclear fission is the breaking apart of atomic nuclei into two or more pieces.
The
energy produced is derived from the difference in masses of the reactants and the
products.
This mass difference is liberated as heat.
The amount of energy per
gram of U-235 is calculated to be 7.5 x 1010 J. In contrast to this, one gram of coal
burnt in air produces about 3 x 104 J. That is, uranium-235 fission produces about 1
million times more energy than burning fossil fuels.
The fuel used in almost all nuclear power reactors is uranium dioxide.
Natural uranium ores consist of approximately 99 % of the U-238 isotope, about
0.71% U-235 and small amounts of U-234. For power production, the amount of U235 in the UO2 is increased to 2 – 4 %, the resulting material being enriched uranium
dioxide.
Quick quiz
1. a; 2. b; 3. c; 4. b; 5. a; 6. c; 7. a; 8. a; 9. b; 10. c; 11. b; 12. c; 13. c;
14. b; 15. a; 16. b; 17. c; 18. b; 19. b; 20. a; 21. b.
Calculations and questions
1.
166
75Re,
(a)
75 protons, 91 neutrons, 166 nucleons; (b)
18
8O,
neutrons, 140 nucleons; (c)
140
56Ba,
56 protons, 84
14
5B,
8 protons, 10 neutrons, 18 nucleons; (d)
5
protons, 9 neutrons, 14 nucleons.
2.
(a)
181
80Hg,
80 protons, 101 neutrons, 181 nucleons; (b)
neutrons, 169 nucleons; (c)
117
53I,
169
77Ir,
77 protons, 92
53 protons, 64 neutrons, 117 nucleons; (d)
98
40Zr,
40 protons, 58 neutrons, 98 nucleons.
3. (a)
+
0
-1e;

27
14Si
(d) 3920Ca 

+
0
1e;
17
8O
+
0
1e.
+
0
-1e;
(b)
+
0
1e.
(d) 179F 
4. (a) 248O 
5. (a)
27
13Al
24
9F
39
19K
243
100Fm
241
94Am
+

0
-1e;
239
98Cf
+
(b)
28
13Al
47
23V
4
2He;
(d) 23792U 

28
14Si
47
22Ti
(b)
241
95Am
237
93Np
+
0
-1e.
6. 5.21 min.
7. 49.8 h.
8. 3.7 x 1010 disintegrations s-1 g-1.
9. 3.83 x 1012 disintegrations s-1 g-1.
10. 8.25 x 1017 disintegrations s-1 g-1.
11. 4.20 h.
12. 13.96 x 103 y.
13. (a) 92.5 nuclei min-1; (b) 17.6 d.
14. (a) 9.5 x 104 atoms dm-3; (b) 12.6 d.
15. (a) 3.11 x 10-10 g m-2; (b) 794.3 Bq.
16. 1.13 x 10-12 J.

+
+
0
-1e;
(c)
24
11Na
0
1e;
(c) 3215P 

237
93Np
+

32
16S
4
2He;
(c)
24
12Mg
+
0
-1e;
241
94Pu
17. 1.27 x 10-12 J.
18. 1.38 x 10-12 J.
19. (a) 4.52 x 10-8 kg Ra; (b) 4.44 x 10-8 kg Rn.
20. 1.614 x 1010 kJ mol-1.
21. (a) 4.10 x 10-12 J; (b) 2.44 x 1037 fusions s-1.
Solutions
1 Write the nuclear symbol, and the number of protons, neutrons and nucleons in:
(a) rhenium-166; (b) barium-140; (c) oxygen-18; (d) boron-14.
(a) 16675Re, 75 protons, 91 neutrons, 166 nucleons;
(b) 14056Ba, 56 protons, 84 neutrons, 140 nucleons;
(c) 188O, 8 protons, 10 neutrons, 18 nucleons;
(d) 145B, 5 protons, 9 neutrons, 14 nucleons.
2 Write the nuclear symbol, and the number of protons, neutrons and nucleons in:
(a) mercury-181; (b) iridium-169; (c) iodine-117; (d) zirconium-98.
(a) 18180Hg, 80 protons, 101 neutrons, 181 nucleons;
(b) 16977Ir, 77 protons, 92 neutrons, 169 nucleons;
(c) 11753I, 53 protons, 64 neutrons, 117 nucleons;
(d) 9840Zr, 40 protons, 58 neutrons, 98 nucleons.
3
Write nuclear equations for the decay of the following radioactive isotopes; the
particles emitted in the decay are given in brackets: (a)
27
14Si
(positron); (b)
28
13Al
(electron); (c) 2411Na (electron); (d) 179F (positron).
(a) 2714Si 
27
13Al
+
0
1e;
(b) 2813Al 
28
14Si
+
0
-1e;
(c) 2411Na 
(d) 179F 
4
24
12Mg
17
8O
+
0
-1e;
0
1e.
+
Write nuclear equations for the decay of the following radioactive isotopes; the
particles emitted in the decay are given in brackets: (a)
24
8O
(electron); (b)
47
23V
(positron); (c) 3215P (electron); (d) 3920Ca (positron).
(a) 248O 
24
9F
+
(b) 4723V 
47
22Ti
(c) 3215P 
32
16S
(d) 3920Ca 
5
0
-1e;
0
1e;
+
+
39
19K
0
-1e;
+
0
1e.
Write nuclear equations for the decay of the following radioactive isotopes; the
particles emitted in the decay are given in brackets: (a)
(alpha); (c) 24194Pu (electron); (d) 23792U (electron).
(a) 243100Fm 
239
98Cf
+
4
2He;
(b) 24195Am 
237
93Np
+
4
2He;
241
94Am
+
0
-1e;
(c) 24194Pu 
243
100Fm
(alpha); (b)
241
95Am
(d) 23792U 
237
93Np
+
0
-1e.
6 Determine the half-life of a radioactive isotope from the variation of the number of
radioactive disintegrations observed, in counts per minute, over a period of time,
given in the Table.
Time / min
0
Activity / counts 3160
2
4
6
8
10
12
14
2512
1778
1512
1147
834
603
579
per min
Plot ln N versus t, slope = - = -ln 2 / t½
As activity is proportional to N, plot activity versus t.
Time / min 0
2
4
6
8
10
12
14
Activity
3160
2512
1778
1512
1147
834
603
579
Ln N
8.085 7.829 7.483 7.321 7.045 6.726 6.402 6.361
slope  (8.058 - 6) / (0 - 15.5)
t½
 ln 2 x 15.5 / 2.058 = 5.2 min
7
A sample of radioactive sodium-24 with a half-life of 15.0 h is used to measure
the diffusion coefficient of Na in NaCl. How long will it take for the activity to drop to
0.1 of its original activity?
ln (N0 / N) = t = (ln 2 / t½)t
ln (1 / 0.1) = (ln 2 / 15) x t
t = 49.8 hr
8 What is the specific activity of radium-226, which has a half-life of 1600 y?
specific activity = ln 2 x NA / (molar mass x t½)
= 4.174 x 1023 / (225 x 1600 x 365 x 24 x 60 x 60)
= 3.68 x 1010 disintegrations s-1 g-1
9 What is the specific activity of plutonium-241, which has a half-life of 14.35 y?
specific activity = ln 2 x NA / (molar mass x t½)
= 4.174 x 1023 / (241 x 14.35 x 365 x 24 x 60 x 60)
= 3.827 x 1012 disintegrations s-1 g-1
10 What is the specific activity of neptunium-233, which has a half-life of 36.2 min?
specific activity = ln 2 x NA / (molar mass x t½)
= 4.174 x 1023 / (233 x 36.2 x 60)
= 8.248 x 1017 disintegrations s-1 g-1
11 A purified sample of a radioactive compound is found to have an activity of 1365
counts per minute at 10am but only 832 counts per minute at 1 pm.
What is the
half-life of the sample?
ln (N0 / N) = t = (ln 2 / t½)t
ln (1365 / 832) = (ln 2 / t½) x 3
t½ = 4.20 hr
12
A 250 mg piece of carbon from an ancient hearth showed 1530 carbon-14
disintegrations in 36 h.
250 mg of fresh carbon from charcoal gave 8280
disintegrations in the same time.
What is the date of the carbon sample?
The
half-life of carbon-14 is 5.73 x 103 y.
ln (N0 / N) = t = (ln 2 / t½)t
ln (8280 / 1530) = (ln 2 / 5.73 x 103) t
t = 13.96x 103 years
13 A cellar of dimensions 2 x 3 x 2.5 m is found to show an activity of 0.37 Bq dm -3,
due to the presence of radon. (a) How many nuclei decay per minute in the cellar?
(b) How long will it take for the activity to fall to 0.015 Bq, if no more radon leaks into
it? The half-life of radon is 3.8 d.
(a) An activity of 0.37 Bq dm-3 means that there are 0.37 disintegrations per second
in each dm3 of the cellar, so the total number of disintegrations is:
(0.37 x 20 x30 x25) / 60 = 92.5 disintegrations per minute
= number of nuclei decaying
(b)
ln (N0 / N) = t = (ln 2 / t½)t
ln (0.37 / 0.015) = (ln 2 / 3.8) t
t = 17.6 days
14
The suggested upper limit for radon concentration in a building is equivalent to
an activity of 200 Bq m-3. (a) How many radon atoms are needed per dm 3 of air to
give this figure?
(b) How long will it take for the activity to fall to one tenth of this
value in a sealed room? The half-life of radon is 3.8 d.
(a)
-dN / dt = N = (ln 2 / t½) N
activity = 200 Bq = 200 disintegratins per sec = dN / dt
hence:
200 = (ln 2 / 3.8 x 24 x 60 x 60) N
N = 9.47 x 107 atoms m-3 = 9.47 x 104 atoms dm-3
(b)
ln (N0 / N) = t = (ln 2 / t½)t
ln (200 / 20) = (ln 2 / 3.8) t
t = 12.6 days
15
In 1986 a nuclear reactor at Chernobyl exploded, depositing caesium-137 over
large areas of northern Europe. An initial activity of 1000 Bq m -2 of vegetation was
found over parts of Scotland.
(a) How many grams of caesium-137 were deposited
per square metre of vegetation?
half-life of caesium-137 is 30.1 y.
(b) What was the activity after 10 years?
The
(a)
-dN / dt = N = (ln 2 / t½) N
activity = 1000 Bq = 200 disintegrations per sec = dN / dt
hence:
1000 = (ln 2 / 30.1 x 365 x 24 x 60 x 60) N
N = 1.369 x 1012 atoms m-2
The mass of 1 atom molar mass / NA = 137 / 6.02214 x 1023
Total mass was = (1.369 x 1012 x 137 / 6.02214 x 1023
= 3.11 x 10-10 g
(b)
ln (N0 / N) = t = (ln 2 / t½)t
Taking (N0 / N) to be proportional to the activity:
ln (1000 / activity) = (ln 2 / 30.1) x 10
activity after 10 years = 794.3 Bq
16 Calculate the binding energy per nucleon for
u; 10n, 1.0087 u;
4
2He,
4
2He.
The masses are 11H, 1.0078
4.0026 u.
The mass difference:
m = mass AZX – [mass of Z protons + (A – Z neutrons)]
211H + 210n 
4
2He
m = 4.0026 u - (2 x 1.0078 u + 2 x 1.0087 u)
= -0.0304 u
1 u = 1.6605 x 10-27 kg;
m = -0.0304 x 1.6605 x 10-27 kg
E = mc2
= 0.0304 x 1.6605 x 10-27 x (2.9978 x 108)2
= 4.537 x 10-12 J
Binding energy per nucleon = 1.134 x 10-12 J
17
Calculate the binding energy per nucleon for
23
12Mg.
The masses are
23
12Mg,
22.9941 u; 11H, 1.0078 u; 10n, 1.0087 u.
The mass difference:
m = mass AZX – [mass of Z protons + (A – Z neutrons)]
1211H + 1110n 
23
12Mg
m = 22.9941 u - (12 x 1.0078 u + 11 x 1.0087 u)
= -0.1952 u
1 u = 1.6605 x 10-27 kg;
m = -0.1952 x 1.6605 x 10-27 kg
E = mc2
= 0.1952 x 1.6605 x 10-27 x (2.9978 x 108)2
= 2.913 x 10-11 J
Binding energy per nucleon = 1.266 x 10-12 J
18
Calculate the binding energy per nucleon for
34
16S.
The masses are
33.967865 u, 11H, 1.0078 u; 10n, 1.0087 u.
The mass difference:
m = mass AZX – [mass of Z protons + (A – Z neutrons)]
1611H + 1810n 
34
16S
m = 33.967865 u - (16 x 1.0078 u + 18 x 1.0087 u)
= -0.31354 u
1 u = 1.6605 x 10-27 kg;
m = -0.31354 x 1.6605 x 10-27 kg
34
16S,
E = mc2
= 0.31354 x 1.6605 x 10-27 x (2.9978 x 108)2
= 4.679 x 10-11 J
Binding energy per nucleon = 1.376 x 10-12 J
19 On decay, an atom of radium-226 emits one -particle and is converted into an
atom of radon-222. A quantity of radium-226 produced 4.48 x 10-6 dm3 of helium at
273 K and 1 atm pressure. Determine (a) the mass of radium-226 that decayed, (b)
the mass of radon-222 produced, if no radon decays.
One mole of helium gas
occupies 22.4 dm3 at 273 K and 1 atm pressure.
Ra 
226
222
Rn + 4He
(a) The amount of He produced = 4.48 x 10-6 / 22.4 = 2 x 10-7 mol
The amount of Ra is thus 2 x 10-7 mol = 0.22603 x 2 x 10-7 = 4.52 x 10-8 kg
(b) The amount of Rn is 2 x 10-7 mol = 0.222 x 2 x 10-7 = 4.44 x 10-8 kg
20 Calculate the energy released per mole in the fission reaction:
235
92U
+
The masses are
1
0n

235
92U,
102
42Mo
+
235.0439 u;
128
50Sn
102
42Mo,
+ 610n
101.91025 u;
128
50Sn,
127.91047 u;
1.0087 u.
m = mass of products – mass of reactants
= mass of [6 x 10n +
128
50Sn
+ 10242Mo] - mass of [23592U +
1
0n]
= [6 x 1.0087 + 127.91047 + 101.91025] - [235.0439 + 1.0087]
= -0.17968 u
m = -0.17968 x 1.6605 x 10-27 kg
1
0n,
E = mc2
= 0.17968 x 1.6605 x 10-27 x (2.9978 x 108)2
= 2.68147 x 10-11 J per atom of U
Energy per mole = 2.68147 x 10-11 x 6.02214 x 1023 = 1.61482 x 1013 J mol-1
= 1.614 x 1010 J mol-1
21
Energy generation in the Sun is by way of the reaction of hydrogen to form
helium. One reaction is:
1
4 H 
4
He + 2 0+1e
(a) Calculate the energy liberated per fusion reaction.
produces approximately 1026 J s-1.
required for this? The masses are
It is estimated that the sun
(b) How many fusion reactions per second are
1
1H,
1.0078 u;
4
2He,
4.0026 u;
u.
(a)
m = mass of products – mass of reactants
4
1
= mass of [ He + 2 0+1e] - mass of [4 H ]
= [4.0026 + 2 x 5.486 x 10-4] - [4 x 1.0078]
= -0.02750 u per reaction
m = -0.02750 x 1.6605 x 10-27 kg
E = mc2
= 0.02750 x 1.6605 x 10-27 x (2.9978 x 108)2
= 4.1040 x 10-12 J
(b) If there are N fusion reactions per sec
N x 4.1040 x 10-12 = 1 x 1026
N = 2.437 x 1037 fusion reactions per sec.
0
+1e,
5.486 x 10-4