Episode 512: Nuclear equations
Now that your students are familiar with different types of radiation, you can look at the processes
by which they are emitted.
Summary
Discussion: Nuclide notation and N-Z plot. (10 minutes)
Student Questions: Practice with notation. (10 minutes)
Worked Examples: Equations for alpha, beta and gamma decay. (20 minutes)
Student Questions: Practice with nuclear equations. (30 minutes)
Discussion:
Nuclide notation
Revise nuclide notation:
A
Z
X
Discuss how A = mass or nucleon number, Z = charge or atomic number and N = neutron
number are related (A = Z + N).
Discuss isotopes (common examples: H, D and T, U-235 and U-238, C-14 and C-12).
Set the task of finding out the name for nuclides having the same A but different Z (isobars), and
the same N but different Z (isotones).
Show an N-Z plot (Segrè plot).
Neutron
number
n
p + - + 
A
+
B
p
n + + + 
Proton number
1
Student questions:
Practice with notation
Set some simple questions involving nuclide notation.
TAP 512-1: Nuclide notation
Grid showing change in A and Z with different emissions
A, Z-1
Nucleon number
A, Z
A, Z+1
(A)
-
+

A-4, Z-2
Proton number (Z)
Neutron number
(N)
+
N+1, Z-1
N, Z
-
N-2, Z-2
N-1, Z+1

2
Proton number (Z)
Worked examples:
Equations for alpha, beta and gamma decay
Nuclear decay processes can be represented by nuclear equations. The word equation implies
that the two sides of the equation must ‘balance’ in some way.
TAP 512-2: Decay processes
You could give examples of equations for the sources used in school and college labs.
 sources are americium-241,
241
95
Am
4
Am  237
93 Np  2 He.
- sources are strontium-90,
90
38
241
95
Sr  90
39 Np 
0
1
90
38
Sr
e.
The underlying process is:
n –> p + e- + 
Here,  is an antineutrino. Your specification may require you to explain why this is needed to
balance the equation.
–> p + e- into the AZ notation:
1
0
n 11 H 
0
1
e.
 sources are cobalt-60
60
60
27
 decay of the 27 Co . The
Co . The  radiation comes from the radioactive daughter 60
28 Ni of the
60
28
Ni is formed in an ‘excited state’ and so almost immediately loses
the energy by emitting a 
 or  decay, and all such  rays
have a well-defined energy. (So a cobalt-60 source which is a pure gamma emitter must be
designed so that betas are not emitted. How? – (by encasing in metal which is thick enough to
absorb the betas but which still allows gammas to escape.)
Student questions:
Practice with nuclear equations
TAP 512-3: Practice with nuclear equations
The more unusual decay processes (positron emission, neutron emission, electron capture) could
be included, and students challenged to write them as nuclear equations.
3
TAP 512-1: Nuclide notation
A periodic table may be needed
A
1 Write down the nuclear notation ( Z X ) for:
(a)
an alpha particle
(b)
a proton
(c)
a hydrogen nucleus
(d)
a neutron
(e)
a beta particle
(f)
a positron.
A
2 Write down the nuclear notation ( Z X ) for
(a)
carbon 13
(b)
nitrogen 14
(c)
neon 22
(d)
tin 118
(e)
iron 54
4
Answers
1
4
2
He
(a)
an alpha particle
(b)
a proton
(c)
a hydrogen nucleus
(d)
a neutron
(e)
a beta particle
(f)
a positron.
1
1
H or 11 p
1
0
1
1
H
n
0
1
0
1
e
e
2
13
6
(a)
carbon-13
(b)
nitrogen-14
(c)
neon-22
(d)
tin-118
118
50
(e)
iron-54
54
26
22
10
C
14
7
N
Ne
Sn
Fe
5
TAP 512-2: Decay processes
Radioactive decay processes
 decay
Z
N

Z–2
N–2
2 fewer protons
2 fewer neutrons
proton number Z
– decay
–
Z
N
Z+ 1
N–1

1 more proton
1 less neutron
proton number Z
+ decay
+
Z–1
N+1

Z
N
1 less proton
1 more neutron
proton number Z
decay

Z
N
same protons
and neutrons
proton number Z
6
External reference
This activity is taken from Advancing Physics chapter 18, 90O
7
TAP 512-3: Practice with nuclear equations
1
The isotope
element?
235
U decays into another element, emitting an alpha particle. What is the
This element decays, and the next, and so on until a stable element is reached. The
complete list of particles emitted in this chain is:
235
92 U  [, , , , , , , , , , ]  X.
What is the stable element X? (You could write down each element in the series, but
there is a quicker way.)
2
The following fission reaction can take place in a nuclear reactor:
235
1
137
95
1
92 U  0 n 55 Cs  [ ] Rb  0 n.
Complete the equation, showing how many neutrons are produced in the reaction. What
is the significance of the number of neutrons produced?
Why are the products of the reaction, caesium-137 and rubidium-95, likely to be
radioactive? What type of decay are these isotopes likely to show?
3
Boron absorbs neutrons with results as follows:
10
5
B  01 n  73 Li  24  .
Why is boron suitable for use in a control rod?
4
27
When the isotope 13 Al is irradiated with alpha particles, the products from each
aluminium nucleus are a neutron, and a nuclide that emits positrons to give the stable
30
isotope 14 Si . Write nuclear equations for these two processes.
8
5
Complete the following nuclear equations. In each case describe the decay process:
131
53 I

Xe 
67
0
31Ga  1e
11
6C

Zn
B  01e

99m
43Tc
6

0
1e
Tc   .
The Manhattan Project, the development of the atomic bomb, led to the discovery of the
transuranic elements (elements beyond uranium in the periodic table). Plutonium,
element 94, is formed by the bombardment of uranium-238 with neutrons. The nuclear
equations are:
238
1
239
92 U 0 n 92 U
239
239
0
92 U 93 Np  1 e
239
239
0
93 Np  94 Pu 1 e
Complete the following nuclear equation for the formation of americium:
239
94 Pu
201 n  Am 
0
1e.
Curium is produced if plutonium-239 is bombarded with alpha particles. If the curium
242
isotope is 96 Cm , complete the equation
239
94 Pu

4
2

If curium is made the target for alpha particle bombardment californium is produced.
Complete the nuclear equation to find the atomic number of californium:
242
96 Cm

4
245
Cf
2 
 01n.
By firing heavier particles such as carbon or boron ions at the target materials heavier
elements can be synthesised. Complete the nuclear equation (Lw is lawrencium)
252
Cf 
9
5B
 Lw  4 01 n.
One of the transuranic elements is commonly found in the home. Which is this and where
is it used?
9
Practical advice
These questions are intended for homework or practice in class.
Answers and worked solutions
The complete decay chain involves the loss of seven alpha particles ( 2  ) and four beta
4
1.
0
e
particles ( 1 ). This represents a loss of 7  4, i.e. 28, in mass number and (7  2 – 4),
i.e. 10, in atomic number. X is therefore an isotope with mass number (235 – 28), i.e.
208
207, and atomic number (92 – 10), i.e. 82. This is lead, 82 Pb .
2.
235
92 U

1
137
0 n 55 Cs

95
37 Rb
 3 01 n.
The reaction produces more neutrons than it absorbs, this will cause a ‘chain reaction’.
To see why the products of the reaction are likely to be radioactive you need to consult
the plot of neutron number against atomic number for the known stable nuclei. For
elements with atomic numbers up to about 30 the number of neutrons in the nucleus is
the same as the number of protons if the nucleus is stable. For higher atomic numbers
the ratio of neutrons to protons gradually increases to 1.5. Look at the position of both
137
95
55 Cs and 37 Rb on the plot and you will see that they both have a considerable excess
of neutrons. They are therefore likely to be radioactive. To become more stable the
nuclides need to decrease the neutron to proton ratio. The emission of a beta particle
does this, increasing the number of protons by one and decreasing the number of
neutrons by one. These isotopes are therefore likely to decay by emitting beta particles.
3.
When boron captures a neutron it is transformed into a stable isotope. If the control rods
are pushed into the reactor more neutrons are absorbed, causing the chain reaction to
slow down. If they are pulled out the chain reaction will proceed more vigorously.
4.
27
13 Al
30
15 P


4
2

30
14 Si
30
15 P


1
0n
0
1e
5.
131
131
53 I 54 Xe

11
11
6 C 5 B

0
1e
0
1e
beta decay
positron emission
67
0
67
electron capture
31 Ga  1e 30 Zn
99m
99
nuclear rearrangem ent
43 Tc  43 Tc
6.
239
1
241
94 Pu  2 0 n 95 Am 
239
4
242
94 Pu  2  96 Cm 
242
4
245
96 Cm  2  98 Cf 
252
98 Cf

9
257
5 B103 Lw
0
1e
1
0n
1
0n
 4 01 n.
Americium is commonly found in homes where it is used in smoke detectors.
External reference
This activity is taken from Advancing Physics chapter 18, 210S
10