phy 310 chapter 8
TRANSCRIPT
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 RahmanSchool 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)
Ndt
dN
teNN 0teAA 0
OR
2ln2/1 T
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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
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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 ()
He42 α4
2OR
Q HePbPo 42
21482
21884
(Parent) ( particle)(Daughter)
XAZ Y4
2
AZ QHe4
2
Q HeRaTh 42
22688
23090
Q HeRnRa 42
22286
22688
Q HeThU 42
23490
23892 4
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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 ()
e01
βOR e01
βORBeta-minus
(electron) :
Beta-plus
(positron) :
(Parent) ( particle)(Daughter)
XAZ Y1
AZ Qe0
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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 01
23491
23490
Q eUPa 01
23492
23491
Q ePoBi 01
21484
21483
(Parent) (Positron)(Daughter)
XAZ Y1
AZ Qe0
1
Qv enp 01
10
11
Neutrino is uncharged
particle with negligible
mass.
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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 42
21482
21884
γ eUPa 0
123492
23491
γ TiTi 20881
20881
Gamma ray
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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 YesTable 8.1 8
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Figures 8.1 and 8.2 show a deflection of , and in electric
and magnetic fields.
Figure 8.1
B
E
αγ β γ
β
α
Figure 8.2
Radioactive
source
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Law of radioactive decay states:
For a radioactive source, the decay rate is directly
proportional to the number of radioactive nuclei Nremaining 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
Ndt
dN
Ndt
dN
Negative sign means the number of
remaining nuclei decreases with time
Decay constant
(8.1)
N
dt
dN
nuclei eradioactiv remaining ofnumber
ratedecay
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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 :
dtN
dN (8.2)
tN
Ndt
N
dN
00
tN
N tN 00ln
λtN
N
0
ln
λteNN 0
Exponential law of
radioactive decay(8.3)
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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.
teNN 0
2
0N
0N
4
0N
16
0N8
0N
2/1T 2/12T 2/13T2/14T
2/15T0t,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.
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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)
teNN 0
2/1Tt 2
; 0NN
2/1
00
2
TeN
N
2/12T
e
2/1
2
1 Te
2/1ln2lnT
e
λλT
693.02ln2/1 Half-life (8.4)
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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
U23892
Po210884
Ra22688
Bi21483
Rn22286
Th23490
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secondper decays1073Ci1 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
secondper decay 1Bq 1
Ndt
dN
dt
dNA
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Thus
00 NA
NA andteNN 0
teNA 0
λteAA 0
Activity at time t Activity at time, t =0
and teN 0
(8.5)
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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
days 20;101.0 days; 0.5 2002/1 tNT
2/1
2ln
T
1days 139.0
teNN 0 20139.020100.1 eN
nuclei 102.6 18
1820 102.6100.1 nuclei 1038.9 19
nuclei 1038.9 1917
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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 :
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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
80NNN
nuclei 2.0 0N
teNN 0 0.4
002.0 eNN 0.42.0 e
0.4ln2.0ln e
1day 402.0
eln0.42.0ln
2ln2/1 T
402.0
2ln2/1 T
days 72.12/1 T 19
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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 :
2ln2/1 T
1717 s 104.0 ;s 106.5 dt
dN
7106.5
2ln
s 1024.1 62/1 T
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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 s 104.0 ;s 106.5 dt
dN
0Ndt
dN
077 106.5100.4 N
nuclei 1014.7 130 N
321002.6
1014.723
13
g 1080.3 9
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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 :
2ln2/1 T
6060243651.913y 913.12/1 T
2ln1003.6 7
18 s 1015.1
s 1003.6 7
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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
10107.30.12Ci 0.12 A
decays/s 1044.4 11
secondper decays1073Ci1 10 .
NA
AN
8
11
1015.1
1044.4
N
nuclei 1086.3 19
2281002.6
1086.323
19
g 1046.1 223
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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
231002.6228
0.15
nuclei 1096.3 22N
228 1096.3 1015.1
Ndt
dNA
secondparticles/ 1055.4 14 αA
Ignored it.
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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)
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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)
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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
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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; 15 102/1 tAT
2ln2/1 T
2ln15
12 h 1062.4
teAA 0
301062.4 2
12000 e
1min 3000 A28
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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; 15 102/1 tAT
VA
11 kVA 22 kVA then and
2
1
2
1
V
V
A
A
2
1
3000
5.0
V
32 cm 6000V
initial final
(8.6)
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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
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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.
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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
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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 TBq 100.8 5
0 A
2ln2/1 T
2ln1091.6 5
16 s 1000.1
0
0
dt
dNA
0
65 1000.1100.8 N00 NA
nuclei 100.8 11
0 N
1311002.6
100.823
11
g 1074.1 10 33
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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 TBq 100.8 5
0 As 108.64360024hr 24 4t
g 1074.1 10
teNN 0 46 1064.81000.1
0
101074.1
eN
46 1064.81000.110
0 1074.1
eN
g 1090.1 10
0
N
s 108.64360024hr 24 4t
teAA 0 46 1064.81000.15100.8
eA
Bq 1034.7 5A34
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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;1025.9100
25.91/20
20
TNNN
2ln2/1 T
2ln5730
14 y 1021.1
teNN 0
teNN41021.1
0021025.9
et ln1021.11025.9ln 42
years 19674t 35
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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
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Next Chapter…
CHAPTER 9 : Nuclear Reactor
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