-
*
is defined as the spontaneous disintegration of certain atomic
nuclei accompanied by the emission of alpha particles, beta
particles or gamma radiation.
Radioactivity
-
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.
*
SF027
-
Alpha particle ()
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:
*
OR
(Parent)
( particle)
(Daughter)
Alpha particles can penetrate a sheet of paper.
SF027
-
*
Examples of decay :
parent
daughter
particle
-
*
Two types :
a) Beta minus , -
b) Beta plus , +
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).
Beta particle ()
*
-
*
Beta minus ( )-negatively charge.
Also called as negatron or electron.
Symbol;
- or
It is produced when one of the neutrons in the parent nucleus
decays into a proton, an electron and an antineutrino.
massless, neutral
*
-
*
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 :
(Parent)
( particle)
(Daughter)
Examples of minus decay :
*
-
*
Beta plus (+)- positively charge.
Also called as positron or antielectron.
Symbol;
+ or
It is produced when one of the protons in the parent nucleus
decays into a neutron, a positron and
a neutrino.
massless,neutral
*
-
*
In beta-plus decay, a positron is emitted, this time the charge
of the parent nucleus decreases by one as shown below :
(Parent)
(Positron)
(Daughter)
Example of plus decay :
*
-
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 :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.
*
Gamma ray
SF027
-
Comparison of the properties between alpha particle, beta
particle and gamma ray.
Table 1 shows the comparison between the radioactive radiations.
*
+2e
1e OR +1e
0 (uncharged)
Yes
Yes
No
Strong
Moderate
Weak
Weak
Moderate
Strong
Yes
Yes
Yes
Yes
Yes
Yes
Table 1
AlphaBetaGammaChargeDeflection by electric and magnetic
fieldsIonization powerPenetration powerAbility to affect a
photographic plateAbility to produce fluorescence
SF027
-
Decay constant ()
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. (1):
Hence the decay constant is defined as the probability that a
radioactive nucleus will decay in one second. Its unit is s1.
*
Negative sign means the number of remaining nuclei decreases
with time
Decay constant
(1)
SF027
- The decay constant is a characteristic of the radioactive
nuclei.Rearrange the eq. (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. (2) from t=0 to time t :
*
Exponential law of radioactive decay
(2)
(3)
SF027
-
*
The number of nuclei N as function of time t
Half-life is the time required for the number of radioactive
nuclei to decrease to half the original number (No)
-
*
From
Hence
*
- 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 2 shows the value of half-life for several isotopes.
*
Table 2
IsotopeHalf-life4.5 109 years1.6 103 years138 days 24 days3.8
days20 minutes
SF027
-
Activity of radioactive sample (A)
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 :
*
SF027
- Thus
*
and
Activity at time t
Activity at time, t =0
and
(4)
SF027
-
*
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 :
SF027
-
*
80% of a radioactive substance decays in 4.0 days. Determine
i. the decay constant,
ii. the half-life of the substance.
Example 2 :
SF027
-
*
Solution :
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
SF027
-
*
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 mol1)
Solution :
a. The decay constant is given by
Example 3 :
SF027
-
*
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
SF027
-
*
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.
SF027
-
*
A sample of radioactive material has an activity of 9.00 x
1012 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 1012 Bq ?
Solution
Example 4 :
-
*
N = 25% , No = 100%
t =34.6 min
(1.72 h)
Example 5 :
-
*
Good luck
For
2nd semester examination
SF027
He
4
2
4
2
X
A
Z
+
Y
4
2
-
-
A
Z
Q
He
4
2
Q
+
+
He
Pb
Po
4
2
214
82
218
84
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
e
0
1
0
1
-
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or
b
Q
e
Pa
Th
0
1
234
91
234
90
+
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Q
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234
92
234
91
+
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-
Q
e
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Bi
0
1
214
84
214
83
+
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-
X
A
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Y
A
1
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+
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0
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or
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C
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1
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6
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1
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0
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+
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He
Pb
Po
4
2
214
82
218
84
+
+
-
*
e
U
Pa
0
1
234
92
234
91
+
*
Ti
Ti
208
81
208
81
dt
dN
N
dt
dN
N
dt
dN
nuclei eradioactiv remaining ofnumber
ratedecay
N
dt
dN
dt
N
dN
tN
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dt
N
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0
0
tN
N
tN
0
0
ln
t
N
N
0
ln
t
eNN
0
t
eNN
0
t
0
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2
1
T
0
0
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2
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2
1
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2
1
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1
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6930
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o
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N
T
t
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and
When
U
238
92
Po
210
884
Ra
226
88
Bi
214
83
Rn
222
86
Th
234
90
dt
dN
secondper decay 1Bq 1
N
dt
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dt
dN
A
secondper decays1073Ci1
10
.
00
NA
NA
t
eNN
0
t
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0
t
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0
t
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0
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B
A
+
days
20
;
10
1.0
days;
0
.
5
20
0
2
/
1
=
=
=
t
N
T
t
e
N
N
l
-
=
0
t
e
N
N
l
-
=
0
l
2
ln
2
/
1
=
T
(
)
60
60
24
365
1.913
y
913
.
1
2
/
1
=
=
T
s
10
03
.
6
7
=
NA
A
N
secondper decays1073Ci1
10
.
N
dt
dN
A
s 174
o
o
A
A
tt
A
A
ln
ln
T
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2
1
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t
t
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2ln
t
