5
Radioactivity and Nuclear Energy
2
Chapter 2 Atomic Structure and Radioactive Decay
Atomic Structure and Radioactive Decay
Practice 2.1 (p. 43)
Practice 2.2 (p. 60)
1
B
1
B
2
A
2
A
3
A
90 hours is equal to 6 half-lives.
4
(a) Electron
1
Expected activity = 500   
 2
(b) Proton
5
(c) Electron
= 7.81 Bq
(d) Neutron
 8 Bq
(e) Electron
3
A
(a) (i)
4
The daughter nuclide is radium-224.
228
224
4
90Th  88 Ra + 2 He
5
Tracers have short half-lives so that their
Number of protons = 90
(ii) Number of neutrons
= 232  90 = 142
(iii) Number of electrons
effect on the system disappears after a short
= number of protons = 90
234
(b) 90Th
6
time.
6
(a) A0 =
Atom P has 88 protons, so it is a radium atom.
Atom Q has 88 electrons. Therefore, it has 88
=
protons and it is a radium atom.
A
e kt
3.2  10 8
e (9.6310
3
)( 200)
= 2.20  109 Bq
Atom R has 38 electrons. Therefore, it has 38
protons and it is a strontium atom.
(b)
Atom S has 90 protons, so it is a thorium atom.
Atoms P and Q are isotopes.
7
6
A = A0ekt
A
ln
= kt
A0
1
A
k =  ln
t A0
(a) When  particles bombarded a thin gold
foil in a vacuum, a few (about 1 in 8000)
=
bounced back.
(b) In Rutherford’s nuclear model, all the
1
3.8  10 7
ln
20  60 3.2  10 8
= 1.78  103 s1
positive charge of the atom and most of
(c)
the mass were concentrated in a tiny core
called the nucleus. When  particles
bombarded the gold foil, a few came close
A = A0ekt
A
ln
= kt
A0
t =
to the nuclei and were repelled back by
strong electrostatic forces.
=
1
A
ln
k A0
1
2.41  10 5
ln
5.2  10 8
7.4  10 9
= 110 000 s
New Senior Secondary Physics at Work
1
 Oxford University Press 2010
5
7
Radioactivity and Nuclear Energy
Chapter 2 Atomic Structure and Radioactive Decay
(a) Decay constant k
ln 2
ln 2
=
=
= 4.81  105 s1
4  60  60
t1
4
(a) Nuclear radiation is dangerous because its
penetrating and ionizing effects can
destroy or damage living cells.
2
(b) If radiation appears outside the body, 
(b) Activity after 2 hours
radiation is the most dangerous.  radiation
= A0ekt
5
= (4  103)  e (4.8110
has the strongest penetrating power, so it
)( 26060)
can damage tissue deep inside the body.
= 2.83  103 Bq
8
 radiation is the least dangerous because
(a) Carbon-14 is radioactive.
it has the weakest penetrating power and it
(b) No. This is because the age of the Hong
can be stopped by the skin.
Kong Year Book is much shorter than the
(c) If radiation appears inside the body, 
half-life of C-14.
radiation is the most dangerous. 
(c) This is because the amount of C-14 left in
radiation has the strongest ionizing power.
the material is too little to give an accurate
9
result.
(a) D1: 238
92 U 
D 2:
234
91 Pa

234
90Th
234
92 U
+
+
It ionizes and damages body tissue.
 radiation is the least dangerous because
4
2 He
0
1 e
it has the weakest ionizing power. It will
penetrate the body and cause relatively
(b) A = 226, Z = 88
10
little damage to body tissue.
Let T be the time needed.
5
N = N0ekT
The annual dose of background radiation is
1
N 0 = N0ekT
5
1
ln = kT
5
about 2 mSv, which is much smaller than the
dose that causes sickness (1 – 10 Sv).
Therefore, background radiation causes very
1
little damage and very few people are affected
ln 5
k
 ln 5
=
k
ln 5
=
k
T=
by it. We can take background radiation as
normal.
Revision exercise 2
Multiple-choice (p. 72)
Therefore, the time needed for the number of
1
ln 5
N 0 is
undecayed nuclei to drop to
.
5
k
Practice 2.3 (p. 68)
1
B
2
D
3
C
New Senior Secondary Physics at Work
Statement C is the most suitable description.
2
1
B
2
B
3
A
4
C
5
A
6
B
 Oxford University Press 2010
5
7
Radioactivity and Nuclear Energy
Chapter 2 Atomic Structure and Radioactive Decay
(ii)
B
k=
ln 2
= 0.131 yr1
5 .3

237
93Np
 42 He
(Correct mass numbers and atomic
numbers.)
A = A0ekt
(1A)
(Correct presentation of equation.)
= 500  e0.131  7.5
(1A)
= 187 Bq
(b)
Alternative method:
7.5
= 1.42
5.3
Number of half-lives passed =
Number of undecayednuclei
Total number of nuclei
=
1.42
1
Expected activity = 500   
 2
8
241
95 Am
= 187 Bq
2.5 10 9 1 1 1
= = 
4 2 2
1010
2 half-lives have passed.
(1M)
Time passed = 2  432
(1M)
B
= 864 years
Let n be the number of half-lives passed.
1
6.25 = 100   
 2
2
n
(a)  radiation should be emitted.
 radiation and so the amount of radiation
received varies with the volume of milk.
It takes 4 half-lives, i.e. 60 hours, for 100 g of
(1A)
sodium-24 to reduce to 6.25 g.
A
10
C
11
A
12
D
13
(HKCEE 2002 Paper II Q40)
14
(HKCEE 2004 Paper II Q42)
15
(HKCEE 2006 Paper II Q43)
16
(HKALE 2006 Paper II Q23)
17
(HKCEE 2007 Paper II Q24)
18
(HKCEE 2007 Paper II Q25)
19
(HKALE 2007 Paper II Q24)
(b) (i)
(a) (i)
The activity of X drops from 400 to
200 disintegrations per second in
10 years. Therefore, the half-life of X
is 10 years.
(1A)
The activity of Y drops from 400 to
200 disintegrations per second in
5 years. Therefore, the half-life of Y
is 5 years.
(1A)
(ii) X is more suitable.
half-life. Its activity remains stable
over a longer period of time.
3
Number of neutrons
= 241 – 95
(1M)
= 146
(1A)
(a)
N = N0e
N
ln
= kt
N0
kt
(1A)
(1M)
1
N
k =  ln
t N0
=
1
7.0  10 10
ln
15  60 3.5  10 11
= 1.79  103 s1
New Senior Secondary Physics at Work
(1A)
This is because it has a longer
Conventional (p. 74)
1
(1A)
The milk and the boxes block some of the
n=4
9
(1A)
3
(1A)
 Oxford University Press 2010
5
Radioactivity and Nuclear Energy
(b) Half life =
=
ln 2
k
(b)
(1M)
ln 2
1.79 10 3
= 387 s
4
Chapter 2 Atomic Structure and Radioactive Decay
(1A)
(a)
(Correct labelled axes.)
(1A)
(Showing background radiation on the
graph.)
(1A)
(Correct half-life of the sample.)
(1A)
(c) Disagree.
(1A)
Since K-42 decays into Ca-42, which has
the same mass as K-42,
(Correct labelled axes.)
(1A)
the balance reading does not change as
(Correct data points.)
(1A)
K-42 decays. The weight-time graph is not
(Correct curve.)
(1A)
similar to the curve in (b), and therefore
the half-life of K-42 cannot be obtained
(b) The points are not on a smooth curve
from the graph.
because radioactive decay is a random
process.
6
(1A)
80 s.
 Half-life = 80 s
42
0
(a) 42
19 K  20 Ca  1e
(1A)
(1M)
not the oil pipes. Therefore radioactivity
(1A)
will be detected only on the ground around
the leaks.
(1A)
An  source is not suitable because 
(Correct atomic numbers and mass
numbers.)
(a) A  source is suitable.
(1A)
 radiation can penetrate the ground but
(c) The count rate drops from 200 to 100 in
5
(1A)
radiation cannot penetrate the ground.
(1A)
A  source is not suitable because  rays
(Correct presentation of equation.) (1A)
penetrate the pipes and the ground without
significant decrease in strength.
(1A)
(b) The source with a half-life of 8 hours is
more suitable.
New Senior Secondary Physics at Work
4
(1A)
 Oxford University Press 2010
5
Radioactivity and Nuclear Energy
Chapter 2 Atomic Structure and Radioactive Decay
This is because the half-life is long enough
Therefore,
to detect the leakage of oil pipes before the
20 min
1200 
 600 20
min
 300
activity of the source varies significantly,
20
min
150 20
min
 75
t = 4  20 + 5 = 85 min
(1A)
and the activity of the source will not stay
The activity of T will drop to 75 counts
high over a long time to cause harmful
per min at t = 85 min.
effects to people.
7
(1A)
9
(a) When a U-238 nucleus decays to a Th-234
nucleus, the mass number decreases by 4
and
(a) Number of neutrons
= 232 – 90
(1M)
= 142
(1A)
(b) 3  particles and
(1A)
2  particles are emitted.
the atomic number decreases by 2. (1A)
When the Th-234 nucleus decays to a
unchanged
(1A)
(1A)
(c) A and D (or B and E) are a pair of isotopes.
Pa-234 nucleus, the mass number remains
(1A)
(1A)
(d) Disagree.
while the atomic number increases by 1.
(1A)
The half-life of A is much longer than the
(1A)
period of time when the three records are
(b) From U to Th:  decay
(1A)
taken.
From Th to Pa: β decay
(1A)
The decrease in count rate is mainly due to
(c) Change in mass number
(1A)
the random nature of radioactive decay.
= 234 – 214
(1A)
= 20
(1M)
10 (a) (i)
Number of  particles emitted
20
=
4
=5
8
(1A)
= 2  15 hours
= 2 half-lives
Activity after 30 hours
1 1
= 2  104  
2 2
(1A)
(a) 30 counts per minute
(1A)
= 5  103 Bq
(b) The corrected count rate at t = 5 min is
–1
600 counts min and that at t = 25 min is
–1
300 counts min .
(1M)
(1M)
(1A)
(ii)
Total volume
activity of body

Volume of sample activity of sample
(1A)
Half-life = 25 – 5
= 20 min
30 hours
Total volume of blood
(1A)
5  10 3
6
(1M)
= 4.17  103 cm3
(1A)
=5
(c) At t = 5 min,
the corrected count rate of T
= 2  (630 – 30) = 1200 counts min–1
(iii) This allows the Na-24 to circulate
(1A)
throughout the body so that it mixes
Half-life of T is the same as that of S as
with the blood thoroughly.
they are made of the same radioisotope.
(1A)
(1A)
New Senior Secondary Physics at Work
5
 Oxford University Press 2010
5
Radioactivity and Nuclear Energy
Chapter 2 Atomic Structure and Radioactive Decay
(b) Number of Na-24 nuclei after 30 hours
1 1
= 108  
2 2
= 2.5  107
12 (a)
(1M)
Number of Mg-24 nuclei formed
= 108 – 2.5  107
= 7.5  107
(1A)
11 (a) In Geiger-Marsden scattering experiment,
 particles from a radioactive source
bombard with a thin gold foil in a vacuum.
(1A)
The amount of  particles at different
positions is measured.
(1A)
It is found that nearly all the  particles
passed straight through the foil,
(1A)
while a very small number were deflected
and a few (about 1 in 8000) bounced back.
(Correct scale.)
(1A)
((0, 192) plotted.)
(1A)
(Rest of the points plotted.)
(1A)
(b) Cannot predict which nuclei will
(1A)
decay/when a particular nucleus will decay.
(b) All the positive charge of the atom and
(1A)
most of the mass are concentrated in a tiny
core called the nucleus,
(c) Cannot predict which children will flip a
(1A)
head/when a particular child will flip a
with electrons orbiting around the nucleus.
head.
(1A)
(d) Time taken for activity/count rate to drop
The rest of the atom is largely empty space.
by half/time taken for half the nuclei to
(1A)
decay.
(c) When  particles bombarded with the gold
(1A)
(e) Yes, if children were told to flip coin at
foil, most of them passed straight through
regular time interval.
the empty space of the gold atoms. (1A)
OR
The few which happened to come close to
Yes, because about half of the children
a nucleus were repelled by strong
flipped a head each time.
electrostatic forces. As a result, they were
deflected or bounced back.
(1A)
OR
(1A)
No, because time is not part of the
experiment.
13
(1A)
(HKCEE 2005 Paper I Q8)
14 (a) The activity drops from 80 to 40 kBq in
3.5 days. Therefore the half-life is 3.5 days.
(1A)
New Senior Secondary Physics at Work
6
 Oxford University Press 2010
5
Radioactivity and Nuclear Energy
Chapter 2 Atomic Structure and Radioactive Decay
(c) A = A0ekt
(b) Record the counter reading over a fixed
(1M)
(3.231010 )( 40365246060)
time on stopclock. Repeat for a period of
= 26  10  e
time.
= 1.73  107 Bq
6
(1A)
Draw a graph of reading against time to
find half-life.
Final activity
(c)
Initial activity
(1A)
The activity of the watch after 40 years is
1.73  107 Bq.
(1A)
(d) The decay products are radioactive. (1A)
They may emit  (or ) radiation.
12 .5 1 1 1 1 1
=
=
=   
200 16 2 2 2 2
(1A)
17 (a) Expected number of decay
= 104  (1  ekt)
4 half-lives are required.
3
= 104  [ 1  e 9.710 1 ]
(1M)
Time required = 4 half-lives
= 96.5
= 4  3.5 days
 102
= 14 days
(b) (i)
Therefore the sample should be prepared
at 9.30 am on May 3.
(1A)
(1A)
15 (a) The time taken for the activity to halve.
(1A)
Final activity
(b)
Initial activity
=
1.75 1 1 1 1 1 1
=
=    
56 .0 32 2 2 2 2 2
It takes 5 half-lives.
(1M)
Time required = 5 half-lives
= 5  8.1 days
= 40.5 days
(1A)
(c) Iodine-135.
returns to safer level by next day.
(d) Iodine-127 should be used.
It is not radioactive.
16 (a)  particles are absorbed
by watch casing.
ln 2
(b) k =
t1
(Decreasing gradient.)
(1A)
(ii) The rate of decay is not constant.
(1A)
18 (a) (i)
(1A)
s
The decay constant is 3.23  10
New Senior Secondary Physics at Work
(1A)

(1A)
(ii) 212;
(1A)
208;
(1A)
(1A)

(1A)
(1A)
(b) Measure the count rate due to background
(1M)
radiation. Then place detector very close
from source.
(1A)
Place a paper absorber between the source
ln 2
68  365  24  60  60
10 1
212;
(1A)
2
= 3.23  10
(1A)
(1A)
Its activity remains high for hours and
=
(Curve starting at same rate.)
and detector. If the count rate decreases, 
is emitted from the source, i.e. the decay
(1A)
10 1
in (a)(ii) occurs.
s .
7
(1A)
 Oxford University Press 2010
5
Radioactivity and Nuclear Energy
Chapter 2 Atomic Structure and Radioactive Decay
Then place an aluminium plate between
2
(a) An  source is used because  radiation
the source and detector. If the count rate
has strong ionizing power.
decreases,
Also, the range of  radiation is short so
(1A)
but still larger than that due to background
(1A)
that it will not affect the people nearby.
radiation,  is emitted from the source, i.e.
the decay in (a)(i) occurs.
(c) (i)
(1A)
(1A)
(b) No, it is not suitable to use radioactive
material emitting  particles.
Half-life
ln 2
=
k
ln 2
=
0.0115
(1A)
This is because the ionizing power of 
radiation is much weaker than  radiation.
The current produced will be very weak.
= 60.3 min
(1A)
(1A)
241
95 Am
(c) (i)
(ii)

237
93Np

4
2 He
(Correct mass numbers and atomic
numbers.)
(1A)
(Correct presentation of equation.)
(1A)
(ii) Number of neutrons
= 237 – 93 = 144
(d) No, it is not harmful to health.
(1A)
(1A)
This is because the range of  radiation is
very short (~5 cm) and so  radiation
cannot reach people from the ceiling.(1A)
(Correct curve.)
(1A)
(e) No, a source with short half-life cannot be
used.
Physics in articles (p. 80)
1
(a)
222
86 Rn

218
84 Po
+
This is because the activity of the source,
4
2 He
and hence the current in the detector, will
change sharply within a short time. This
(Correct mass numbers and atomic
numbers.)
(1A)
will lead to false alarm.
(1A)
(1A)
(Correct presentation of equation.) (1A)
(b) Radon in the body decays to emit 
radiation which has strong ionizing power.
(1A)
Also, the decay product polonium is a
radioactive solid which will stay inside the
body and undergo further radioactive
decay.
(1A)
(c) Lung cancer
(1A)
New Senior Secondary Physics at Work
8
 Oxford University Press 2010