Nuclear physics
Chapter 12.2
Exercises
Exercises
Chapter 12
Unit 12.2
Unit 12.2
Exercises
1 / 28
Outline I
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Q11
Q12
Q13
Q14
Q15
Q16
Q17
Unit 12.2
Exercises
2 / 28
Q1
An alpha particle is fired head-on at a stationary gold nucleus from far
away. Calculate the initial speed of the particle so that the distance of
closest approach is 8.5 × 10−15 m. (Take the mass of the alpha particle to
be 6.64 × 10−27 kg.)
Unit 12.2
Exercises
3 / 28
Q1
An alpha particle is fired head-on at a stationary gold nucleus from far
away. Calculate the initial speed of the particle so that the distance of
closest approach is 8.5 × 10−15 m. (Take the mass of the alpha particle to
be 6.64 × 10−27 kg.)
By conservation of energy
1 2 k(2e)(79e)
mv =
2
r d
2k(2e)(79e)
v=
md
s
=
(2)(8.99 × 109 )(2)(1.6 × 10−19 )(79)(1.6 × 10−19 )
(6.64 × 10−27 )(8.5 × 10−15 )
= 3.7 × 107 m s−1
Unit 12.2
Exercises
3 / 28
Q2
A particle of mass m and charge e is directed from very far away toward a
massive (M >> m) object of charge +Ze with a velocity v , as shown in
the diagram. The distance of closest approach is d. Sketch (on the same
axes) a graph to show the variation with separation of:
a. the particle’s kinetic energy
b. the particle’s electric potential energy.
v
d
Ze
Unit 12.2
Exercises
4 / 28
Q2
The idea
is that
since
the nucleus
is very
massive
will not
A particle
of mass
m and
charge
e is directed
from
very faritaway
toward a
massive
(M >>
m)atobject
of charge
+Ze with
a velocity
as shown in
recoil.
Then
the point
of closest
separation
thev ,kinetic
the diagram.
Thebe
distance
of closest
is d. as
Sketch
(on the same
energy will
a minimum
andapproach
will increase
the separation
axes)increases.
a graph to The
showpotential
the variation
withis separation
energy
given by of:
a. the particle’s kinetic energy
kZe 2
b. the particle’s electric potential
energy.
E =
P
r
and so will be a maximum at the point of closet separation
and will tend to zero as the separation increases.
v
d
Ze
Unit 12.2
Exercises
4 / 28
Q2
A particle of mass m and charge e is directed from very far away toward a
massive (M >> m) object of charge +Ze with a velocity v , as shown in
the diagram. The distance of closest approach is d. Sketch (on the same
axes) a graph to show the variation with separation of:
a. the particle’s kinetic energy
b. the particle’s electric potential energy.
v
d
Ze
Unit 12.2
Exercises
4 / 28
Q2
A particle of mass m and charge eEnergy/J
is directed from very far away toward a
massive (M >> m) object of charge +Ze with a velocity v , as shown in
the diagram. The distance of closest approach is d. Sketch (on the same
axes) a graph to show the variation with separation of:
a. the particle’s kinetic energy
b. the particle’s electric potential energy.
v
Seperation (to left)
Minimum
Seperation
d
Seperation (to right)
Ze
Unit 12.2
Exercises
4 / 28
Q2
A particle of mass m and charge eEnergy/J
is directed from very far away toward a
massive (M >> m)KE
object of charge +Ze with a velocity v , as shown in
the diagram. The distance of closest approach is d. Sketch (on the same
axes) a graph to show the variation with separation of:
a. the particle’s kinetic energy
b. the particle’s electric potential energy.
v
Seperation (to left)
Minimum
Seperation
d
Seperation (to right)
Ze
Unit 12.2
Exercises
4 / 28
Q2
A particle of mass m and charge e is directed from very far away toward a
massive (M >> m) object of charge +Ze with a velocity v , as shown in
the diagram. The distance of closest approach is d. Sketch (on the same
axes) a graph to show the variation with separation of:
a. the particle’s kinetic energy
b. the particle’s electric potential energy.
v
d
Ze
Unit 12.2
Exercises
4 / 28
Q2
A particle of mass m and charge eEnergy/J
is directed from very far away toward a
massive (M >> m)KE
object of charge +Ze with a velocity v , as shown in
the diagram. The distance of closest approach is d. Sketch (on the same
axes) a graph to show the variation with separation of:
a. the particle’s kinetic energy
b. the particle’s electric potential energy.
v
Seperation (to left)
Minimum
Seperation
d
Seperation (to right)
Ze
Unit 12.2
Exercises
4 / 28
Q2
A particle of mass m and charge eEnergy/J
is directed from very far away toward a
massive (M >> m)KE
object of charge +Ze with a velocity v , as shown in
the diagram. The distance of closest approach is d. Sketch (on the same
axes) a graph to show the variation with separation of:
a. the particle’s kinetic energy
b. the particle’s electric potential energy.
PE
v
Seperation (to left)
Minimum
Seperation
d
Seperation (to right)
Ze
Unit 12.2
Exercises
4 / 28
Q3
a. Deviations from Rutherford scattering are expected when the alpha
particles reach large energies. Suggest an explanation for this
observation.
b. Some alpha particles are directed at a thin foil of gold (Z = 79) and
some others at a thin foil of aluminium (Z = 13). Initially, all alpha
particles have the same energy. This energy is gradually increased.
Predict in which case deviations from Rutherford scattering will first
be observed.
Unit 12.2
Exercises
5 / 28
Q3
a. Deviations from Rutherford scattering are expected when the alpha
particles reach large energies. Suggest an explanation for this
observation.
Unit 12.2
Exercises
6 / 28
Q3
a. Deviations from Rutherford scattering are expected when the alpha
particles reach large energies. Suggest an explanation for this
observation.
As the energy increases the alpha particle can approach
closer and closer to the nucleus. Eventually it will be
within the range of the strong nuclear force and some
alphas will be absorbed by the nucleus and will not scatter.
Unit 12.2
Exercises
6 / 28
Q3
b. Some alpha particles are directed at a thin foil of gold (Z = 79) and
some others at a thin foil of aluminium (Z = 13). Initially, all alpha
particles have the same energy. This energy is gradually increased.
Predict in which case deviations from Rutherford scattering will first
be observed.
Unit 12.2
Exercises
6 / 28
Q3
b. Some alpha particles are directed at a thin foil of gold (Z = 79) and
some others at a thin foil of aluminium (Z = 13). Initially, all alpha
particles have the same energy. This energy is gradually increased.
Predict in which case deviations from Rutherford scattering will first
be observed.
Since the nuclear charge of aluminium is smaller than that
of gold the alphas will get closer to aluminium and so will
experience the nuclear force first. Hence deviations will
first be seen for aluminium.
Unit 12.2
Exercises
6 / 28
Q4
Show that the nuclear density is the same for all nuclei. (Take the masses
of the proton and neutron to be the same.)
Unit 12.2
Exercises
7 / 28
Q4
Show that the nuclear density is the same for all nuclei. (Take the masses
of the proton and neutron to be the same.)
The radius of a nucleus of mass number A is
1
R = 1.2 × A 3 × 10−15 m
and its mass is
M = Amn (where mn is the mass of a nucleon)
The density is therefore
ρ=
M
4
3
3 πR
=
Amn
4
3 π(1.2
×A
1
×10−15
3
)3
=
4
3 π(1.2
mn
× 10−15 )3
and so is independent of A. An estimate of this density is
Unit 12.2
Exercises
7 / 28
Q4
Show that the nuclear density is the same for all nuclei. (Take the masses
of the proton and neutron to be the same.)
An estimate of this density is
ρ=
Unit 12.2
1.67 × 10−27
≈ 1017 kg m−3
−15
3
× 10 )
4
3 π(1.2
Exercises
7 / 28
Q5
a. State the evidence in support of nuclear energy levels.
Radium’s first excited nuclear level is 0.0678 MeV above the ground state.
b. Write down the reaction that takes place when radium decays from
the first excited state to the ground state.
c. Find the wavelength of the photon emitted.
Unit 12.2
Exercises
8 / 28
Q5
a. State the evidence in support of nuclear energy levels.
Unit 12.2
Exercises
9 / 28
Q5
a. State the evidence in support of nuclear energy levels.
The main evidence is the discrete energies of alpha
particles and gamma particles in alpha and gamma decay.
Unit 12.2
Exercises
9 / 28
Q5
Radium’s first excited nuclear level is 0.0678 MeV above the ground state.
b. Write down the reaction that takes place when radium decays from
the first excited state to the ground state.
Unit 12.2
Exercises
9 / 28
Q5
Radium’s first excited nuclear level is 0.0678 MeV above the ground state.
b. Write down the reaction that takes place when radium decays from
the first excited state to the ground state.
226
88
Unit 12.2
0
Ra →226
88 Ra +0 γ
Exercises
9 / 28
Q5
Radium’s first excited nuclear level is 0.0678 MeV above the ground state.
c. Find the wavelength of the photon emitted.
Unit 12.2
Exercises
9 / 28
Q5
Radium’s first excited nuclear level is 0.0678 MeV above the ground state.
c. Find the wavelength of the photon emitted.
hf =
hc
= ∆E
λ
hc
λ=
∆E
(6.63 × 10−34 )(3.0 × 108 ))
=
(0.0678 × 106 )(1.6 × 10−19 )
= 1.83 × 10−11 m
Unit 12.2
Exercises
9 / 28
Q6
238
Plutonium (242
94 Pu) decays into uranium ( 92 U) by alpha decay. The energy
of the alpha particles takes four distinct values: 4.90 MeV, 4.86 MeV,
4.76 MeV and 4.60 MeV. In all cases a gamma ray photon is also emitted
except when the alpha energy is 4.90 MeV. Use this information to
suggest a possible nuclear energy level diagram for uranium.
Unit 12.2
Exercises
10 / 28
Q6
238
Plutonium (242
94 Pu) decays into uranium ( 92 U) by alpha decay. The energy
of the alpha particles takes four distinct values: 4.90 MeV, 4.86 MeV,
4.76 MeV and 4.60 MeV. In all cases a gamma ray photon is also emitted
except when the alpha energy is 4.90 MeV. Use this information to
suggest a possible nuclear energy level diagram for uranium.
We are told that no photon is emitted when the alpha particle has an
energy of 4.9 MeV, so the difference between the plutonium and uranium
ground state energies must be ∆E = 4.9 MeV. If the alpha energy is less
than 4.9 MeV and a photon is emitted that means there must be an
excited state of the uranium nucleus, and the difference between the alpha
energy and 4.9 MeV tells you the energy of the excited state.
Unit 12.2
Exercises
10 / 28
Q6
238
Plutonium (242
94 Pu) decays into uranium ( 92 U) by alpha decay. The energy
of the alpha particles takes four distinct values: 4.90 MeV, 4.86 MeV,
4.76 MeV and 4.60 MeV. In all cases a gamma ray photon is also emitted
except when the alpha energy is 4.90 MeV. Use this information to
suggest a possible nuclear energy level diagram for uranium.
You have a two step process:
1. the plutonium nucleus emits an alpha particle and
forms a uranium nucleus
2. the uranium nucleus may be formed in an excited
state - if so it will relax to the ground state by
emitting a gamma ray photon
Unit 12.2
Exercises
10 / 28
Q6
242 Pu
94
α emmision
< 4.9 MeV
∆E =
4.9 MeV
α emmision
= 4.9 Mev
γ
emmsion
238 U
92
Unit 12.2
Eα <
4.9 MeV
Exercises
Excited
State
Eγ
Ground State
11 / 28
Q7
The diagram shows a few nuclear energy levels for
40 Ar, 40 K
18
19
and
40 Ca.
20
40 K
19
II
III
I
40 Ca
20
IV
40 A
18
Identify the four indicated transitions.
Unit 12.2
Exercises
12 / 28
Q7
The diagram shows a few nuclear energy levels for
40 Ar, 40 K
18
19
and
40 Ca.
20
40 K
19
II
III
I
40 Ca
20
IV
40 A
18
Identify the four indicated transitions.
I
Unit 12.2
Exercises
12 / 28
Q7
The diagram shows a few nuclear energy levels for
40 Ar, 40 K
18
19
and
40 Ca.
20
40 K
19
II
III
I
40 Ca
20
IV
40 A
18
Identify the four indicated transitions.
I
Unit 12.2
40
19 K
+
→40
18 Ar + e + v (beta plus decay)
Exercises
12 / 28
Q7
The diagram shows a few nuclear energy levels for
40 Ar, 40 K
18
19
and
40 Ca.
20
40 K
19
II
III
I
40 Ca
20
IV
40 A
18
Identify the four indicated transitions.
II
Unit 12.2
Exercises
12 / 28
Q7
The diagram shows a few nuclear energy levels for
40 Ar, 40 K
18
19
and
40 Ca.
20
40 K
19
II
III
I
40 Ca
20
IV
40 A
18
Identify the four indicated transitions.
II
Unit 12.2
40
19 K
→40
19 K + γ(gamma decay)
Exercises
12 / 28
Q7
The diagram shows a few nuclear energy levels for
40 Ar, 40 K
18
19
and
40 Ca.
20
40 K
19
II
III
I
40 Ca
20
IV
40 A
18
Identify the four indicated transitions.
III
Unit 12.2
Exercises
12 / 28
Q7
The diagram shows a few nuclear energy levels for
40 Ar, 40 K
18
19
and
40 Ca.
20
40 K
19
II
III
I
40 Ca
20
IV
40 A
18
Identify the four indicated transitions.
III
Unit 12.2
40
19 K
−
→40
20 Ca+e +v (beta minus decay)
Exercises
12 / 28
Q7
The diagram shows a few nuclear energy levels for
40 Ar, 40 K
18
19
and
40 Ca.
20
40 K
19
II
III
I
40 Ca
20
IV
40 A
18
Identify the four indicated transitions.
IV
Unit 12.2
Exercises
12 / 28
Q7
The diagram shows a few nuclear energy levels for
40 Ar, 40 K
18
19
and
40 Ca.
20
40 K
19
II
III
I
40 Ca
20
IV
40 A
18
Identify the four indicated transitions.
IV
Unit 12.2
40
19 K
+
→40
18 Ar + e + v (beta plus decay)
Exercises
12 / 28
Q8
a. Find the decay constant for krypton-92, whose half-life is 3.00 s.
1
b. Suppose that you start with 100
mol of krypton. Estimate how many
undecayed atoms of krypton there are after
i. 1 s
ii. 2 s
iii. 3 s
Unit 12.2
Exercises
13 / 28
Q8
a. Find the decay constant for krypton-92, whose half-life is 3.00 s.
λ=
ln 2
ln 2
=
= 0.231 s−1
T1
3.00
2
1
mol of krypton. Estimate how many
b. Suppose that you start with 100
undecayed atoms of krypton there are after
i. 1 s
ii. 2 s
iii. 3 s
Unit 12.2
Exercises
13 / 28
Q8
a. Find the decay constant for krypton-92, whose half-life is 3.00 s.
λ=
ln 2
ln 2
=
= 0.231 s−1
T1
3.00
2
1
mol of krypton. Estimate how many
b. Suppose that you start with 100
undecayed atoms of krypton there are after
i. 1 s
ii. 2 s
iii. 3 s
We start with
1
× 6.02 × 1023 = 6.02 × 1021
100
nuclei and so:
Unit 12.2
Exercises
13 / 28
Q8
a. Find the decay constant for krypton-92, whose half-life is 3.00 s.
λ=
ln 2
ln 2
=
= 0.231 s−1
T1
3.00
2
1
mol of krypton. Estimate how many
b. Suppose that you start with 100
undecayed atoms of krypton there are after
i. 1 s, N = 6.02 × 1021 × e (−0.231)(1) = 4.78 × 1021
ii. 2 s
iii. 3 s
We start with
1
× 6.02 × 1023 = 6.02 × 1021
100
nuclei and so:
Unit 12.2
Exercises
13 / 28
Q8
a. Find the decay constant for krypton-92, whose half-life is 3.00 s.
λ=
ln 2
ln 2
=
= 0.231 s−1
T1
3.00
2
1
mol of krypton. Estimate how many
b. Suppose that you start with 100
undecayed atoms of krypton there are after
i. 1 s, N = 6.02 × 1021 × e (−0.231)(1) = 4.78 × 1021
ii. 2 s
iii. 3 s
We start with
1
× 6.02 × 1023 = 6.02 × 1021
100
nuclei and so:
Unit 12.2
Exercises
13 / 28
Q8
a. Find the decay constant for krypton-92, whose half-life is 3.00 s.
λ=
ln 2
ln 2
=
= 0.231 s−1
T1
3.00
2
1
mol of krypton. Estimate how many
b. Suppose that you start with 100
undecayed atoms of krypton there are after
i. 1 s, N = 6.02 × 1021 × e (−0.231)(1) = 4.78 × 1021
ii. 2 s, N = 6.02 × 1021 × e (−0.231)(2) = 3.79 × 1021
iii. 3 s
We start with
1
× 6.02 × 1023 = 6.02 × 1021
100
nuclei and so:
Unit 12.2
Exercises
13 / 28
Q8
a. Find the decay constant for krypton-92, whose half-life is 3.00 s.
λ=
ln 2
ln 2
=
= 0.231 s−1
T1
3.00
2
1
mol of krypton. Estimate how many
b. Suppose that you start with 100
undecayed atoms of krypton there are after
i. 1 s, N = 6.02 × 1021 × e (−0.231)(1) = 4.78 × 1021
ii. 2 s, N = 6.02 × 1021 × e (−0.231)(2) = 3.79 × 1021
iii. 3 s
We start with
1
× 6.02 × 1023 = 6.02 × 1021
100
nuclei and so:
Unit 12.2
Exercises
13 / 28
Q8
a. Find the decay constant for krypton-92, whose half-life is 3.00 s.
λ=
ln 2
ln 2
=
= 0.231 s−1
T1
3.00
2
1
mol of krypton. Estimate how many
b. Suppose that you start with 100
undecayed atoms of krypton there are after
i. 1 s, N = 6.02 × 1021 × e (−0.231)(1) = 4.78 × 1021
ii. 2 s, N = 6.02 × 1021 × e (−0.231)(2) = 3.79 × 1021
iii. 3 s, N = 6.02 × 1021 × e (−0.231)(3) = 3.01 × 1021
Unit 12.2
Exercises
13 / 28
Q9
a. State the probability that a radioactive nucleus will decay during a
time interval equal to a half-life.
b. Calculate the probability that it will have decayed after the passage of
three half-lives.
c. A nucleus has not decayed after the passage of four half-lives. State
the probability it will decay during the next half-life.
Unit 12.2
Exercises
14 / 28
Q9
a. State the probability that a radioactive nucleus will decay during a
time interval equal to a half-life.
Unit 12.2
Exercises
15 / 28
Q9
a. State the probability that a radioactive nucleus will decay during a
time interval equal to a half-life.
The probability of decay within a half-life is always 12 .
Unit 12.2
Exercises
15 / 28
Q9
b. Calculate the probability that it will have decayed after the passage of
three half-lives.
Unit 12.2
Exercises
15 / 28
Q9
b. Calculate the probability that it will have decayed after the passage of
three half-lives.
The probability that the nucleus will not decay after the
passage of three half – lives is
1
2
×
no decay
1
2
×
no decay
1
2
no decay
=
1
8
Hence the probability that the nucleus will decay some
time within three half – lives is
1−
Unit 12.2
7
1
= = 0.875
8
8
Exercises
15 / 28
Q9
c. A nucleus has not decayed after the passage of four half-lives. State
the probability it will decay during the next half-life.
Unit 12.2
Exercises
15 / 28
Q9
c. A nucleus has not decayed after the passage of four half-lives. State
the probability it will decay during the next half-life.
The probability of decay in any one half – life interval is
0.5. More mathematically, we want to find P(D|N) where
we use the notation of conditional probability and the
events
D = decay in the next half-life
N = no decay in the first 4 half lives.
Then
P(D|N) =
Unit 12.2
P(D ∩ N)
P(N)
Exercises
15 / 28
Q9
c. A nucleus has not decayed after the passage of four half-lives. State
the probability it will decay during the next half-life.
Now
P(N) =
1
1
=
4
2
16
and
P(D ∩ N) =
1
32
Hence
P(D|N) =
Unit 12.2
1
16
1
×
=
32
1
2
Exercises
15 / 28
Q10
Estimate the activity of 1.0 g of radium-226 (molar mass =
226.025 g mol–1 ). The half-life of radium-226 is 1600 yr.
Unit 12.2
Exercises
16 / 28
Q10
Estimate the activity of 1.0 g of radium-226 (molar mass =
226.025 g mol–1 ). The half-life of radium-226 is 1600 yr.
The half-life is so long so that what we are really asked to find is the initial
activity of 1.0 g of pure radium. We have that A = λN0 e −kt so that the
initial activity is λN0 . A mass of 1.0 g of radium corresponds to
1
= 0.0044243 moles
226.025
and hence
N0 = (0.0044243)(6.02 × 1023) = 2.6634 × 1021 nuclei
Since
λ=
ln 2
ln 2
=
= 1.3737 × 10−11 s−1
T1
(1600)(365)(24)(60)(60)
2
Unit 12.2
Exercises
16 / 28
Q10
Estimate the activity of 1.0 g of radium-226 (molar mass =
226.025 g mol–1 ). The half-life of radium-226 is 1600 yr.
We find an activity of
A = −λN
−11
21
(2.6634 × 10 )(1.3737 × 10
Unit 12.2
Exercises
) = 3.66 × 1010 Bq
16 / 28
Q11
The half-life of an unstable element is 12 days. Find the activity of a given
sample of this element after 20 days, given that the initial activity was
3.5 MBq.
Unit 12.2
Exercises
17 / 28
Q11
The half-life of an unstable element is 12 days. Find the activity of a given
sample of this element after 20 days, given that the initial activity was
3.5 MBq.
The decay constant is
λ=
ln 2
ln 2
= 0.0578 day−1
=
T1
12
2
and so
A = λN0 e −kt = 3.5e (−0.0578)(20) = 1.1 MBq
Unit 12.2
Exercises
17 / 28
Q12
A radioactive isotope of half-life 6.0 days used in medicine is prepared 24 h
prior to being administered to a patient. The activity must be 0.50 MBq
when the patient receives the isotope. Estimate the number of atoms of
the isotope that should be prepared.
Unit 12.2
Exercises
18 / 28
Q12
A radioactive isotope of half-life 6.0 days used in medicine is prepared 24 h
prior to being administered to a patient. The activity must be 0.50 MBq
when the patient receives the isotope. Estimate the number of atoms of
the isotope that should be prepared.
The decay constant is
λ=
ln 2
ln 2
= 1.34 × 10−6 s−1
=
T1
(6)(24)(60)(60)
2
From
A = λN0 e −λt
−6 )(24)(60)(60)
0.50 × 106 = 1.34 × 10−6 × N0 e (−1.34×10
N0 = 4.2 × 1011
Unit 12.2
Exercises
18 / 28
Q13
The age of very old rocks can be found from uranium dating. Uranium is
suitable because of its very long half-life: 4.5 × 109 yr. The final stable
product in the decay series of uranium-238 is lead-206. Find the age of
rocks that are measured to have a ratio of lead to uranium atoms of 0.80.
You must assume that no lead was present in the rocks other than that
due to uranium decaying.
Unit 12.2
Exercises
19 / 28
Q13
The age of very old rocks can be found from uranium dating. Uranium is
suitable because of its very long half-life: 4.5 × 109 yr. The final stable
product in the decay series of uranium-238 is lead-206. Find the age of
rocks that are measured to have a ratio of lead to uranium atoms of 0.80.
You must assume that no lead was present in the rocks other than that
due to uranium decaying.
After time t the number of uranium atoms remaining in the rocks is
N = N0 e −λt and so the number that decayed (and hence eventually
became lead) is
N − N0 = N0 (1 − e −λt )
Hence we have that
N0 (1 − e −kt )
= 0.80
N0 e −λt
Unit 12.2
Exercises
19 / 28
Q13
The age of very old rocks can be found from uranium dating. Uranium is
suitable because of its very long half-life: 4.5 × 109 yr. The final stable
product in the decay series of uranium-238 is lead-206. Find the age of
rocks that are measured to have a ratio of lead to uranium atoms of 0.80.
You must assume that no lead was present in the rocks other than that
due to uranium decaying.
This means that
1 − e −λt = 0.80e −λt
1 = 1.80e −λt
e −λt = 1.8
Hence
λt = ln(1.80) = 0.5878
Unit 12.2
Exercises
19 / 28
Q13
The age of very old rocks can be found from uranium dating. Uranium is
suitable because of its very long half-life: 4.5 × 109 yr. The final stable
product in the decay series of uranium-238 is lead-206. Find the age of
rocks that are measured to have a ratio of lead to uranium atoms of 0.80.
You must assume that no lead was present in the rocks other than that
due to uranium decaying.
Since
λ=
ln 2
ln 2
=
= 1.54 × 10−10 yr−1
T1
4.5 × 109
2
t=
Unit 12.2
0.5858
= 3.8 × 109 yr
1.54 × 10−10
Exercises
19 / 28
Q14
9
The isotope 40
19 K of potassium is unstable, with a half-life of 1.37 × 10 yr.
40
It decays into the stable isotope 18 Ar. Moon rocks were found to contain a
ratio of potassium to argon atoms of 1 : 7. Find the age of the Moon
rocks.
Unit 12.2
Exercises
20 / 28
Q14
9
The isotope 40
19 K of potassium is unstable, with a half-life of 1.37 × 10 yr.
40
It decays into the stable isotope 18 Ar. Moon rocks were found to contain a
ratio of potassium to argon atoms of 1 : 7. Find the age of the Moon
rocks.
One half life gives a ratio of
1
2
parent to
1
2
daughter or 1:1.
Two half lives halve the parent again so we have a ratio of
daughter or 1:3.
Three half lives means
1
8
parent, and
7
8
1
4
parent to
3
4
daughter, or 1:7.
So a ratio of 1 : 7 corresponds to three half-lives. The age is therefore
about
t = 3 × 1.37 × 109 = 4.1 × 109 yr
Unit 12.2
Exercises
20 / 28
Q15
Two unstable isotopes are present in equal numbers (initially). Isotope A
has a half-life of 4 min and isotope B has a half-life of 3 min. Calculate
the ratio of the activity of A to that of B after:
a. 0 min
b. 4 min
c. 12 min
Unit 12.2
Exercises
21 / 28
Q15
Two unstable isotopes are present in equal numbers (initially). Isotope A
has a half-life of 4 min and isotope B has a half-life of 3 min. Calculate
the ratio of the activity of A to that of B after:
a. 0 min
Unit 12.2
Exercises
22 / 28
Q15
Two unstable isotopes are present in equal numbers (initially). Isotope A
has a half-life of 4 min and isotope B has a half-life of 3 min. Calculate
the ratio of the activity of A to that of B after:
a. 0 min
The activity is given by A = λN = λN0 e −λt where λ =
ln 2
T1
is the
2
decay constant.
AA
λA N0A
3
3
=
= × 1 = = 0.75
AB
λB N0b
4
4
Unit 12.2
Exercises
22 / 28
Q15
Two unstable isotopes are present in equal numbers (initially). Isotope A
has a half-life of 4 min and isotope B has a half-life of 3 min. Calculate
the ratio of the activity of A to that of B after:
b. 4 min
Unit 12.2
Exercises
22 / 28
Q15
Two unstable isotopes are present in equal numbers (initially). Isotope A
has a half-life of 4 min and isotope B has a half-life of 3 min. Calculate
the ratio of the activity of A to that of B after:
b. 4 min
The activity is given by A = λN = λN0 e −λt where λ =
ln 2
T1
is the
2
decay constant.
ln 2
λA N0A e −λA ×4
3 e − 4 ×4
AA
=
=
×
= 0.95
AB
4 e − ln32 ×4
λB N0B e −λB ×4
Unit 12.2
Exercises
22 / 28
Q15
Two unstable isotopes are present in equal numbers (initially). Isotope A
has a half-life of 4 min and isotope B has a half-life of 3 min. Calculate
the ratio of the activity of A to that of B after:
c. 12 min
Unit 12.2
Exercises
22 / 28
Q15
Two unstable isotopes are present in equal numbers (initially). Isotope A
has a half-life of 4 min and isotope B has a half-life of 3 min. Calculate
the ratio of the activity of A to that of B after:
c. 12 min
The activity is given by A = λN = λN0 e −λt where λ =
ln 2
T1
is the
2
decay constant.
ln 2
λA N0A e −λA ×12
3 e − 4 ×12
AA
=
=
×
= 1.5
AB
4 e − ln32 ×12
λB N0B e −λB ×12
Unit 12.2
Exercises
22 / 28
Q16
A sample contains two unstable isotopes. A counter placed near it is used
to record the decays. Discuss how you would determine each of the
half-lives of the isotopes from the data.
Unit 12.2
Exercises
23 / 28
Q16
A sample contains two unstable isotopes. A counter placed near it is used
to record the decays. Discuss how you would determine each of the
half-lives of the isotopes from the data.
This is a very diffcult question and many different possibilities must
be considered.
Essentially we must be able to determine form a graph of activity
versus time the initial activities of the two isotopes and their
respective half – lives.
Unit 12.2
Exercises
23 / 28
Q16
One possibility is represented by the following graph in which a short (S)
and a long (L) half – life isotopes are present.
The shape of the curve is not a pure exponential.
A × 1010 Bq
2
1.5
1
0.5
0
1
Unit 12.2
2
Exercises
3
4
t/min
24 / 28
Q16
A × 1010 Bq
2
1.5
1
0.5
0
1
2
3
4
t/min
We see that after about 1 minute we have a smooth exponential curve
which implies that one of the isotopes has essentially decayed away,
leaving behind just one isotope. This is justified by estimating a half-life
for times greater than 1 minute. We get consistently a half-life of 1
minute for the long half – life isotope. Extending smoothly the exponential
curve backwards, we intercept the vertical axis at about 1 × 1010 Bq.
Unit 12.2
Exercises
25 / 28
Q16
A × 1010 Bq
2
1.5
1
0.5
0
1
2
3
4
t/min
t
Thus the activity of isotope L is given by AL = 1 × 1010 × 0.5 1 . This
means that the initial activity of the short lifetime isotope S is also
1 × 1010 Bq.
Unit 12.2
Exercises
25 / 28
Q16
Subtracting from the data points of the given graph the activity of this
isotope we get the following graph.
A × 1010 Bq
1
0.8
0.6
0.4
0.2
0
0
0.2 0.4 0.6 0.8
1
t/min
This represents the decay of just isotope S. From this graph we find a
half-life of about 0.1 minute. Obviously, this analysis gets more
complicated when the half-lives are not so different or when the initial
activities are very different.
Unit 12.2
Exercises
26 / 28
Q17
The half-life of an isotope with a very long half-life cannot be measured by
observing its activity as a function of time, since the variation in activity
over any reasonable time interval would be too small to be observed. Let
m be the mass in grams of a given isotope of long half-life.
a. Show that the number of nuclei present in this quantity is N0 = m
µ NA
–1
where µ is the molar mass of the isotope in g mol and NA is the
Avogadro constant.
−λt , how that the initial activity is
b. From A = − dN
dt = N0 eλe
mNA
A0 = µ λ and hence that the half-life can be determined by
measuring the initial activity (in Bq) and the mass of the sample (in
grams).
Unit 12.2
Exercises
27 / 28
Q17
The half-life of an isotope with a very long half-life cannot be measured by
observing its activity as a function of time, since the variation in activity
over any reasonable time interval would be too small to be observed. Let
m be the mass in grams of a given isotope of long half-life.
a. Show that the number of nuclei present in this quantity is N0 = m
µ NA
–1
where µ is the molar mass of the isotope in g mol and NA is the
Avogadro constant.
Unit 12.2
Exercises
28 / 28
Q17
The half-life of an isotope with a very long half-life cannot be measured by
observing its activity as a function of time, since the variation in activity
over any reasonable time interval would be too small to be observed. Let
m be the mass in grams of a given isotope of long half-life.
a. Show that the number of nuclei present in this quantity is N0 = m
µ NA
–1
where µ is the molar mass of the isotope in g mol and NA is the
Avogadro constant.
If the mass (in grams) is m and the molar mass is µ, the
number of moles of the radioactive isotope is m
µ . The
initial number of nuclei is then
N0 =
m
× NA
µ
since one mole contains Avogadro’s number of molecules.
Unit 12.2
Exercises
28 / 28
Q17
The half-life of an isotope with a very long half-life cannot be measured by
observing its activity as a function of time, since the variation in activity
over any reasonable time interval would be too small to be observed. Let
m be the mass in grams of a given isotope of long half-life.
−λt , how that the initial activity is
b. From A = − dN
dt = N0 eλe
A
A0 = mN
µ λ and hence that the half-life can be determined by
measuring the initial activity (in Bq) and the mass of the sample (in
grams).
Unit 12.2
Exercises
28 / 28
Q17
The half-life of an isotope with a very long half-life cannot be measured by
observing its activity as a function of time, since the variation in activity
over any reasonable time interval would be too small to be observed. Let
m be the mass in grams of a given isotope of long half-life.
−λt , how that the initial activity is
b. From A = − dN
dt = N0 eλe
mNA
A0 = µ λ and hence that the half-life can be determined by
measuring the initial activity (in Bq) and the mass of the sample (in
grams).
The activity is
A = λN = λN0 e −λt = λ
m
NA e −λt
µ
and the initial activity is thus
A0 = λ
Unit 12.2
m
NA x
µ
Exercises
28 / 28
Q17
The half-life of an isotope with a very long half-life cannot be measured by
observing its activity as a function of time, since the variation in activity
over any reasonable time interval would be too small to be observed. Let
m be the mass in grams of a given isotope of long half-life.
−λt , how that the initial activity is
b. From A = − dN
dt = N0 eλe
A
A0 = mN
µ λ and hence that the half-life can be determined by
measuring the initial activity (in Bq) and the mass of the sample (in
grams).
Measuring the initial activity then allows determination of
2
the decay constant and hence the half-life from λ = ln
T1
2
Unit 12.2
Exercises
28 / 28