F1
α- and β-radiation consist of α- and β-particles. An α-particle is the nucleus of the helium isotope 24 He .
There are two types of β-particle. A β−-particle is an electron, a β+-particle is a positron (the antiparticle of the
electron).
γ-radiation is electromagnetic radiation of very short wavelength.
FLAP P9.2
Radioactive decay
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F2
The two principles are: the conservation of linear momentum and the conservation of (relativistic) energy.
FLAP P9.2
Radioactive decay
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F3
The equation is:
137 Cs
55
→
137 Ba
56
+ e − + νe
The equation exhibits the conservation of nucleon number; there are 137 nucleons on each side. It also exhibits
charge conservation; there is a charge of +55e from the 55 protons before decay, and (+56 − 1)e from the 56
protons and one electron afterwards. β −-decay always involves the emission of an antineutrino, as well as an
electron, which carries some of the energy released in the decay.
FLAP P9.2
Radioactive decay
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F4
The equation for the decay of a radionuclide is:
N (t ) = N0 e − λ t
The decay constant λ is related to the half-life by:
log e 2
0. 693
λ =
=
τ
τ
The activity is proportional to N, the number of nuclei present. The time T for N(t) to drop from N 0 to N 0 /100,
and hence for the activity to drop to 1% of its initial value, is given by:
N0
eλ T =
= 100
N (T )
Therefore
T =
log e 100
log e 100
4. 605
=
τ =
× 8 min
λ
log e 2
0. 693
which gives T = 53 min.
1
FLAP P9.2
Radioactive decay
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S570 V1.1
Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route
through the module and to proceed directly to Ready to study? in Subsection 1.3.
Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the
Closing items.
If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it
here.
FLAP P9.2
Radioactive decay
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R1
In 222
88 Ra the mass number A = 222 and the atomic number Z = 88, so there are Z = 88 protons and
(A − Z) = (222 − 88) = 134 neutrons.
FLAP P9.2
Radioactive decay
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R2
The mass of the proton is 1.672 × 10−27 kg. According to the Einstein’s mass–energy equation E = mc2 its energy
equivalent is
1
1.672 × 10−27 kg × (2.998 × 108 m s−1)2 = 1.5028 × 10−10 J
1
1
1
1
Since 1 eV = 1.602 ×
the mass energy of the proton is 1.503 × 10−10 J/1.602 × 10−19 J (eV)−1
= 938.1 MeV. The mass of the proton is therefore 938.1 MeV/c2.
1
10−191 J,
1
FLAP P9.2
Radioactive decay
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1
1
S570 V1.1
1
1
R3
y = e ax , so ax = loge y and hence x =
1
1
log e y .
a
dy
d ax
=
(e ) = ae ax
dx
dx
(The maths strand of FLAP gives further discussion of logarithms, exponential functions and derivatives.)
FLAP P9.2
Radioactive decay
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T1
For the decay
238 U
92
→
234 Th
90
+ 24 He + 4. 30 MeV we
may use Equations 2 and 3.

4 
Kα ≈ Q 1 −

AP 

(Eqn 2)
 4 
KD ≈ Q 

 AP 
(Eqn 3)
The kinetic energies are therefore
4 
Kα ≈ 4. 30  1 −
MeV = 4. 23 MeV

238 
and
4 
K D ≈ 4. 30 
MeV = 0. 07 MeV
 238 
FLAP P9.2
Radioactive decay
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T2
In a perfect vacuum, α-particles would have infinite range. The absence of matter means that they would travel
unimpeded, with no loss of energy.
FLAP P9.2
Radioactive decay
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The ‘decay line’ P →
is from left to right downwards at 45° to the
horizontal, and the ‘decay line’ P → Dβ+ is from right to left upwards at
45° to the horizontal. They are shown in Figure 13. Only a nucleus
below the stability line, say at P, can achieve greater stability via
β+-decay (whereas one above the line can achieve greater stability by
β−-decay).
Dβ−
number of neutrons in the
nucleus, N = A − Z
T3
the stability line
Dβ+
P
Dβ−
number of protons in the nucleus, Z
Figure 13
FLAP P9.2
Radioactive decay
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S570 V1.1
See Answer T3.
3
T4
In atomic mass units the electron mass is
9.11 × 10 −31 kg
= 5.49 × 10–4 u
1. 66 × 10 −27 kg u −1
1
Using Einstein’s mass–energy equation E = mc2 with m = 9.11 × 10–31 kg gives electron rest energy E as
1
E = 9.11 × 10–31 kg × (3.00 × 108 m s–1)2 = 8.20 × 10–14 J
8. 20 × 10 −14 J
=
= 0. 512 MeV
1. 60 × 10 −19 J (eV) −1
1
1
1
1
Electron rest mass is 0.512 MeV/c2 .
1
FLAP P9.2
Radioactive decay
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T5
For β−-decay we can calculate Q directly from the atomic masses. From the data, the missing mass is
∆m = (14.003 242 – 14.003 074) u = 1.68 × 10–4 u
1
1
1
1
Q1 = (1.68 × 10–4 × 931.5) MeV = 0.156 MeV.
1
1
β+-decay
For
we need to subtract two electron rest energies when we calculate Q2 from the atomic masses.
Using the result from Answer T4, the electron rest energy is 0.512 MeV and we find from the data,
1
Q2 = 931.5(13.005 739 – 13.003 355) MeV – 2(0.512) MeV
1
1
1
1
= 2.22 MeV – 1.02 MeV = 1.20 MeV
1
1
1
FLAP P9.2
Radioactive decay
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T6
If we substitute numerical values into Planck’s formula (Equation 8) we get:
E
10 6 eV × 1. 602 × 10 −19 J (eV) −1
f =
=
6. 626 × 10 −34 J s
h
= 2. 418 × 10 20 s −1 = 2. 418 × 10 20 Hz
FLAP P9.2
Radioactive decay
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T7
energy/MeV half-life/ps
We will denote the three transitions L(left), C(centre) and R(right)
according to their position in Figure 8. The values of the energies and
frequencies of the γ-ray photons are:
(i) EL = (1.637 − 0.862) MeV = 0.775 MeV
0. 775 MeV × 1. 602 × 10 −13 J ( MeV) −1
fL =
6. 626 × 10 −34 J s
1
1.637
3.1
1.317
8.7
0.862
3.5
1
= 1.87 × 1020 Hz.
1
(ii) EC = (1.317 − 0) MeV = 1.317 MeV, fC = 3.18 × 1020 Hz.
1
1
1
(iii) E R = (0.862 − 0) MeV = 0.862 MeV, fR = 2.08 × 1020 Hz.
1
Figure 8
3
1
Part of the energy level diagram of
1
72 Se
34
0
showing possible γ-decays and the half-lives of individual states. Each
vertical arrow represents a strong transition leading to the emission of a γ-ray. The half-life here is equal to the time for half a
sample of excited state nuclei to leave the excited state by decay to a lower state. The meaning of half-life will be discussed
more fully in Section 3 of this module.
FLAP P9.2
Radioactive decay
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T8
(a) The thickness x necessary to reduce the intensity to 10% of its initial value may be obtained from
Equation 10
I(x) = I0e−µx
(Eqn 10)
I0
= I0 e − µx . If both sides are divided by I 0 and reciprocals are taken, we find eµ x = 10.
10
log e 10
Then if natural logs are taken of both sides and then divided by µ we have x =
.
µ
If we now substitute µ = 0.055 mm−1 (as given in the text) we find x = 42 mm.
with I = I0/10:
1
(b) If we repeat the calculation for 1% we find x =
1
log e 100
, so x = 84 mm.
µ
1
(c) For water, the value of µ is much smaller (0.055 mm−1 ) and the value of x proportionately larger
(i.e. 0.055 mm −1/0.005 mm −1 = 11 times larger) than the value in (a) for lead, so the thickness of water is
11 × 42 mm = 0.46 m.
1
1
1
1
1
FLAP P9.2
Radioactive decay
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T9
(a) It is part of the neptunium series, because A = 4n + 1.
(b) The mass number A changes by 8. As only α-decay can change A and each α-decay reduces A by 4, there
must be two α-decays. However, each α-decay also reduces the atomic number Z by 2, so two α-decays alone
would reduce Z by 4, but we want only to reduce it by 2. Each β −-decay increases Z by 1, so there must be two
β−-decays. Therefore in total, there are two α-decays and two β−-decays.
FLAP P9.2
Radioactive decay
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T10
(a) The decay constant is given by Equation 16
log e 2
0. 693
λ =
=
τ
τ
0. 693
λ =
= 0.087 day−1
8. 0 day
(Eqn 16)
1
Alternatively, λ =
0. 693
= 1. 0 × 10 −6 s −1
8. 0 × 24 × 3600 s
(b) We have R(t) = R0 /10, and the algebra is the same as in Question T8. The time taken is therefore:
log e 10
2. 303
t =
=
= 26.5 days
λ
0. 087 day −1
1
Alternatively, t =
2. 303
= 2.3 × 106 s
1. 0 × 10 −6 s −1
1
FLAP P9.2
Radioactive decay
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T11
From the answer to Question T10(a):
λ = 1.0 × 10−6 s−1
1
As in the example in the text, we can estimate N0 as we know the mass number to be 131
1. 00 × 10 −9 kg
N0 =
= 4. 60 × 1015
(131 u) × (1. 66 × 10 −27 kg u −1 )
Therefore R0 = λN0 = (1.0 × 10−6 s−1) × 4.60 × 1015 = 4.6 × 109 s−1 = 4.6 GBq = 0.12 Ci.
1
FLAP P9.2
Radioactive decay
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1
S570 V1.1
1
1
T12
The age in terms of half-lives of
14 C
6
is given by Equation 18,
 log e ( R0 R) 
t = τ

 log e 2 
(Eqn 18)
i.e.
 log e (100 55) 
t = 5730 
 years = 4942 years
log e 2


FLAP P9.2
Radioactive decay
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T13
We substitute the expected age into Equation 19,
t =

τ
N (t ) 
log e 1 + D
log e 2
N P (t ) 

(Eqn 19)
and put X = N D/NP.
1. 3 × 10 9
3. 5 × 10 9 =
log e (1 + X )
log e 2
loge (1 + X) = 3.5 × 0.693/1.3 = 1.87. So 1 + X = e1.87 = 6.5, so X = 5.5 and ND/NP = 5.5.
1
FLAP P9.2
Radioactive decay
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S570 V1.1