Download Unit 1: Area of study 1—Nuclear physics and radioactivity

Task 1
Unit 1: Area of study 1—Nuclear physics and radioactivity
Test: Nuclear physics and radioactivity
Total marks 45
Name: __________________________ Class: ____________ Date:
Section A
(1 mark per question)
Select the best answer for each question.
1 Beta radiation is:
A
B
C
D
energy emitted from an electron of an atom.
an electron ejected from an atom.
a helium nucleus emitted from an atom.
an electron emitted from the nucleus of an atom.
2 Which of the following travels at the greatest speed?
A
B
C
D
gamma radiation.
beta radiation.
alpha radiation.
They all have the same speed.
3 The heaviest form of radiation is:
A
B
C
D
gamma.
beta.
alpha.
none of the above.
4 The type of radiation that is most ionising is:
A
B
C
D
gamma.
beta.
alpha.
none of the above.
5 The most penetrating form of radiation is:
A
B
C
D
gamma.
beta.
alpha.
none of the above.
6 The type of radiation we are exposed to from the Sun is:
A
B
C
D
gamma.
beta.
alpha.
none of the above.
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Page 1
7 The value of x in the decay equation:
A
B
C
D
238
92
U 90x Th  42 α  γ is:
238
236
234
232
Questions 8 and 9 refer to the following information.
Strontium-90 is one of the radioisotopes that was released during the Fukushima nuclear
disaster in Japan. Strontium-90 has a half-life of 28.8 years.
8 How many neutrons are there in each nucleus of Sr-90?
A
B
C
D
90
38
128
52
9 If 1.8 × 1010 atoms of Sr-90 were released during the accident, how many of the original
Sr-90 nuclides will still be in existence in 144 years?
A
B
C
D
none
3.6 × 109
2.8 × 108
5.6 × 108
10 A radioactive sample of oxygen-15 has a half-life of 110 minutes. If the amount
remaining after 5.5 h is 0.80 g, the original sample had a mass of:
A
B
C
D
2.4 g
2.8 g
6.4 g
9.6 g
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5457 3
Page 2
Section B
1 Complete the following decay equations by replacing  :
C147 N  
a
14
6
b
214
82
c
60
27
(1)
Pb 214
83 Bi  
(1)
60
Co* 27
Co  
(1)
2 When bombarded with neutrons, gold (Au-197) undergoes neutron absorption to become
the radioactive isotope gold-198. Given that gold has an atomic number of 79, write a
balanced equation for:
a the absorption of a neutron by a gold-197 atom
(2)
b the beta decay of a radioactive isotope of gold.
(2)
3 Cobalt-60 has a half-life of 5.3 years. If a sample has an initial activity of 2  109 Bq, find
the activity after:
a 10.6 years
(2)
b 15.9 years.
(2)
4 Why is alpha radiation not used as a source in external radiotherapy treatment?
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5457 3
(2)
Page 3
5 A 78 kg man is exposed to 200 mJ of gamma radiation. Calculate:
a his absorbed dose
(1)
b his dose equivalent
(1)
c his dose equivalent if he had been exposed to 200 mJ of alpha radiation instead.
(1)
d the energy of the radiation in electronvolts (eV).
(1)
6 As a doctor you prescribe a course of radiotherapy for a patient with breast cancer. The
procedure involves inserting a radioactive source into the affected region. Explain which
type of radiation source—alpha, beta or gamma—you would prefer to use and why. (4)
7 A scientist uses a Geiger counter to measure the radiation of a radioactive sample. She
records the count as 60 000 emissions per minute.
a Find the time it takes for a 20 g sample to decay to 5 g.
(1)
b Find the half-life of the sample.
(2)
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5457 3
Page 4
8 Give two reasons why technetium-99—a gamma-emitter with a half-life of 6 hours—is
preferred over radon-222—an alpha-emitter with a half-life of 3.8 days—as a radioactive
tracer for diagnostic purposes.
(4)
9 The radioactive decay of a particular isotope is shown on the graph below.
a Find the time it takes for a 20 g sample to decay to 5 g.
(1)
b Find the half-life of the sample.
(1)
c How much of the original radioisotope remains after 10 hours?
(2)
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5457 3
Page 5
10 The table below shows how one isotope of polonium can undergo several decay reactions
to become another isotope of polonium. Determine what type of decay process occurs for
each transmutation a, b and c.
(3)
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5457 3
Page 6
Solutions
Total marks 45
Section A
Radiation is always emitted from a nucleus. A beta particle is a high
1 D
speed electron ejected from a nucleus when a neutron decays into a
proton and an electron within the nucleus of a radioactive atom.
A
Gamma radiation is electromagnetic radiation and therefore travels at
2
the speed of light: 3  108 m s–1.
3 C
An alpha () particle consists of 2 protons and 2 neutrons (i.e. a helium
nucleus). Beta () is one electron. Gamma () has no mass.
Alpha, since it has the highest charge of all radiation types: 2+.
4 C
Gamma; it can penetrate through several metres of concrete.
5 A
6 D
The Sun produces several different types of radiation, but not ,  or 
in large enough quantities to reach the Earth’s surface.
7 C
234; since the nucleus has ejected an -particle the mass number must
decrease by 4.
90 (neutrons + protons) – 38 protons = 52 neutrons
8 D
144 years is 5 half-lives, so amount remaining is: 1.8×1010 / 25 =
9 C
2.8×108
10 C
5.5 hours = 330 minutes = 3  110 minutes = 3 half-lives
Therefore the mass has halved 3 times from its original amount.
Hence the original amount = 2  2  2  0.80 = 23  0.80 = 6.4 g.
Section B
1 a 146 C147 N  01 β
2
3
4
5
214
82
c
60

60
27 Co  27 Co 
a
197
79
b
198
79
1
1
1
1
1
1
1
1
1
1
0
Pb 214
83 Bi  1 β
b
1
γ
Au 01n198
79 Au
0
Au 198
80 Hg  1 β
a
10.6 years = 2 half-lives
Activity = 2  109  2  2 = 0.5  109 = 5  108 Bq
b 15.9 years = 3 half-lives
Activity = 2  109  2  2  2 = 0.25  109 = 2.5  108 Bq
Alpha has very weak penetration and would not pass through the skin. It is
absorbed in several centimetres of air.
200  10 3
energy
a Absorbed dose =
=
= 0.002 56 Gy = 2.6 mGy
mass
78
b Dose equivalent = absorbed dose  quality factor = 2.56  10–3  1 = 2.6
mSv
c The quality factor (QF) for  = 20, QF for  = 1, so dose equivalent would
be 20 times greater.
d Energy E = 200 mJ = 200 × 10-3 / 1.6×10-19 = 1.3×1018 eV
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Beta would be the best type of radiation.
Beta radiation is able to penetrate short distances, so it will reach all the
surrounding cancer cells, but few, if any, of the healthy cells further away.
Alpha radiation lacks penetrating power and would not reach all cancer cells.
Gamma is too penetrating and would affect too many healthy cells.
a
Activity is emissions per second =
b
60000
= 30 000 (1 half-life)
2
30000
= 15 000 (2 half-lives)
2
15000
= 7500 (3 half-lives)
2
7500
= 3750 (4 half-lives).
2
60000
60
= 1000 Bq
Diagnostic procedures aim to create an image of a part of the body.
Gamma radiation is penetrating enough to escape the body and be detected by
the camera; alpha radiation would not be able to create an image.
A shorter half-life means the activity will decrease more rapidly, thus
lessening the overall exposure.
a 5 g is 25% of 20 g, therefore, reading off the graph, the time taken is 8
hours.
b The half-life is the time for 50% of the sample to decay = 4 hours.
c After 10 hours 19% remains; 19% of 20 g  3.8 g (accept any value
between 17% and 20%).
a is an alpha decay as the mass number drops by 4 and the atomic number
drops by 2.
b & c are both beta decays since the atomic number increases by 1 and the
mass number does not change. This only occurs when a neutron changes to a
proton and an electron is ejected from the nucleus.
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