UNIT VIII : NUCLEAR REACTIONS (Honours)
A. Introduction
Definitions: NUCLEON = a particle found in the nucleus; specifically, a proton or neutron.
NUCLIDE = a general term applied to each unique isotope of an atom.
The following atomic particles are referred to frequently in the following notes.
1
0n
= a neutron. Also given the symbol “n”.
1
1p
= a proton (actually 1H ). Also given the symbol “p”.
4
2 He
= an alpha particle (actually 2 He
0
–1
= a beta particle. (It is actually a high–energy e–)
0
+1
= a positron. (It is a positively charged electron; an anti–electron)
0
0
= a gamma particle; also “” (high energy radiation emitted by the nucleus)
1

4
2
). Also given the symbol “”.
Shorthand: An isotope such as 239
94 Pu is sometimes also written as Pu–239 (all plutonium isotopes have
an atomic number of 94, so the “94” can be omitted in the shorthand version).
B. Balancing Nuclearreactions
During a nuclear reaction the total charge and the total mass are conserved. The sum of the atomic
masses on the reactant side equals the sum of the atomic masses on the product side. Similarly, the sum
of the atomic numbers (that is, charges) on both sides of a nuclear reaction must be equal.
EXAMPLES:
(a)
239
94Pu
(b)
239
93 Np
4
2 He
+
242
96 Cm
239
94Pu
+
1
0n
Adding atomic masses: 239 + 4 = 242 + 1
Adding atomic numbers: 94 + 2 = 96 + 0
+?
To find the atomic mass of the unknown particle: 239 = 239 + X
and solve to find X = 0; that is, the particle has a mass of 0.
To find the atomic number of the unknown particle: 93 = 94 + Y
and solve to find Y = –1; that is, the particle has a charge of –1.
0
–1
Hence, the particle is:
(c)
2
1H
1
0n
+?
1
+ 1H
A quick check shows that the unknown particle has
a mass of 0 and a charge of 0, so it is a gamma, .
EXERCISES:
1. Determine the symbol for the missing particle, including its mass and charge.
(a)
27
13 Al
(b)
137
55 Cs
(c)
54
27 Co
(d)
14
6C
(e)
12
6C
30
15 P
+ ____
0
–1
54
26 Fe
14
7N
+
12
6C
1
(f)
197
79 Au
+ 0  + ____
(g)
220
86Rn
4
2 He
+ ____
(h)
140
56Ba
0
–1
(i)
50
23V
+ 0n
0
+ ____
24
12 Mg
+ ____
+
12
6C
+ ____
204
85 At
+ ____
+ ____
50
22Ti
+ 5 ____
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C. Energy From Nuclear Reactions (A digression)
One of the reactions proposed to supply nuclear energy for the future is:
2
1H
3
4
2 He
+ 1H
1
+ 0 n + energy
The Handbook of Chemistry and Physics lists the following molar masses:
2
1H  2.0140 g
3
1H  3.0161 g



 total mass of reac tants  5.0301 g


4
2 He  4.0026 g
1
0 n  1.0087 g


 total mass of products  5.0113 g


So that: mass lost = 5.0301 - 5.0113 = 0.0188 g
But the Einstein equation states: E = m•c2

where:
E = energy given off
m = mass lost
c = velocity of light
which means: a loss of mass is accompanied by a loss of energy.
(Note: It is not correct to say that mass is converted to energy, which implies that as mass is used up,
energy is produced. What is true is that if a reaction decreases in energy, it also decreases the amount of
mass present. Therefore, adding energy to a particle increases the particle’s mass.)
2
3
Assume 2.0140 g of 1H and 3.0161 g of 1H are reacted
mass lost = m = 0.0188 g
c = 3.00 x 1010 cm/s
Hence:
E = (0.0188 g) x (3.00 x 1010 cm/s)2 = 1.69 x 1019
g cm2
s2
g cm2
(this is a conversion factor)
s2

1 kJ
g cm2
and:
E = 1.36 x 1019
x
= 1.69 x 109 kJ
2
2
2
g cm / s
s

To put this number in the proper perspective: when 18.8 mg of mass is lost by the reaction it gives off
energy equivalent to 470 tonnes of TNT!


but:
1 kJ = 1010
D. The Types of Nuclear Reactions
1. Atomic Fusion
——————————————————————————————————————————————
In this nuclear reaction, two small particles are fused (i.e. "welded") together to form a
heavier particle. Some low-mass "nuclear debris" may also be formed.
EXAMPLES:
12
6C
2
1H
+
3
12
6C
+ 1H
24
12 Mg
4
2 He
1
+ 0n
1
+ 0n
The reacting nuclei must possess extremely high energies before they will react; otherwise they will
just bounce off each other. The temperatures required are 107 to 108 oC, and therefore these
reactions can be termed THERMONUCLEAR REACTIONS.
UNIT VIII : NUCLEAR REACTIONS (HONOURS)
3
————————————————————————————————————————————————
2. Induced Fission
——————————————————————————————————————————————
During induced fission, a small high-energy particle (usually a neutron) hits a nucleus and causes
(“induces”) the nucleus to split into two smaller fragments of roughly equal mass.
EXAMPLES:
1
0n
+
239
94Pu
1
0n
+
235
92 U
94
36 Kr
91
36 Kr
+
+
144
58 Ce
142
56Ba
1
+ 2 0n
1
+ 3 0n
An "atomic explosion" can occur when an incoming high-energy neutron causes a fission reaction
which in turn creates more high energy neutrons. The energy given off in the reaction accelerates the
products to very high velocities. Most neutrons produced then start their own fission reaction. The
cycle repeats itself in a "chain reaction" until the total energy given off creates an explosion.
3. Annihilation
——————————————————————————————————————————————
In an annihilation reaction, a particle and its antiparticle react and destroy each other, producing
gamma rays. All the mass of the particles is converted to energy.
0
–1
EXAMPLE:
+
0
+1
2 0
0
4. Transmutation
——————————————————————————————————————————————
Transmutation involves bombarding a large nucleus with a small high-energy particle, causing a
different nucleus to be formed and a different small particle to be given off.
EXAMPLES:
239
94Pu
+ 2 He
4
226
88Ra
+ 0n
1
0n
0
0
1
+
+
242
96 Cm
227
88 Ra
The purpose of carrying out a transmutation reaction is to produce a new nuclide, starting with a
previously-available nuclide. By focussing on the heavy nuclides involved in a transmutation, the
standard nuclear reaction equation is re-written as a "condensed nuclear equation". The equation:
239
94Pu
4
1
0n
+ 2 He
+
242
96 Cm
can be re-written in condensed nuclear form as:
239
94Pu
(,n)
242
96 Cm
where the small particles involved are shown inside the parentheses, the main reacting nuclide is
shown before the left parenthesis, and the main product nuclide is shown after the right parenthesis.
The comma inside the parentheses separates the incoming small particle (on the left) from the
outgoing small particle (on the right).
226
88Ra
EXAMPLE:
0
0
1
+ 0n
+
227
88 Ra
can be re-written as
226
88Ra
(n,)
227
88 Ra
EXERCISES:
2. Write condensed nuclear equations for the following.
(a)
4
2 He
10
+
14
7N
1
17
8O
7
1
+ 1H
4
(c)
240
94Pu
1
0
0
1
+ 0n
12
11
+
241
94Pu
1
(b)
(d) 0 n + 6 C
5B + 0 n
3 Li + 2 He
6C + 2 0 n
3. Write the complete nuclear reaction corresponding to the following. You will have to
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supply the missing species indicated with a ?.
(a)
238
92 U
2
(b)
128
53 I
(n,) ?
(c)
12
6C
(p,pn) ?
( 1H , 2 ?)
(d) ? (p,)
238
93 Np
(e)
60
28 Ni
60
27 Co
(?,2n)
62
30 Zn
4. Classify each of the following as one of: fusion, induced fission, annihilation, or
transmutation.
(a)
2
1H
+
(b)
1
0n
+ 0n
(c)
1
0n
+
(d) 2
51
23V
1
1H
1
0
0
2
232
90Th
14
7N
+
52
23V
(e)
4
2 He
+
12
6C
1
(f)
27
13 Al
+
28
14 Si
(g)
240
95 Am
(h)
19
9F
( 0 n = anti–neutron)
139
53 I
28
14 Si
+
91
37 Rb
1
+ 3 0n
+ 0
0
4
+ 2 He
4
+ 2 He
15
8O
1
+ 0n
54
27 Co
1
+0n
101
41Nb
22
11Na
+
143
56Ba
1
+ 0n
E. Radioactive Decay
1. Types of Radioactive Decay
——————————————————————————————————————————————
Radioactive decay of a nuclide occurs as a result of an unstable nucleus. This instability is due to the
presence of too much energy, too few or too many neutrons, or too few or too many protons. The first
three types of decay below are discussed further in Section 2 (The Band of Nuclear Stability), which
follows.
Radioactive decay (other than electron capture) is distinguished from the previous nuclear reactions
by a simple fact: a nuclide undergoes radioactive decay by itself without having another particle collide
and react with it. In other words, all radioactive decay processes are SPONTANEOUS.
i) Alpha Decay
When a nucleus decays by alpha decay, it simultaneously ejects 2 protons and 2 neutrons in
the form of an alpha particle.
EXAMPLE:
243
96 Cm
239
94Pu
4
+ 2 He
ii) Beta Decay
In this form of decay, the nucleus ejects a beta particle (high energy electron). It should be
noted that the electron did not exist within the nucleus; as is shown in Section 2, which
follows, the electron is formed at the moment a neutron within the nucleus undergoes decay.
EXAMPLE:
215
83 Bi
215
84Po
+
0
–1
iii) Positron Decay
This is similar to beta decay, except that a positron (anti–electron) is ejected from the nucleus.
EXAMPLE:
iv) Gamma Decay
200
81Tl
200
80 Hg
+
0
+1
UNIT VIII : NUCLEAR REACTIONS (HONOURS)
5
————————————————————————————————————————————————
Some radioactive nuclides are formed as "daughter nuclides" or decay products of another
nuclide, and possess too much energy to be stable. If a newly formed nuclide gives off its
excess energy directly, a gamma particle is formed.
EXAMPLE:
27m
13 Al
27
13 Al
+ 0
0
(the "m" indicates a high energy nucleus, also
known as a “metastable” nucleus)
v) Electron Capture
Electron capture is a process in which the nucleus captures one of its own “inner” electrons
existing just outside the nucleus. To distinguish electron capture from a reaction in which a high
0
0
energy electron (beta particle) hits a nucleus, the symbol –1e is used instead of –1 .
EXAMPLES:
201
80Hg
+
0
–1e
201
79 Au
73
33 As
+
0
–1e
73
32 Ge
vi) Spontaneous Fission
Some heavy nuclei are so unstable that they simply split apart as a result of the proton-proton
repulsion within the nuclei. The result of such “spontaneous fission” is similar to that from
induced fission; the main difference is that a small particle has to collide with a nuclide to
produce induced fission, whereas the nuclide splits apart on its own in spontaneous fission.
EXAMPLES:
242
98 Cf
2
255
105 Ha
121
49In
111
50Sn
+
142
55Cs
1
+ 2 0n
EXERCISES:
5. Write a balanced nuclear reaction equation for each of the following.
(a)
(b)
(c)
(d)
(e)
(f)
6.
Alpha emission from Pu–242
Beta emission from Mg–28
Positron emission from Si–26
Electron capture by Ar–37
Gamma emission from Pt–193m
Beta emission from Al–30
(g)
(h)
(i)
(j)
(k)
(l)
Electron capture by Fe–55
Positron emission by Ru–93
Gamma emission by Pd–107m
Alpha emission by Cf–251
Beta emission by O–20
Electron capture by V–49
The nuclides below form by the decay modes specified. Write the symbols of the
radionuclide parents, giving both the atomic number and mass number.
(a)
(b)
(c)
(d)
(e)
Rb–80 (by electron capture)
Sb–121 (by beta emission)
Cr–50 (by positron emission)
Cl–34 (by gamma emission)
Cf–253 (by alpha emission)
(f)
(g)
(h)
(i)
(j)
Fm–257 (by electron capture)
Bi–211 (by beta emission)
Nd–141 (by positron emission)
Sm–143 (by gamma emission)
Ta–179 (by electron capture)
2. The Band of Nuclear Stability
——————————————————————————————————————————————
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A plot of the number of neutrons vs. the number of protons for all the known isotopes produces the
following graph.
band in which stable (that is,
non-radioactive) isotopes
are found
number of
neutrons
line for 1 : 1 ratio of neutrons:protons
number of protons
Neutrons "dilute" the repulsion between adjacent positively-charged protons in the nucleus. The
number of neutrons has to increase faster than the number of protons because the number of p-p
repulsions increases very rapidly and threatens to blow the nucleus apart. (At high atomic numbers,
the huge numbers of p-p repulsions make all isotopes unstable.)
A magnified portion of the band might look as shown below..
S
A
S
S
S
S
increasing
number of
neutrons
S
S
S
S = stable isotope
S
S
S
B
S
A & B & blank = unstable isotopes
S
increasing number of protons
As can be seen from the magnified portion of the band, the stable isotopes are in the middle of the
band while the unstable isotopes are on the outer part of the band.
Fact:
Isotopes ABOVE the centre of the band of nuclear stability undergo BETA DECAY.
Reason: Look at unstable isotope "A" In the diagram above. It has more neutrons than any of the
stable isotopes with the same number of protons. To become stable (move toward the
centre of the belt), it has to increase its number of protons and decrease its number of
neutrons. That is, the ratio:
# of n
# of p
must decrease.
EXAMPLE:
20
9F
undergoes BETA DECAY to lower the n:p ratio.
20
9F
20
10 Ne
+
0
–1
UNIT VIII : NUCLEAR REACTIONS (HONOURS)
7
————————————————————————————————————————————————
# of n = 11
# of p = 9
# of n = 10
# of p = 10
(note that the n:p ratio is decreased in the products)
The process that occurs in the nucleus is the conversion of a neutron to a proton:
1
0n

1
1p
+
0
–1
(Strictly speaking, the conversion of a neutron to a proton also emits a weird particle
called an anti-neutrino; go look up “neutrino” if you like to make your head hurt.)

The number of neutrons decreases by 1 and the number of protons increases by 1.
Fact:
Isotopes BELOW the centre of the band of nuclear stability undergo POSITRON DECAY.
Reason: Look at unstable isotope "B" in the diagram on the previous page. It has less neutrons than
any of the stable isotopes with the same number of protons. To become stable it has to
decrease its number of protons and increase its number of neutrons. That is, the following
ratio increases.
# of n
# of p
EXAMPLE:
17
9F
undergoes POSITRON DECAY to increase the n:p ratio.
17
9F
# of n = 8
# of p = 9
Fact:
17
8O
+
# of n = 9
# of p = 8
0
+1
(note that the n:p ratio is increased in the products)
Most unstable HEAVY NUCLIDES undergo ALPHA DECAY.
Reason: In this case, whether or not the n:p ratio changes is not important. It is more important that
the nucleus quickly get rid of as many as possible of the extra p-p repulsions that are
causing instability.
EXAMPLE:
252
99 Es
248
97 Bk
4
+ 2 He
EXERCISES:
7. The stable isotopes of K are: K–39, K–40, and K–41. If an atom of potassium–38 has the
option of decaying by positron emission or beta emission, which route will it likely take and
why? Write the nuclear equation.
8. The stable isotopes of Ar are: Ar–36, Ar–38 and Ar–40. Suppose that an atom of argon–41
can decay by either beta emission or electron capture. Which route will it likely take and why?
Write the nuclear equation.
9. What decay particle is emitted from a nuclide of low to intermediate atomic number and a
relatively high neutron : proton ratio? Explain why this particular kind of emission is most
likely.
10. What decay particle is emitted from a nuclide of low to intermediate atomic number and a
relatively low neutron : proton ratio? How does the emission of this particle benefit the
nucleus?
11. What does electron capture do to the neutron : proton ratio of a nucleus? Which kinds of
radionuclides are more likely to undergo this change: those above or those below the band of
stability?
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12. What is the most likely form of decay process which will be undergone by No–257?
3. Half Life
——————————————————————————————————————————————
When a radioactive isotope undergoes decay, not all of the nuclei decay at the same time. In fact, a
collection of unstable nuclei will decay on a random basis. It is impossible to predict exactly when an
individual atom will decay but it is possible to predict how long it will take for half of the atoms in a very
large sample to decay.
Definition: The HALF LIFE (t1/2) is the time required for one-half of the atoms in a sample
of a radioactive isotope to decay.
Note: The half life of each radioactive isotope is an unchangeable constant. Nothing can change the
half life, including temperature, pressure, chemical bonding or chemical reactions ... nothing!
EXAMPLES:
t1/2
=
45 µs for Rn–216. Since the half life is very short, many atoms decay in a given
time interval and the sample has a high radioactivity.
t1/2 = 1.41 x 1010 y for Th–232. Since the half life is so long, very few atoms decay in a
given time interval and the sample has a low radioactivity.
Assume a sample of a radioactive nuclide has an original mass of 1 g and a half-life of one hour.
After one hour (one half-life), one half of the atoms in the sample decay and form a different nuclide.
At this point, only half of the original sample remains. After a second hour passes (another half-life),
only half of the amount existing at the end of the first hour is left. In other words, at the end of two
half-lives only 1/4 (1/2 of 1/2) of the original amount remains.
If the Original mass of the nuclide sample is Ao, then
Ao
A
 1o
mass left after 1 half life
=
2
2
Ao
Ao
 2
mass left after 2 half lives =
4
2
Ao
Ao
 3
mass left after 3 half lives =
8
2
Ao
mass left after N half lives = N
2
This series of expressions suggests the following radioactive decay formula.
At =
where:
Ao
2N
N = the number of half lives which have passed
Ao = the original mass of the isotope
At = the mass left after an elapsed time “t”
The radioactive decay formula also involves an auxiliary formula:
N=
where:
tTOT
t1/ 2
t1/2 = the half life
tTOT = the total time

UNIT VIII : NUCLEAR REACTIONS (HONOURS)
9
————————————————————————————————————————————————
EXAMPLES:
a) A sample of pure Th–225 (t1/2 = 8.7 min) has a mass of 4.8 mg. What mass of Th–225
remains after 34.8 min?
First, list the available values in a “shopping list”:
Ao
At
N
t1/2
tTOT
= 4.8 mg
= unknown
=
= 8.7 min
= 34.8 min
There is sufficient information to immediately solve the auxiliary equation:
N=
34.8 min
tTOT
=
=4
8.7 min
t1/ 2
Now the radioactive decay equation is used:
A
4.8 mg
4.8 mg
=
= 0.30 mg
 At = No =
4
2
16
2
b) A sample of Np–239 originally has a mass of 19.2 mg. After 11.5 days only 0.60 mg of
Np–239 remains. What is the half life of Np–239?


The data is: Ao = 19.2 mg
At = 0.60 mg
N =
t1/2 = unknown
tTOT = 11.5 d
A
The radioactive decay formula; At = No is rearranged to give:
2
19.2 mg
A
2N = o =
= 32
0.60 mg
At
and since: 25 = 32 , then N = 5
Now:
 N =
t TOT
t1/ 2
is rearranged to give:
t1/2 =
tTOT
11.5 d
=
= 2.3 d
N
5
c) A sample of Th–225 (t1/2 = 8.7 min) has a mass of 3.0 mg after 43.5 min. What was the
original mass of Th–225 in the sample?


The data is: Ao = unknown
At = 3.0 mg
N =
t1/2 = 8.7 min
tTOT = 43.5 min
First, the number of half lives is found:
t TOT
43.5 min
=
=5
t1/ 2
8.7 min
A
At = No
is rearranged to solve for Ao :
2
N=
and the equation:
Ao = At x 
2N = 3.0 mg x 25 = 3.0 mg x 32 = 96 mg
EXERCISES:
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13. Ag–115 has a half life of 20.0 min. What mass of Ag–115 remains after 2.0 h if the original
sample contains 10.5 g of Ag–115?
14. A 2.50 µg sample of Mo–105 (t1/2 = 40 s) is allowed to decay for 160 s. What mass of
Mo–105 decays by the time 160 s elapse?
15. After 5.1 h only 4.2 mg of a 33.6 mg sample of Ru–95 remain. What is the half life of
Ru–95?
16. Tc–101 has t1/2 = 14 min. One hour and 24 min after a sample of Tc–101 is prepared,
3.00 µg of the sample is left. What was the original sample mass?
17. A chemist obtains 54.4 mg of pure I–134. After 208 min, 51.0 mg of the I–134 has
decayed. What is the half life of the I–134?
18. Cs–132 has t1/2 = 6.5 d. If a sample of Cs–132 originally has a mass of 0.030 00 mg, what
mass of Cs–132 is present after 39 d?
19. A freshly prepared sample of O–15, having a mass of 4.480 mg is left for 868 s, at which
point 4.445 mg of the O–15 decays. What is the half life of the O–15?
20. I–125 has a half life of 60 d. What mass of I–125 remains after 6 months (180 d) if the
original sample contains 12.0 g of I–125?
21. A 25.6 µg sample of Sb–132 has a half life of 2.0 min. What mass of Sb–132 decays
by the time 16 min elapses?
22. After 30 h, only 3.6 mg of a 230.4 mg sample of Ga–73 remains. What is the half-life of
Ga–73?
23. F–20 has t1/2 = 12.0 s. What is the original mass of a sample if only 3.00 mg of F–20
remain after 84.0 s?
24. In 1983 a chemist stored 448 g of pure Pb–210. It is rediscovered by one of his successors in
2083, and found to contain 14 g of Pb–210. What is Pb–210’s half-life?
25. K–40 decays to Ar–40 by positron emission with a half life of 1.4 x 109 y. A geologist
determines that a particular rock sample contains 7.0 mg of K–40 and 49.0 mg of Ar–40; that
is, 49.0 mg of K–40 have already decayed. How old is the rock?
26. A freshly prepared sample of Cu–64 having a mass of 296 mg is left for 52 h, at which time
277.5 mg have decayed. What is the half life of the Cu–64?
4. Radioactive Decay Series
——————————————————————————————————————————————
When a radioactive isotope decays, the resulting "daughter nuclides" may themselves be radioactive,
and can produce still more nuclides that may also be radioactive, and so on. Each of the isotopes
involved generally have different half lives, so that some time after the production of the original
radioactive isotope the sample will consist of a mixture of all the isotopes involved. Such a series of
decays is called a radioactive decay series.
Consider what happens when Ac–228 decays.
228
89 Ac
0
–1
+
228
90Th
UNIT VIII : NUCLEAR REACTIONS (HONOURS)
11
————————————————————————————————————————————————
This decay can be shown in a shorthand notation as:
228
89 Ac


228
90 Th
However, the Th–228 is also radioactive, so that the overall decay series is:


  
228
228
224
220
216
212




89 Ac  90 Th  88 Ra  86 Rn  84 Po  82 Pb
212
208
208



 83 Bi  81Tl  82 Pb (stable)











Lead is the final, stable decay product formed by many such radioactive decay series. It is believed
that substantial
amounts
of uranium
present at the time the earth was formed, but that lead was


 were 


much less common. The relative scarcity of uranium and relative abundance of lead which is
presently found today is thought to be the direct result of slow radioactive decay of uranium to lead.
EXERCISES:
27. Th–225 decays in a series of reactions involving the emission of an alpha, another
alpha, another alpha, another alpha, and a beta. Identify all the species formed during the
decay series and state the final stable product.
28. Es–250 decays in a series of reactions involving electron capture, alpha emission,
another alpha emission, and another alpha emission. Identify all the species formed during
the decay sequence and state the final product.
29. No–257 decays in a series of reactions involving alpha emission, electron capture,
alpha emission, beta decay, alpha emission, alpha emission, beta decay, alpha emission,
alpha emission, and beta decay. Identify all the species formed during the decay sequence
and state the final product.
F. Nuclear Reactors
1. The CANDU (CANadian Deuterium Uranium) Reactor
——————————————————————————————————————————————
A schematic diagram of the CANDU reactor is shown below.
PRIMARY SYSTEM
SECONDARY SYSTEM
12
Hebden : Chemistry 11 Honours
————————————————————————————————————————————————
steam
fu el chan nels
he avy wa ter
mo derator
ge nerator
he at e xchan ger
he avy wa ter
pu mp
ord inary water
pu mp
river or la ke
wate r
Powdered uranium oxide “pitchblende” (U3O8, a mixture of 2 parts UO3 and 1 part UO2) is pressed
into a pellet and several pellets are assembled in a metal tube to form a single fuel element:
Several fuel elements are then assembled into a fuel bundle:
The fuel bundle is 50 cm long, has a mass of 25 kg and remains in the reactor about a year.
Each fuel bundle has an energy equivalent to 450 000 L of oil or 500 tonnes of coal.
As nuclear fission occurs in the fuel channels, the heat energy is absorbed by heavy water in a metal
box surrounding the reactor core. High-energy neutrons from the fission reaction bounce off D2O
molecules.
path of
neutron
D2O
During these collisions the neutrons give off their excess energy in the form of heat. D2O is used
since neutrons rebound off D2O, leaving the D2O unaltered, whereas H2O absorbs neutrons more
easily and can undergo an unwanted nuclear reaction.
The D2O surrounding the core is called a moderator because the D2O "moderates" or slows down
the neutrons and transforms their energy into heat.
UNIT VIII : NUCLEAR REACTIONS (HONOURS)
13
————————————————————————————————————————————————
The heat from the reactor core is carried away by heavy water in the primary cooling system. The
coolant D2O is kept under pressure to prevent it from boiling. Eventually, the D2O in the primary
system becomes radioactive and must be replaced.
The primary system is cooled by a secondary cooling system. The water in the secondary system
boils and the steam drives a turbine-generator, which in turn produces electricity. Because the
secondary system has no contact with radioactive material, ordinary water can be used.
The excess heat from the secondary system is carried away by cooling water from a lake or river.
The amount of heat which enters the river/lake is often sufficient to raise the river/lake water
temperature by 2-3oC, giving rise to a certain amount of "thermal pollution".
2. Fast-Breeder Reactors
——————————————————————————————————————————————
A FISSILE substance will undergo FISSION when hit by a neutron. For example, U–235 is fissile, but
U–238 will not undergo fission and is not fissile.
On the other hand, U–238 is said to be "FERTILE". A FERTILE substance can be changed into a
FISSILE substance when it absorbs a neutron. Therefore, when U–238 absorbs a neutron it "gives
birth" (it is "fertile") to fissile Pu–239 (after passing through an intermediate, Np–239):
238
92 U
+ 01n
239
92 U
 (23.5 min)

239
93 Np
  (2.36 d)


239
94 Pu
The Pu–239 produced is highlyradioactive and highlytoxic (as well as being fissile) and remains





dangerous for a long time since it has a half life of 24 110 y. Also, Pu–239 is easily made into a
nuclear bomb so that elaborate safeguards must be set up to protect against theft of Pu–239 by
terrorists.
Since only fissile materials, not fertile, can be used in a nuclear reactor, the "fast-breeder reactor" was
developed to "breed" more fissile material than it consumes. It is referred to as "fast" since it uses no
moderator to slow down the neutrons. The core of fissile material is surrounded by a blanket of fertile
material. The neutrons produced by the core are absorbed by the blanket, causing the fertile blanket
to slowly become fissile. Since an atom of U–235 produces many neutrons when it undergoes fission,
and each neutron is capable of producing a new, fissile atom, a breeder reactor produces more fuel
than it uses. After being converted into fissile material, the "blanket" is processed to allow the newlybred fissile material to be recovered.
Besides the dangerous nature of the Pu–239 formed, there is another serious drawback to the fast
breeder reactor: it uses liquid sodium as a coolant. The liquid sodium is itself cooled by pipes filled
with water. Unfortunately, sodium is a corrosive metal which corrodes through metal pipes if not
regularly inspected, and explodes when it contacts water (liquid sodium burns when it contacts air.)
3. The Problem of Nuclear Waste Disposal
——————————————————————————————————————————————
The fuel in a nuclear reactor is replaced after most of the fissile material has been exhausted.
However, the "spent" fuel still contains some fissile material and virtually all of the radioactive decay
products are themselves highly radioactive. The highly radioactive nuclear "waste" remains
dangerous for many tens of thousands of years and hence must be permanently stored in protective
containers away from earthquake zones, underground water, surface erosion, etc. We cannot
assume that future generations will be able to deal with leaking containers and therefore we must find
disposal methods which will keep the waste "safe" for hundreds of thousands of years. Concrete or
steel containers deteriorate in a few years and as a result are only temporary. Unfortunately, most
nuclear waste is presently stored in concrete or steel. Some of the disposal methods currently being
investigated are:
a) mixing solid radioactive material into special ceramic glass which does not allow
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Hebden : Chemistry 11 Honours
————————————————————————————————————————————————
radioactivity to be "leached" out of the glass.
b) burial of the radioactive material in deep salt deposits, possibly in the form of a ceramic
glass. The salt deposits are dry and being millions of years old should remain undisturbed for
another million years. Burial in deep down-thrusting geological faults has also been suggested,
since this would allow the material to be carried deep into the earth's mantle.
G. Some Applications of Radioactivity
1. Radiological Dating
——————————————————————————————————————————————
Virtually everything in nature contains at least a small amount of radioactive atoms. This natural
radioactivity exists in the rocks, soil and air around us, and even within us. Because the half life of
radioactive nuclides cannot vary, and because the half lives of elements is known with great precision,
it is possible to estimate the age of materials.
(a) Radiocarbon Dating: Cosmic rays often generate high energy neutrons which collide
with the nitrogen in the atmosphere to form C–14:
14
7N
1
14
6C
+ 0n
1
+ 1H
The C–14 then decays by beta emission:
14
6C
14
7N
+
0
–1
The production of C–14 proceeds at a more or less steady rate (small rate corrections are made
by comparison with tree-ring studies). As a result, all living things, including trees, contain a
constant small percentage of radioactive carbon in addition to the non-radioactive C–12 present.
As long as the tree is alive, there is a constant intake of fresh C–14, and the percentage of
C–14 present is constant. When the tree dies the amount of C–14 starts to decrease with a half
life of 5715 y. By comparing the amount of radioactivity in an ancient sample of wood with the
radioactivity in a fresh sample of wood, it is possible to determine the time that has passed since
the tree was living. Other samples such as bone, cloth, pollen, etc. can also be dated.
To measure the amount of C–14 present, a measuring device records the number of beta
particles emitted as C–14 nuclei decay (each decay is called a "disintegration"). The number of
disintegrations in a gram of carbon is called the material’s specific activity. Fresh organic
material has a specific activity of 918 disintegrations/second per gram of carbon. A 1.0 g sample
of carbon which has 459 disintegrations/second is then one half–life old; that is, 5715 y old.
EXAMPLE: A sample of bone is found in an ancient fire pit excavated during an archaeological
investigation. The bone’s C–14 has a specific activity of 115 disintegrations/s.
What is the bone's approximate age?
At =
Ao
2N
2N =
Now:


is rearranged to give:
918 di sintegrations / s
Ao
=
= 8.0 , and: N = 3
115 di sintegrations / s
At
N=
t TOT
t1/ 2
is rearranged to give:
tTOT = N x t1/2 = 3 x 5715 y = 1.7 x 104 y
UNIT VIII : NUCLEAR REACTIONS (HONOURS)
15
————————————————————————————————————————————————
(b) Geological Dating: Modern theories about the formation of the solar system suggest that the
ratio of lead to uranium present was very small when the earth's crust formed. The half life of U–
238 is known to be 4.468 x 109 y. Calculating the age of a rock sample requires finding the
amount of U–238 present relative to the amount of lead. (Strictly speaking, the lead nuclide
measured is Pb–206 because it is the decay product of U–238. If other lead isotopes are present
then the rock originally contained some natural lead and a suitable correction can be made.)
The decay of K–40 to Ar–40 by positron emission is also used to date rocks.
40
19 K
40
18 Ar
+
0
+1
This process has a half life of 1.277 x 109 y. When a sample of rock melts, any gas bubbles
trapped in the rock are released. After the rock solidifies, any K–40 present slowly undergoes
decay to Ar–40, which is an inert gas. The Ar–40 accumulates as microscopic gas bubbles within
the rock. In order to determine the age of the rock from the time it solidified, the rock is powdered
and heated to release the Ar–40 present. By measuring the ratio of K–40 to Ar–40, the number of
elapsed half lives can then be found.
2. Tracer Analysis
——————————————————————————————————————————————
Tracer analysis is a technique in which radioactive isotopes are added to a chemical or biological
system. By deliberately introducing a radioactive nuclide into a living system, it becomes possible to
“trace” exactly where the radioactive nuclide accumulates by simply sampling various parts of the
system and monitoring the parts for the presence of radioactivity.
For example, H–3 (tritium) can be introduced into an organic drug molecule by an appropriate
chemical reaction. Injecting the drug into an animal allows a researcher to see how the drug acts in
the body and what part of the body accumulates the drug. If, for example, a radioactive drug reacts
primarily with white blood cells and an animal is injected with the drug, then a sample of blood taken
from the animal can be separated into red blood cells, white blood cells and plasma. Only the white
blood cells will be found to be radioactive. Similarly, I–131 (t1/2 = 8.04 d) is administered to people
with suspected thyroid gland problems. If the thyroid gland does not accumulate much of the I–131,
thyroid malfunctioning is effectively proven. A very important radioisotope is Tc–99m. Brain tumours
–
–
strongly concentrate any TcO 4 present, so that an injection of TcO 4 directly into the brain tissue
makes any tumours show up as concentrated regions of radioactivity.
3. Neutron Activation Analysis
——————————————————————————————————————————————
Neutron activation analysis is a technique in which a sample is bombarded by neutrons, causing the
different elements in the sample to become radioactive and emit gamma radiation. The intensity and
frequency of the radiation is then used to identify the amounts and types of the elements present.
Many stable nuclei can be changed into gamma emitters by bombarding the nuclei with neutrons. For
example:
68
30 Zn
+ 01n
69m
30 Zn
69
30 Zn
+ 00 
The gamma radiation emitted by each type of isotope is unique, so that by measuring the frequency
of the gamma radiation emitted by a sample after neutron activation has occurred, the particular


 Moreover, by measuring the intensity of each type

isotopes present
in the sample
can beidentified.
of radiation it is possible to measure the amount of each isotope present.
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Hebden : Chemistry 11 Honours
————————————————————————————————————————————————
This incredibly sensitive technique is able to measure concentrations as low as 10–9 %. It is possible
to determine if a historical figure was poisoned by arsenic if a single hair from the person is available.
The minute amount of arsenic in the hair can be measured easily. An added feature of this method is
that the testing procedure does not destroy the sample.
EXERCISE:
30. Iridium is believed to have been deposited in a thin layer of rock found around the
world as a result of the impact of a comet or asteroid having a diameter of 10 km. During one
neutron activation analysis of a rock sample taken from the iridium-rich layer, a rock sample
was accidentally contaminated
"due to the platinum wedding or engagement ring worn by a technician who had prepared
the samples for analysis. Platinum used in jewelry contains about 10 percent iridium ... If a
platinum ring loses 10 percent of its mass in 30 years, the average loss per minute, if it all
deposits on a sample, is about [a hundred times] higher than our sensitivity of
measurement."
Walter Alvarez, Frank Asaro, Helen V. Michel, Luis W. Alvarez, "Iridium Anomaly
Approximately Synchronous with Terminal Eocene Extinctions", Science, Volume 216,
p.886, 1982, as quoted by Carl Sagan, Ann Druyan, "Comet", Random House, New
York, 1985, p.282.
If a platinum wedding ring has a mass of 20 g and a technician handles the sample for
1 minute, what must be the "sensitivity" of the neutron activation analysis measurement; that
is, the minimum number of grams of iridium that can be detected? How many atoms of
iridium are lost from the ring each minute?