Phy 310 chapter 8

is defined as the spontaneous
disintegration of certain atomic
nuclei accompanied by the
emission of alpha particles,
beta particles or gamma
radiation.
CHAPTER 8: Radioactivity
(3 Hours)
Dr Ahmad Taufek Abdul Rahman
School of Physics & Material Studies, Faculty of Applied Sciences, Universiti Teknologi MARA Malaysia, Campus of Negeri Sembilan
1

At the end of this chapter, students should be able to:
 Explain α, β+, βˉ and γ decays.
 State decay law and use
 Define activity, A and decay constant, .
 Derive and use
 Define half-life and use
Learning Outcome:
8.1 Radioactive decay (2 hours)
N
dt
dN



t
e
N
N 

 0
t
e
A
A 

 0
OR

2
ln
2
/
1 
T
2
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY

 The radioactive decay is a spontaneous reaction
that is unplanned, cannot be predicted and
independent of physical conditions (such as
pressure, temperature) and chemical changes.
 This reaction is random reaction because the
probability of a nucleus decaying at a given instant
is the same for all the nuclei in the sample.
 Radioactive radiations are emitted when an unstable
nucleus decays. The radiations are alpha particles,
beta particles and gamma-rays.
8.1 Radioactive decay
3

 An alpha particle consists of two protons and two neutrons.
 It is identical to a helium nucleus and its symbol is
 It is positively charged particle and its value is +2e with mass
of 4.002603 u.
 When a nucleus undergoes alpha decay it loses four nucleons,
two of which are protons, thus the reaction can be represented
by general equation below:
 Examples of  decay :
8.1.1 Alpha particle ()
He
4
2 α
4
2
OR
Q


 He
Pb
Po 4
2
214
82
218
84
(Parent) ( particle)
(Daughter)
X
A
Z 
Y
4
2


A
Z
  Q
He
4
2
Q


 He
Ra
Th 4
2
226
88
230
90
Q


 He
Rn
Ra 4
2
222
86
226
88
Q


 He
Th
U 4
2
234
90
238
92 4

 Beta particles are electrons or positrons (sometimes is called
beta-minus and beta-plus particles).
 The symbols represent the beta-minus and beta-plus (positron)
are shown below:
 Beta-minus particle is negatively charged of 1e and its mass
equals to the mass of an electron.
 Beta-plus (positron) is positively charged of +1e (antiparticle
of electron) and it has the same mass as the electron.
 In beta-minus decay, an electron is emitted, thus the mass
number does not charge but the charge of the parent
nucleus increases by one as shown below:
8.1.2 Beta particle ()
e
0
1


β
OR e
0
1

β
OR
Beta-minus
(electron) :
Beta-plus
(positron) :
(Parent) ( particle)
(Daughter)
X
A
Z 
Y
1
A
Z 
  Q
e
0
1

5

 Examples of  minus decay:
 In beta-plus decay, a positron is emitted, this time the charge of
the parent nucleus decreases by one as shown below:
 For example of  plus decay is
Q


  e
Pa
Th 0
1
234
91
234
90
Q


  e
U
Pa 0
1
234
92
234
91
Q


  e
Po
Bi 0
1
214
84
214
83
(Parent) (Positron)
(Daughter)
X
A
Z 
Y
1
A
Z 
  Q
e
0
1
Q
v 


 e
n
p 0
1
1
0
1
1
Neutrino is uncharged
particle with negligible
mass.
6

 Gamma rays are high energy photons (electromagnetic
radiation).
 Emission of gamma ray does not change the parent nucleus
into a different nuclide, since neither the charge nor the
nucleon number is changed.
 A gamma ray photon is emitted when a nucleus in an excited
state makes a transition to a ground state.
 Examples of  decay are :
 It is uncharged (neutral) ray and zero mass.
 The differ between gamma-rays and x-rays of the same
wavelength only in the manner in which they are produced;
gamma-rays are a result of nuclear processes, whereas x-
rays originate outside the nucleus.
8.1.3 Gamma ray ()
γ




He
Pb
Po 4
2
214
82
218
84
γ


 

e
U
Pa 0
1
234
92
234
91
γ



Ti
Ti 208
81
208
81
Gamma ray
7

 Table 8.1 shows the comparison between the radioactive
radiations.
8.1.4 Comparison of the properties between alpha
particle, beta particle and gamma ray.
Alpha Beta Gamma
Charge
Deflection by
electric and
magnetic fields
Ionization power
Penetration power
Ability to affect a
photographic plate
Ability to produce
fluorescence
+2e 1e OR +1e 0 (uncharged)
Yes Yes No
Strong Moderate Weak
Weak Moderate Strong
Yes Yes Yes
Yes Yes Yes
Table 8.1 8

 Figures 8.1 and 8.2 show a deflection of ,  and  in electric
and magnetic fields.
Figure 8.1
B
















 E

α
γ β γ
β
α
Figure 8.2
Radioactive
source
9

 Law of radioactive decay states:
For a radioactive source, the decay rate is directly
proportional to the number of radioactive nuclei N
remaining in the source.
i.e.
 Rearranging the eq. (8.1):
Hence the decay constant is defined as the probability that a
radioactive nucleus will decay in one second. Its unit is s1.
8.1.5 Decay constant ()







dt
dN
N
dt
dN








N
dt
dN



Negative sign means the number of
remaining nuclei decreases with time
Decay constant
(8.1)
N
dt
dN



nuclei
e
radioactiv
remaining
of
number
rate
decay


10

 The decay constant is a characteristic of the radioactive nuclei.
 Rearrange the eq. (8.1), we get
At time t=0, N=N0 (initial number of radioactive nuclei in the
sample) and after a time t, the number of remaining nuclei is
N. Integration of the eq. (8.2) from t=0 to time t :
dt
N
dN


 (8.2)

 

t
N
N
dt
N
dN
0
0

   t
N
N t
N 0
0
ln 


λt
N
N


0
ln
λt
e
N
N 
 0
Exponential law of
radioactive decay
(8.3)
11

 From the eq. (8.3), thus the graph of N, the number of remaining
radioactive nuclei in a sample, against the time t is shown in
Figure 8.3.
t
e
N
N 

 0
2
0
N
0
N
4
0
N
16
0
N
8
0
N
2
/
1
T 2
/
1
2T 2
/
1
3T 2
/
1
4T 2
/
1
5T
0
t
,
time
N
life
half
:
2
/
1 
T
Figure 8.3
Stimulation 8.1
Note:
From the graph (decay curve),
the life of any radioactive
nuclide is infinity, therefore to
talk about the life of radioactive
nuclide, we refer to its half-life.
12

 is defined as the time taken for a sample of radioactive
nuclides disintegrate to half of the initial number of nuclei.
 From the eq. (8.3), and the definition of half-life,
when , thus
 The half-life of any given radioactive nuclide is constant, it
does not depend on the number of remaining nuclei.
8.1.6 Half-life (T1/2)
t
e
N
N 

 0
2
/
1
T
t 
2
; 0
N
N 
2
/
1
0
0
2
T
e
N
N 


2
/
1
2 T
e

2
/
1
2
1 T
e 


2
/
1
ln
2
ln T
e

λ
λ
T
693
.
0
2
ln
2
/
1 

Half-life (8.4)
13

 The units of the half-life are second (s), minute (min), hour
(hr), day (d) and year (y). Its unit depend on the unit of decay
constant.
 Table 8.2 shows the value of half-life for several isotopes.
Table 8.2
Isotope Half-life
4.5  109 years
1.6  103 years
138 days
24 days
3.8 days
20 minutes
U
238
92
Po
210
884
Ra
226
88
Bi
214
83
Rn
222
86
Th
234
90
14

second
per
decays
10
7
3
Ci
1 10
. 

 is defined as the decay rate of a radioactive sample.
 Its unit is number of decays per second.
 Other units for activity are curie (Ci) and becquerel (Bq) – S.I.
unit.
 Unit conversion:
 Relation between activity (A) of radioactive sample and time t :
 From the law of radioactive decay :
and definition of activity :
8.1.7 Activity of radioactive sample (A)






dt
dN
second
per
decay
1
Bq
1 
N
dt
dN



dt
dN
A 
15

 Thus
0
0 N
A 


N
A 

 and
t
e
N
N 

 0
 
t
e
N
A 
 

 0
λt
e
A
A 
 0
Activity at time t Activity at time, t =0
and
  t
e
N 
 

 0
(8.5)
16

A radioactive nuclide A disintegrates into a stable nuclide B. The
half-life of A is 5.0 days. If the initial number of nuclide A is 1.01020,
calculate the number of nuclide B after 20 days.
Solution :
The decay constant is given by
The number of remaining nuclide A is
The number of nuclide A that have decayed is
Therefore the number of nuclide B formed is
Example 1 :
Q
B
A 

0
.
5
2
ln


days
20
;
10
1.0
days;
0
.
5 20
0
2
/
1 


 t
N
T
2
/
1
2
ln
T


1
days
139
.
0 

t
e
N
N 

 0     
20
139
.
0
20
10
0
.
1 

 e
N
nuclei
10
2
.
6 18


18
20
10
2
.
6
10
0
.
1 



nuclei
10
38
.
9 19


nuclei
10
38
.
9 19
 17

a. Radioactive decay is a random and spontaneous nuclear
reaction. Explain the terms random and spontaneous.
b. 80% of a radioactive substance decays in 4.0 days. Determine
i. the decay constant,
ii. the half-life of the substance.
Solution :
a. Random means that the time of decay for each nucleus
cannot be predicted. The probability of decay for each
nucleus is the same.
Spontaneous means it happen by itself without external
stimuli. The decay is not affected by the physical conditions
and chemical changes.
Example 2 :
18

Solution :
b. At time
The number of remaining nuclei is
i. By applying the exponential law of radioactive decay, thus the
decay constant is
ii. The half-life of the substance is
days,
0
.
4

t







 0
0
100
80
N
N
N
nuclei
2
.
0 0
N

t
e
N
N 

 0
 
0
.
4
0
0
2
.
0 

 e
N
N
 
0
.
4
2
.
0 

 e
 
0
.
4
ln
2
.
0
ln 

 e
1
day
402
.
0 


  e
ln
0
.
4
2
.
0
ln 



2
ln
2
/
1 
T
402
.
0
2
ln
2
/
1 
T
days
72
.
1
2
/
1 
T 19

Phosphorus-32 is a beta emitter with a decay constant of 5.6  107
s1. For a particular application, the phosphorus-32 emits 4.0  107
beta particles every second. Determine
a. the half-life of the phosphorus-32,
b. the mass of pure phosphorus-32 will give this decay rate.
(Given the Avogadro constant, NA =6.02  1023 mol1)
Solution :
a. The half-life of the phosphorus-32 is given by
Example 3 :

2
ln
2
/
1 
T
1
7
1
7
s
10
4.0
;
s
10
6
.
5 






dt
dN

7
10
6
.
5
2
ln



s
10
24
.
1 6
2
/
1 

T
20

Solution :
b. By using the radioactive decay law, thus
6.02  1023 nuclei of P-32 has a mass of 32 g
7.14  1013 nuclei of P-32 has a mass of
1
7
1
7
s
10
4.0
;
s
10
6
.
5 







dt
dN

0
N
dt
dN



  0
7
7
10
6
.
5
10
0
.
4 N






nuclei
10
14
.
7 13
0 

N
32
10
02
.
6
10
14
.
7
23
13










g
10
80
.
3 9



21

A thorium-228 isotope which has a half-life of 1.913 years decays
by emitting alpha particle into radium-224 nucleus. Calculate
a. the decay constant.
b. the mass of thorium-228 required to decay with activity of
12.0 Ci.
c. the number of alpha particles per second for the decay of 8.0 g
thorium-228.
Solution :
a. The decay constant is given by
Example 4 :

2
ln
2
/
1 
T
 
60
60
24
365
1.913
y
913
.
1
2
/
1 




T

2
ln
10
03
.
6 7


1
8
s
10
15
.
1 




s
10
03
.
6 7


22

Solution :
b. By using the unit conversion ( Ci decay/second ),
the activity is
Since then
If 6.02  1023 nuclei of Th-228 has a mass of 228 g thus
3.86  1019 nuclei of Th-228 has a mass of
  10
10
7
.
3
0
.
12
Ci
0
.
12 





A
decays/s
10
44
.
4 11



second
per
decays
10
7
3
Ci
1 10
. 

N
A 



A
N 

 
8
11
10
15
.
1
10
44
.
4






N
nuclei
10
86
.
3 19


228
10
02
.
6
10
86
.
3
23
19










g
10
46
.
1 2



23

Solution :
c. If 228 g of Th-228 contains of 6.02  1023 nuclei thus
8.0 g of Th-228 contains of
Therefore the number of emitted alpha particles per second is
given by
 
23
10
02
.
6
228
0
.
15







nuclei
10
96
.
3 22


N
  22
8
10
96
.
3
10
15
.
1 


 
N
dt
dN
A 



second
particles/
10
55
.
4 14


 α
A
Ignored it.
24

At the end of this chapter, students should be able to:
 Explain the application of radioisotopes as tracers.
Learning Outcome:
8.2 Radioisotope as tracers (1 hour)
25

8.2.1 Radioisotope
 is defined as an isotope of an element that is radioactive.
 It is produced in a nuclear reactor, where stable nuclei are
bombarded by high speed neutrons until they become
radioactive nuclei.
 Examples of radioisotopes:
a.
b.
c.
8.2 Radioisotope as tracers
Q



 
P
n
P 32
15
1
0
31
15
Q


  e
S
P 0
1
32
16
32
15
Q



 
Na
n
Na 24
11
1
0
23
11
Q


  e
Mg
Na 0
1
24
12
24
11
Q



 
Al
n
l
A 28
13
1
0
27
13
Q


  e
Si
Al 0
1
28
14
28
13
(Radio phosphorus)
(Radio sodium)
(Radio aluminum)
26

 Since radioisotope has the same chemical properties as the
stable isotopes then they can be used to trace the path made
by the stable isotopes.
 Its method :
 A small amount of suitable radioisotope is either
swallowed by the patient or injected into the body of the
patient.
 After a while certain part of the body will have absorbed
either a normal amount, or an amount which is larger than
normal or less than normal of the radioisotope. A detector
(such as Geiger counter ,gamma camera, etc..) then
measures the count rate at the part of the body
concerned.
 It is used to investigate organs in human body such as kidney,
thyroid gland, heart, brain, and etc..
 It also used to monitor the blood flow and measure the blood
volume.
8.2.2 Radioisotope as tracers
27

A small volume of a solution which contained a radioactive isotope
of sodium had an activity of 12000 disintegrations per minute when
it was injected into the bloodstream of a patient. After 30 hours the
activity of 1.0 cm3 of the blood was found to be 0.50 disintegrations
per minute. If the half-life of the sodium isotope is taken as 8 hours,
estimate the volume of blood in the patient.
Solution :
The decay constant of the sodium isotope is
The activity of sodium after 30 h is given by
Example 5 :
h
30
;
min
12000
h;
15 1
0
2
/
1 

 
t
A
T

2
ln
2
/
1 
T

2
ln
15
1
2
h
10
62
.
4 




t
e
A
A 

 0
    
30
10
62
.
4 2
12000



 e
1
min
3000 

A
28

Solution :
In the dilution tracing method, the activity of the sample, A is
proportional to the volume of the sample present, V.
thus the ratio of activities is given by
Therefore the volume of the blood is
h
30
;
min
12000
h;
15 1
0
2
/
1 

 
t
A
T
V
A
1
1 kV
A  2
2 kV
A 
then and
2
1
2
1
V
V
A
A

2
1
3000
5
.
0
V

3
2 cm
6000

V
initial final
(8.6)
29

In medicine
 To destroy cancer cells by gamma-ray from a high-activity
source of Co-60.
 To treat deep-lying tumors by planting radium-226 or caesium-
137 inside the body close to the tumor.
In agriculture
 To enable scientists to formulate fertilizers that will increase the
production of food.
 To develop new strains of food crops that are resistant to
diseases, give high yield and are of high quality.
 To increase the time for food preservation.
31.2.3 Other uses of radioisotope
30

In industry
 To measure the wear and tear of machine part and the
effectiveness of lubricants.
 To detect flaws in underground pipes e.g. pipes use to carry
natural petroleum gas.
 To monitor the thickness of metal sheet during manufacture by
passing it between gamma-ray and a suitable detector.
In archaeology and geology
 To estimate the age of an archaeological object found by
referring to carbon-14 dating.
 To estimate the geological age of a rock by referring to
potassium-40 dating.
31

Radioactive iodine isotope of half-life 8.0 days is used for
the treatment of thyroid gland cancer. A certain sample is required
to have an activity of 8.0  105 Bq at the time it is injected into the
patient.
a. Calculate the mass of the iodine-131 present in the sample to
produce the required activity.
b. If it takes 24 hours to deliver the sample to the hospital, what
should be the initial mass of the sample?
c. What is the activity of the sample after 24 hours in the body of the
patient?
Example 6 :
 
I
131
53
32

Solution :
The decay constant of the iodine isotope is
a. From the relation between the decay rate and activity,
If 6.02  1023 nuclei of I-131 has a mass of 131 g thus
8.0  1011 nuclei of I-131 has a mass of
  s;
10
91
.
6
60
60
24
0
.
8 5
2
/
1 




T
Bq
10
0
.
8 5
0 

A

2
ln
2
/
1 
T

2
ln
10
91
.
6 5


1
6
s
10
00
.
1 




0
0 






dt
dN
A
  0
6
5
10
00
.
1
10
0
.
8 N




0
0 N
A 


nuclei
10
0
.
8 11
0 

N
131
10
02
.
6
10
0
.
8
23
11










g
10
74
.
1 10


 33

Solution :
b. Given
Let N : mass of I-131 after 24 hours
N0 : initial mass of I-131
By applying the exponential law of radioactive decay, thus
c. Given
The activity of the sample is
  s;
10
91
.
6
60
60
24
0
.
8 5
2
/
1 




T
Bq
10
0
.
8 5
0 

A
s
10
8.64
3600
24
hr
24 4





t
g
10
74
.
1 10



t
e
N
N 

 0
  
4
6
10
64
.
8
10
00
.
1
0
10
10
74
.
1 


 

 e
N
    
4
6
10
64
.
8
10
00
.
1
10
0 10
74
.
1 

 

 e
N
g
10
90
.
1 10
0



N
s
10
8.64
3600
24
hr
24 4





t
t
e
A
A 

 0     
4
6
10
64
.
8
10
00
.
1
5
10
0
.
8 

 

 e
A
Bq
10
34
.
7 5


A 34

An archeologist on a dig finds a fragment of an ancient basket
woven from grass. Later, it is determined that the carbon-14 content
of the grass in the basket is 9.25% that of an equal carbon sample
from the present day grass. If the half-life of the carbon-14 is 5730
years, determine the age of the basket.
Solution :
The decay constant of carbon-14 is
The age of the basket is given by
Example 7 :
years
5730
;
10
25
.
9
100
25
.
9
1/2
0
2
0 








 
T
N
N
N

2
ln
2
/
1 
T

2
ln
5730
1
4
y
10
21
.
1 




t
e
N
N 

 0
 t
e
N
N
4
10
21
.
1
0
0
2
10
25
.
9






    e
t ln
10
21
.
1
10
25
.
9
ln 4
2 





years
19674

t 35

Exercise 8.1 :
Given NA =6.021023 mol1
1. Living wood takes in radioactive carbon-14 from the
atmosphere during the process of photosynthesis, the ratio of
carbon-14 to carbon-12 atoms being 1.25 to 1012. When the
wood dies the carbon-14 decays, its half-life being 5600
years. 4 g of carbon from a piece of dead wood gave a total
count rate of 20.0 disintegrations per minute. Determine the
age of the piece of wood.
ANS. : 8754 years
2. A drug prepared for a patient is tagged with Tc-99 which has a
half-life of 6.05 h.
a. What is the decay constant of this isotope?
b. How many Tc-99 nuclei are required to give an activity of
1.50 Ci?
c. If the drug of activity in (b) is injected into the patient 2.05 h
after it is prepared, determine the drug’s activity.
(Physics, 3rd edition, James S. Walker, Q27&28, p.1107)
ANS. : 0.115 h1; 1.7109 nuclei; 1.19 Ci 36

Next Chapter…
CHAPTER 9 :
Nuclear Reactor
37

Phy 310 chapter 8

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