is defined as the spontaneous
disintegration of certain atomic
nuclei accompanied by the
emission of alpha particles,
beta particles or gamma
radiation.
CHAPTER 28: Radioactivity
(2 Hours)
1
Learning Outcome:
28.1
Radioactive decay (1 1/2 hours)
At the end of this chapter, students should be able to:

Explain α, β+, βˉ and γ decays.
State decay law and use

dN
 N
dt
Define activity, A and decay constant, .

Derive and use

N  N 0 e t OR

A  A0 e t
Define half-life and use
T1/ 2 
ln 2

2
28.1 Radioactive decay
Radioactivity is a phenomenon in which an unstable nuclei
undergoes spontaneous decay as a result of which a
new nucleus is formed and energy in the form of
radiation is released
 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.
3
28.1.1 Alpha particle ()


An alpha particle consists of two protons and two neutrons.
It is identical to a helium nucleus and its symbol is
4
4
2 He OR 2
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:
α



A 4
4
A
X

2 He 
Z
Z 2
(Parent) (Daughter) ( particle)
Y
Q
• Alpha particles can penetrate a sheet of paper.
4
α particle
parent
daughter
Examples of  decay :
218
214
4
Po

Pb

84
82
2 He  Q
230
226
4
90Th 88 Ra 2 He  Q
226
222
4
Ra

Rn

88
86
2 He  Q
238
234
4
U

Th

92
90
2 He  Q
5
28.1.2 Beta particle (β)
•
Two types : a) Beta minus , β
+
b) Beta plus , β
•
•
A beta particle has the same mass and charge as an
electron.
Beta particles can penetrate a few mm of Al and their
velocity is high (v ~ c).
6
Beta minus (β )-negatively charge.
•
•
Also called as negatron or electron.
Symbol;
-
β or
•
0
1
 or
0
1
e
It is produced when one of the neutrons in the parent
nucleus decays into a proton, an electron and an
antineutrino.
massless, neutral
7
• 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 :
A
Z
X
 Y
(Parent)
• Examples of
A
Z 1

(Daughter)
0
1
e  Q
( particle)
 minus decay :
0
Th234
Pa

91
1 e  Q
0
Pa234
U

92
1 e  Q
0
Bi214
Po

84
1 e  Q
234
90
234
91
214
83
8
+
Beta plus (β )- positively charge.
•
•
Also called as positron or antielectron.
Symbol;
+
β
•
or
 or e
0
0
1
1
It is produced when one of the protons in the parent
nucleus decays into a neutron, a positron and
a neutrino.
massless,neutral
9
• In beta-plus decay, a positron is emitted, this time the
charge of the parent nucleus decreases by one as shown
below :
A
Z
X
(Parent)

Y

(Daughter)
• Example of
12
7
A
Z 1
0
1
e  Q
(Positron)
 plus decay :
N 126C10e  v  Q
10
28.1.3 Gamma ray ()






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 :
218

214
4
Po

Pb

84
82
2 He  γ
234

234
0
Pa

U

91
92
1e  γ
208

208
Ti

81
81Ti  γ
Gamma ray
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 x11
rays originate outside the nucleus.
28.1.4 Comparison of the properties between alpha
particle, beta particle and gamma ray.

Table 28.1 shows the comparison between the radioactive
radiations.
Alpha
Beta
Gamma
1e OR +1e 0 (uncharged)
Charge
+2e
Deflection by
electric and
magnetic fields
Yes
Yes
No
Strong
Moderate
Weak
Penetration power
Weak
Moderate
Strong
Ability to affect a
photographic plate
Yes
Yes
Yes
Yes
Yes
Yes
Ionization power
Ability to produce
Table 28.1 fluorescence
12
28.1.5 Decay constant ()

Law of radioactive decay states:
 dN 
 is directly
For a radioactive source, the decay rate  

dt 
proportional to the number of radioactive nuclei N
remaining in the source.
i.e.
 dN 
Negative sign means the number of

 N
 dt 
dN
 N
dt

remaining nuclei decreases with time
(28.1)
Decay constant
Rearranging the eq. (28.1):
dN
   dt
N
decay rate

number of remaining radioactiv e nuclei
Hence the decay constant is defined as the probability that a
radioactive nucleus will decay in one second. Its unit is s1.
13

The decay constant is a characteristic of the radioactive nuclei.

Rearrange the eq. (28.1), we get
dN
(28.2)
 dt
N
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. (28.2) from t=0 to time t :
N dN
t
  dt
N0 N
0

ln N NN

  t 0
t
0
N
ln
  λt
N0
N  N 0 e  λt
Exponential law of
radioactive decay
(28.3)
14
The number of nuclei N as function of time t
N  N 0e
 λt
Half-life is the time required for the number of
radioactive nuclei to decrease to half the original
number (No)
15
From
N  N 0 e  t
No
When t  T1 and N 
2
2
Hence
 T1
N0
 N0 e 2
2
 T
1
1
e 2
2
T1
2e 2
ln 2  ln e
T 1
2
ln 2 0.693
T1 

λ
λ
2
16


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 28.2 shows the value of half-life for several isotopes.
Isotope
Half-life
238
92 U
226
88 Ra
4.5  109 years
210
884 Po
234
90Th
222
86 Rn
138 days
214
83 Bi
20 minutes
1.6  103 years
24 days
3.8 days
Table 28.2
17
28.1.6 Activity of radioactive sample (A)
 dN 

 of a radioactive sample.
 dt 

is defined as the decay rate

Its unit is number of decays per second.
Other units for activity are curie (Ci) and becquerel (Bq) – S.I.
unit.
Unit conversion:


1 Ci  3.7 1010 decays per second
1 Bq  1 decay per second

Relation between activity (A) of radioactive sample and time t :

dN
From the law of radioactive decay :
 N
dN dt
and definition of activity : A 
dt
18

Thus
A  N

and
N  N 0e
A   N 0e t
 t

  N 0 e t and A0  N 0
A  A0 e
Activity at time t
 λt
(28.4)
Activity at time, t =0
19
Example 28.1.1 :
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.
20
Solution : T1/ 2  5.0 days; N0  1.010 ; t  20 days
A BQ
The decay constant is given by
The number of remaining nuclide A is
N  N0e t
The number of nuclide A that have decayed is
Therefore the number of nuclide B formed is
20
Example 28.1.2 :
80% of a radioactive substance decays in
4.0 days. Determine
i. the decay constant,
ii. the half-life of the substance.
21
Solution :
At time
The number of remaining nuclei is
i. By applying the exponential law of radioactive decay, thus the
decay constant is
t
N  N0e
ii. The half-life of the substance is
22
Example 28.1.3 :
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 15.0 g
thorium-228.
(Given the Avogadro constant, NA =6.02  1023 mol1)
Solution : T1/ 2  1.913 y  1.913 365  24  60  60
 6.03  107 s


a. The decay constant is given by
T1/ 2 
ln 2

23
Solution :
b. By using the unit conversion ( Ci decay/second ),
1 Ci  3.7 1010 decays per second
the activity is
Since
A  N then
A
N 

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
24
Solution :
c. If 228 g of Th-228 contains of 6.02  1023 nuclei thus
15.0 g of Th-228 contains of
Therefore the number of emitted alpha particles per second is
given by
Ignored it.
dN
A
dt
 N
25
Example 28.1.4 :
A sample of radioactive material has an activity of 9.00 x
10
12
Bq. The material has a half-life of 80.0 s. How long
will it take for the activity to fall to 2.00 x 10
12
Bq ?
Solution
A  A0 e  λt
A
 e  t
Ao
ln 2
T1 
λ
2
ln 2

t1/ 2
 A
ln 
Ao
 A

  t  t 
ln 

 Ao 


  174 s
26
Example 28.1.5 :
N  N 0e
 λt
ln 2

t1/ 2
N = 25% , No = 100%
t =34.6 min
(1.72 h)
27
Learning Outcome:
28.2 Radioisotope as tracers (1/2 hour)
At the end of this chapter, students should be
able to:

Explain the application of radioisotopes as
tracers.
28
28.2 Radioisotope as tracers
• Radioisotope (unstable isotope) is an isotope
which is exhibits radioactivity (known as radioactive
isotope).
a) Blood volume
• The volume of blood in the bloodstream, V2 can
be determined by using dilution method as given
below.
 A2 
V2   V1
 A1 
A1 A2

V1 V2
29
where
A1 A2

V1 V2
A1 = activity of the blood drawn from the patient
A2 = activity of the blood in the bloodstream
V1 = volume of the blood drawn from the patient
V2 = volume of blood in the bloodstream of the patient
A1
 activity per unit volume of the blood draw nfrom the patient
V1
A2
 activity per unit volume of the blood in the blood stream
V2
30
Example 28.2.1
A small volume of a solution which contains a radioactive
4
isotope of sodium Na-24 has an activity of 1.5 x 10 Bq.
The solution is injected into the bloodstream of a patient.
The half-life of the sodium isotope
is 15 hours. After 30
3
hours, the activity of 1.0 cm of blood is measured and
found to be 0.50 Bq. Estimate the volume of blood in the
patient.
31
Solution 28.2.1
A2  Ao e  t
A2  (1.5 104 )e (ln 2 /15 )(30 )
A2  3.75103 Bq
Activity per unit volume
of blood in the patient
 Activity per unit volume
of blood draw nf rom the patient
A2 A1

V2 V1
3.75103
0.5

V2
1106
V2  7.5 103 m3
32
b) Detecting leaks in underground pipes.
The exact position of an underground pipe can be located
if a small quantity of radioactive liquid is added to the
liquid being carried by the pipe.
Geiger counter can be used to detect the leaks.
Any leaks would be detected by an increase in radiation
reading .
The soil close to the leak becomes radioactive.
The short-lived
radioisotope is used to avoid from the
permanent contamination of the soil.
33
c) Detecting brain tumors.
• Technitium-99 is a gamma emitter (half-life 6
hours) and is
used as a medical tracer.
• When injected into the blood stream, 99 Tc will not be
absorbed by the brain, because of the blood-brain barrier.
• However, tumors do not have this barrier.
• Thus, brain tumors readily absorb the 99 Tc.
• These tumors then show as gamma-ray
emitters on detectors external to the body.
• The short-lived radioisotope is used so that it
can quickly eliminate from the body.
34
Good luck
For
2nd semester examination
35