1.
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
2.
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
eNN
0
t
eAA
0
OR
2ln
2/1 T
2
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
3.
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
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
4.
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 ()
He4
2 α4
2OR
Q HePbPo 4
2
214
82
218
84
(Parent) ( particle)(Daughter)
XA
Z Y4
2
A
Z QHe4
2
Q HeRaTh 4
2
226
88
230
90
Q HeRnRa 4
2
222
86
226
88
Q HeThU 4
2
234
90
238
92 4
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
5.
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 ()
e0
1
βOR e0
1
βOR
Beta-minus
(electron) :
Beta-plus
(positron) :
(Parent) ( particle)(Daughter)
XA
Z Y1
A
Z Qe0
1
5
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
6.
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 ePaTh 0
1
234
91
234
90
Q eUPa 0
1
234
92
234
91
Q ePoBi 0
1
214
84
214
83
(Parent) (Positron)(Daughter)
XA
Z Y1
A
Z Qe0
1
Qv enp 0
1
1
0
1
1
Neutrino is uncharged
particle with negligible
mass.
6
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
7.
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
HePbPo 4
2
214
82
218
84
γ
eUPa 0
1
234
92
234
91
γ
TiTi 208
81
208
81
Gamma ray
7
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
8.
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
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
9.
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
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
10.
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 s1.
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
nucleieradioactivremainingofnumber
ratedecay
10
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
11.
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)
tN
N
dt
N
dN
00
tN
N tN 00
ln
λt
N
N
0
ln
λt
eNN
0
Exponential law of
radioactive decay
(8.3)
11
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
12.
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
eNN
0
2
0N
0N
4
0N
16
0N8
0N
2/1T 2/12T 2/13T 2/14T 2/15T0
t,time
N
lifehalf: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
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
13.
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
eNN
0
2/1Tt
2
; 0N
N
2/1
0
0
2
T
eN
N
2/1
2 T
e
2/1
2
1 T
e
2/1
ln2ln T
e
λλ
T
693.02ln
2/1 Half-life (8.4)
13
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
14.
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
U238
92
Po210
884
Ra226
88
Bi214
83
Rn222
86
Th234
90
14
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
15.
secondperdecays1073Ci1 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
secondperdecay1Bq1
N
dt
dN
dt
dN
A
15
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
16.
Thus
00 NA
NA and
t
eNN
0
t
eNA
0
λt
eAA
0
Activity at time t Activity at time, t =0
and t
eN
0
(8.5)
16
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
17.
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 1.01020,
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 :
QBA
0.5
2ln
days20;101.0days;0.5 20
02/1 tNT
2/1
2ln
T
1
days139.0
t
eNN
0 20139.020
100.1
eN
nuclei102.6 18
1820
102.6100.1
nuclei1038.9 19
nuclei1038.9 19
17
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
18.
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
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
19.
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,0.4t
00
100
80
NNN
nuclei2.0 0N
t
eNN
0
0.4
002.0
eNN
0.4
2.0
e
0.4
ln2.0ln
e
1
day402.0
eln0.42.0ln
2ln
2/1 T
402.0
2ln
2/1 T
days72.12/1 T 19
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
20.
Phosphorus-32 is a beta emitter with a decay constant of 5.6 107
s1. 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 mol1)
Solution :
a. The half-life of the phosphorus-32 is given by
Example 3 :
2ln
2/1 T
1717
s104.0;s106.5
dt
dN
7
106.5
2ln
s1024.1 6
2/1 T
20
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
21.
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
1717
s104.0;s106.5
dt
dN
0N
dt
dN
0
77
106.5100.4 N
nuclei1014.7 13
0 N
32
1002.6
1014.7
23
13
g1080.3 9
21
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
22.
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.
(Given the Avogadro constant, NA =6.02 1023 mol1)
Solution :
a. The decay constant is given by
Example 4 :
2ln
2/1 T
6060243651.913y913.12/1 T
2ln
1003.6 7
18
s1015.1
s1003.6 7
22
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
23.
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
107.30.12Ci0.12 A
decays/s1044.4 11
secondperdecays1073Ci1 10
.
NA
A
N
8
11
1015.1
1044.4
N
nuclei1086.3 19
228
1002.6
1086.3
23
19
g1046.1 2
23
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
24.
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
1002.6
228
0.15
nuclei1096.3 22
N
228
1096.31015.1
N
dt
dN
A
secondparticles/1055.4 14
αA
Ignored it.
24
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
25.
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
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
26.
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 PnP 32
15
1
0
31
15
Q eSP 0
1
32
16
32
15
Q NanNa 24
11
1
0
23
11
Q eMgNa 0
1
24
12
24
11
Q AlnlA 28
13
1
0
27
13
Q eSiAl 0
1
28
14
28
13
(Radio phosphorus)
(Radio sodium)
(Radio aluminum)
26
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
27.
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
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
28.
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 :
h30;min12000h;15 1
02/1
tAT
2ln
2/1 T
2ln
15
12
h1062.4
t
eAA
0
301062.4 2
12000
e
1
min3000
A
28
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
29.
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
h30;min12000h;15 1
02/1
tAT
VA
11 kVA 22 kVA then and
2
1
2
1
V
V
A
A
2
1
3000
5.0
V
3
2 cm6000V
initial final
(8.6)
29
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
30.
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
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
31.
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
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
32.
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?
(Given the Avogadro constant, NA =6.02 1023 mol1)
Example 6 :
I131
53
32
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
33.
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;1091.66060240.8 5
2/1 T
Bq100.8 5
0 A
2ln
2/1 T
2ln
1091.6 5
16
s1000.1
0
0
dt
dN
A
0
65
1000.1100.8 N
00 NA
nuclei100.8 11
0 N
131
1002.6
100.8
23
11
g1074.1 10
33
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
34.
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;1091.66060240.8 5
2/1 T
Bq100.8 5
0 A
s108.64360024hr24 4
t
g1074.1 10
t
eNN
0
46
1064.81000.1
0
10
1074.1
eN
46
1064.81000.110
0 1074.1
eN
g1090.1 10
0
N
s108.64360024hr24 4
t
t
eAA
0 46
1064.81000.15
100.8
eA
Bq1034.7 5
A 34
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
35.
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 :
years5730;1025.9
100
25.9
1/20
2
0
TNNN
2ln
2/1 T
2ln
5730
14
y1021.1
t
eNN
0
t
eNN
4
1021.1
00
2
1025.9
et ln1021.11025.9ln 42
years19674t 35
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
36.
Exercise 8.1 :
Given NA =6.021023 mol1
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 h1; 1.7109 nuclei; 1.19 Ci 36
DR.ATAR @ UiTM.NS PHY310 RADIOACTIVITY
Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.
Be the first to comment