Download Pearson Exam 1

Student Name:
Heinemann Physics 3rd Edition Enhanced
Practice Examination 1
Reading time: 15 minutes
Writing time: 1 hour 30 minutes
Section A – Core Areas of Study
Students are to attempt ALL of the questions for BOTH areas of study.
Number of questions
Number of questions
to be answered
Number of
marks
1. Nuclear physics and radioactivity
17
17
33
2. Electricity
12
12
33
Area of study
Section B – Detailed Studies
Students are to attempt ALL of the questions for ONE detailed study only.
Detailed study
Number of questions
Number of questions
to be answered
Number of
marks
Astronomy OR
12
12
24
Astrophysics OR
12
12
24
Energy from the nucleus
12
12
24
Notes to students:
• Write your name in the space above.
• A formula sheet has been provided.
• You are allowed to bring a periodic table into the exam.
• Unless otherwise indicated, the diagrams in this book are not drawn to scale.
• All written responses must be in English.
Disclaimer:
This is a practice examination. It represents Pearson Australia’s view only of what
would be useful preparation material for the externally assessed examination.
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5457 3
Page 1
Section A
Instructions for Section A
Answer all questions for both Areas of Study in this section in the spaces provided. Write using
blue or black pen.
• Where an answer box has a unit printed in it, give your answer in that unit.
• In questions where more than one mark is available, appropriate working should be shown.
Area of Study 1 – Nuclear physics and radioactivity
Questions 1 to 6 relate to the following information.
63
28 Ni is
a radioisotope of the element nickel. It is a β-particle emitter with a half-life of 100 years.
Question 1
The number of protons in the nucleus of a
63
28 Ni
atom is:
A. 28
B. 35
C. 63
D. 91
(2 marks)
Question 2
The number of neutrons in the nucleus of a
63
28 Ni
atom is:
A. 28
B. 35
C. 63
D. 91
(2 marks)
Question 3
58
28 Ni
and
63
28 Ni
are both isotopes of nickel. Explain what the term isotope means.
(2 marks)
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Page 2
Question 4
63
28 Ni is
a radioisotope of nickel. Explain what the term radioisotope means. Explain why Ni-63 is radioactive
and Ni-58 is not radioactive.
(2 marks)
Question 5
Complete the nuclear decay equation for Ni-63 when it emits a β-particle by placing the appropriate numbers
and symbols in the boxes below.
63
Ni
→
+
β
+ energy
28
(2 marks)
Question 6
Explain where the emitted β-particle has come from.
(2 marks)
Questions 7 to 11 relate to the following information.
• Polonium-215 atoms decay into atoms of lead-211 by emitting an α-particle.
• The average kinetic energy of the emitted α-particle is 7.39 MeV.
• Polonium-215 has a half-life of 1.78 ×10−3 seconds.
• 1.00 eV = 1.6 × 10
–19
J
Question 7
What is the average kinetic energy of an emitted α-particle in joules?
J
(2 marks)
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Page 3
As an α-particle travels through air it will ionise about 100 000 atoms for every centimetre travelled. Each time
an α-particle ionises an atom, the α-particle will lose about 34 eV of its kinetic energy.
Question 8
Approximately how much energy would an α-particle lose as it passes through 1.0 cm of air?
eV
(2 marks)
Question 9
Calculate the approximate distance that an α-particle with a kinetic energy of 7.39 MeV will travel in air before
it loses all of its energy. Give your answer to the nearest millimetre.
cm
(2 marks)
Question 10
–3
What percentage of the original sample of polonium-215 would be left after 4.13 × 10 seconds?
(2 marks)
Question 11
Along which of the paths A–C shown in Figure 1 would an α-particle emitted by a polonium-215 atom travel?
+ + + + +
A
C
lead box
B
- - - - Figure 1
(1 mark)
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Page 4
Questions 12 to 14 refer to the following information:
On 20 August 1977 the unmanned space probe Voyager 2 (see Figure 2 below), was launched. Since its launch it
has visited all four of the outer planets – Jupiter in 1979, Saturn in 1981, Uranus in 1986 and Neptune in 1989.
Currently it is about 13.4 billion kilometres from the Sun and heading out of the solar system.
Figure 2
To operate its electronic instruments and be able to send radio signals back to Earth, Voyager 2 was equipped
with a radioisotope thermoelectric generator (RTG). RTGs use the heat generated from radioactive decay to
produce electricity.
The radioisotope chosen for Voyager 2’s RTG needed to:
• be strongly ionising in order to release enough energy to generate the power needed by the probe’s electronic
instruments and radio.
• be easily shielded so that any emitted radiation would not damage the probes sensitive electronic instruments.
• have a working lifetime of 35 years. To achieve this, its activity cannot drop below 40% of its initial
activity.
The table below is a list of five possible radioisotopes that could be used for Voyager 2’s RTG and the daughter
radioisotopes resulting from their initial decay.
Parent radioisotope
108
47 Ag
63
28 Ni
γ
β
β,γ
α
α
720 k
70 k
510 k
5.59 M
6.13 M
418
100
30.2
88
13.1
108
47 Ag
63
29 Cu
137
56 Ba
none
none
Energy released per decay (eV)
–
Half-life
–
Type of radiation emitted
Energy released per decay (eV)
Half-life (years)
Daughter radioisotope
Type of radiation emitted
137
55 Cs
238
94 Pu
250
98 Cf
234
92 U
246
96 Cm
γ
α
α
–
660 k
4.86 M
5.48 M
–
2.6 min
2.46 × 105
years
4.70 × 103
years
Question 12
Which of the five parent radioisotopes do you think would have been the most appropriate choice for use in
Voyager 2’s RTG?
(1 mark)
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5457 3
Page 5
Question 13
With respect to each of the three criterion provided on the previous page, suggest reasons as to why the
radioisotope you chose as your answer for Question 12 is the most suitable for Voyager 2’s RTG and why the
other radioisotopes are unsuitable. Where necessary you should support your answer with appropriate
calculations.
Ionising ability:
Shielding:
Working lifetime:
Calculations
(6 marks)
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Page 6
Questions 14 to 17 relate to the following information.
Boron Neutron Capture Therapy (BNCT) is a specialised form of radiotherapy that
provides a way of destroying the cancer cells in a tumour without injuring adjacent
normal cells.
The process involves injecting the patient with the non-radioactive isotope boron-10.
The cancer cells in the tumour take up the boron-10 and the patient’s tumour is then
irradiated with a beam of slow moving neutrons as shown in Figure 3.
The slow moving neutrons are strongly absorbed by the boron-10 atoms, which will
then decay by emitting α-particles with an average kinetic energy of 1.47 MeV. The
daughter isotope created typically has 0.84 MeV of kinetic energy.
Figure 3
Question 14
Complete the nuclear equation for the absorption of a neutron by boron-10 and its subsequent decay into an
alpha particle by placing the appropriate numbers and symbols in the empty boxes below.
10
0
B
+
n
→
B
→
+
α
5
γ
+
0
(2 marks)
Questions 15 to 17 relate to the following additional information.
A cancer patient with a brain tumour with an estimated mass of 70.0 g was treated using BNCT. During one
particular session the patient’s tumour cells received 0.50 J of radioactive energy.
Question 15
Calculate the absorbed dose received by the tumour in grays.
Gy
(2 marks)
Question 16
What would be the dose equivalent in sieverts received by the tumour if it absorbed 0.50 J of radioactive energy?
Sv
(1 mark)
Question 17
What effect would the emitted α-particles and the daughter nuclei have on the tumour’s cells?
(2 marks)
End of Area of Study 1
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Page 7
Area of Study 2 – Electricity
Questions 1 to 3 relate to the following information.
Figure 1 shows the X-26 Taser that is used by police in the UK. A Taser is a device designed to subdue a person
by making them part of an electric circuit and then sending pulses of electrical current into their body that cause
uncontrolled muscle spasms.
Figure 1
• When operating, the X-26 Taser sends 10 pulses of electric current into the offender each second.
–4
• Each pulse of current lasts for 1.00 × 10
–3
• During each pulse 1.90 × 10
seconds.
A of current flows through the offender’s body.
• During each pulse the potential difference across the offender’s body at the points of contact is 350 V.
Question 1
How much charge passes through the offender’s body during a single pulse from the Taser X-26?
C
(2 marks)
Question 2
How many electrons would flow through the wires of the X-26 Taser during a single pulse?
electrons
(2 marks)
Question 3
How much electrical energy is delivered into the offender’s body in 20 seconds?
W
(2 marks)
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Page 8
Questions 4 to 8 relate to the following information.
As part of their studies in electricity, Paul and Natasha investigated the behaviour of two different types of
conductors X and Y.
The graph in Figure 2 shows the voltage and current characteristics for each of the two conductors.
V (V)
X
Y
8.0
6.0
4.0
2.0
0
20
40
60
80
I (mA)
Figure 2
Questions 4 to 6 relate to the following additional information.
The students then set up the circuit shown in Figure 3. The internal resistance of the variable power supply can
be ignored.
The variable power supply was set so that the current flowing in the ammeter was 50 mA.
A
Y
X
Figure 3
Question 4
What was the emf supplied to the circuit shown in Figure 2?
V
(2 marks)
Question 5
Determine the resistance of conductor X when 50 mA of current is flowing through it.
Ω
(2 marks)
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Page 9
Question 6
Is conductor X an example of an ohmic, or a non-ohmic conductor? Explain your answer and support it with an
appropriate calculation.
(3 marks)
Questions 7 and 8 relate to the following additional information.
The students then set up the circuit shown in Figure 4.
A
The variable power supply was set at 6.0 volts.
X
Y
Figure 4
Question 7
What current now flows through the ammeter?
mA
(2 marks)
Question 8
What power is dissipated in conductor X?
W
(2 marks)
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Page 10
Questions 9 and 10 relate to the circuit shown in Figure 5.
6.0 V
650 Ω
200 Ω
600 Ω
Figure 5
Question 9
Show that the total resistance of the circuit in Figure 5 is 800 Ω.
(3 marks)
Question 10
What current flows through the 600 Ω resistor?
A
(3 marks)
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Page 11
Question 11
Figure 6 shows a diagram of a household power point.
An electrician is to use an AC voltmeter to test the power
point’s wiring when switched on. If the power point is wired
correctly, what should each of the following readings be?
Figure 6
VXY
=
V
VXZ
=
V
VYZ
=
V
(2 marks)
Questions 12 and 13 relate to the following information.
A student wanted to investigate the differences in energy usage and running costs between two different types of
light globes that emit the same amount of light. The results of her investigation are shown below.
Globe 1 – Incandescent light globe
Globe 2 – Compact fluorescent light (CFL) globe
V = 240 volts
P = 100 watts
V = 240 volts
P = 20 watts
Average globe life = 1000 hours
Purchase cost = $1.20
Average globe life = 10 000 hours
Purchase cost = $5.60
Question 12
How much energy would each globe use in 1.0 hour?
Globe 1
J
Globe 2
J
(2 marks)
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Page 12
Question 13 relates to the following additional information.
Electrical energy costs $0.208 for every kWh of energy used.
Question 13
Determine which of the two globes is the most economical to run. Remember to include the purchase cost of
each globe in your calculations.
(6 marks)
End of Area of Study 2
End of Section A
Please turn to the next page.
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Page 13
Section B
Instructions for Section B
Select one Detailed study.
Answer all questions for the Detailed study in the spaces provided.
Choose the response that is correct or that best answers the question.
• A correct answer scores 2, an incorrect answer scores 0.
• Marks will not be deducted for incorrect answers.
• No marks will be given if more than one answer is given for any question.
Detailed Study – Astronomy
Question 1
Which of the following diagrams, A–D in Figure 1, best represents Tyco Brahe’s view of the universe?
A
B
C
D
Figure 1
(2 marks)
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Page 14
Questions 2 to 4 relate to the following information.
In 1610, Galileo used a telescope to observe the planet Jupiter. He discovered four small objects that appeared to
revolve around it. He called these objects moons. Figure 2 shows diagrams of what he observed on the nights of
the 7, 8, 12 and 15 January 1610.
Figure 2
Question 2
As a result of his observations, Galileo concluded that:
A. Larger planets had more moons than smaller planets.
B. All planets have moons.
C. Ptolemy’s geocentric model of the solar system was physically possible.
D. Copernicus’ heliocentric model of the solar system was physically possible.
(2 marks)
Question 3
Which of the following ideas was proved wrong by Galileo’s observations of Jupiter?
A. Many objects in space travel in circular orbits.
B. All objects in space orbit Earth.
C. The planets orbit the Sun.
D. All planets are spheres.
(2 marks)
A diagram of the telescope that Galileo used to make his observations is shown below in Figure 3. The
telescope’s objective lens had a diameter of 37 mm and a focal length of 980 mm. Its eyepiece had a diameter of
22 mm and a focal length of 50 mm.
objective lens
eyepiece
Figure 3
Question 4
What was the magnification provided by Galileo’s telescope?
A. 1.68
B. 0.05
C. 19.60
D. 0.05
(2 marks)
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Page 15
Question 5 relates to the following information.
The International Space Station (ISS) takes 90 minutes to orbit Earth. The path of the ISS, during one orbit, is
marked on the map in Figure 4 by the letter X. During the next orbit, the satellite’s path over Earth’s surface will
shift due to Earth’s rotation.
Figure 4
Question 5
Which of the paths, A–D, shows the satellite’s next orbit, after X?
(2 marks)
Question 6
Astronomers have calculated that Sirius, a star, is 8.6 light-years away from Earth. Given that the speed of light
8
is 3 × 10 ms-1 the distance in kilometres from Earth to Sirius is:
6
A. 2.58 × 10 km
9
B. 2.58 × 10 km
13
km
16
km
C. 8.14 × 10
D. 8.14 × 10
(2 marks)
Question 7
If a star rises at 11:00 p.m. on the 21 May, which of the times A–D, is the best estimate of the time that it will
rise on the 30 May?
A. midnight
B. 11:00 p.m.
C. 11:36 p.m.
D. 10:24 p.m.
(2 marks)
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Page 16
Questions 8 and 9 relate to the following information.
The tilt of Earth’s axis is 23.5° and the city of Melbourne’s latitude is 37.5° South.
Question 8
What would be the maximum altitude reached by the Sun in the sky for an observer in Melbourne?
A. 52.5° above the horizon.
B. 61.0° above the horizon.
C. 76.0° above the horizon.
D. 90.0°, which is directly overhead.
(2 marks)
Question 9
At what time, and on which day, would the Sun reach its maximum altitude in the sky over Melbourne?
A. local midday on the day of the spring equinox.
B. local midday on the day of the summer solstice.
C. local midday on the day of the autumn equinox.
D. local midday on the day of the winter solstice.
(2 marks)
Question 10
The latitude for the city of Hobart in Tasmania is 43° South. This would mean that when viewed from Hobart the
South Celestial Pole would appear to be:
A. 47° above the horizon.
B. 43° above the horizon.
C. directly overhead.
D. on the horizon.
(2 marks)
Question 11
The right ascension–declination coordinate system for celestial bodies equates to
A. declination = latitude, RA = longitude.
B. declination = longitude, RA = latitude.
C. declination = altitude, RA = azimuth.
D. declination = azimuth, RA = altitude.
(2 marks)
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Page 17
Question 12
Copernicus used the geometry of the diagram below to help him calculate the radius of Venus’s orbit. If the
angle Venus-Earth-Sun = 46° and the Earth orbits the Sun at a distance of 1.00 AU, at what distance does Venus
orbit the Sun?
A. 0.43 AU
B. 0.90 AU
C. 0.69 AU
D. 0.72 AU
(2 marks)
End of Detailed Study – Astronomy
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Page 18
Detailed Study – Astrophysics
Question 1
The energy released by the Sun comes mainly from:
A. radioactive decay.
B. exothermic chemical reactions.
C. nuclear fission reactions.
D. nuclear fusion reactions.
(2 marks)
Question 2
Based on the available evidence, astronomers believe that:
A. our universe is expanding.
B. our universe is in a steady state.
C. our universe is contracting.
D. we do not yet have sufficient evidence to determine what the universe is doing.
(2 marks)
Question 3 relates to the following information.
In 1838 the German mathematician and astronomer Friedrich Bessel made the first recorded observations of
stellar parallax. Over a six-month period Bessel measured the parallax shift (or angle) p for 61 Cygni, a star in
the constellation of Cygnus the swan, to be 0.314 arc seconds. Figure 1 shows a diagram illustrating the apparent
shift in 61 Cygni’s position as Earth moved from one side of its orbit to the other.
Figure 1
Question 3
Based on his measurements Bessel would have calculated in parsecs, the distance to 61 Cygni, to be about:
A. 0.314 pc.
B. 3.14 pc.
C. 3.18 pc.
D. 6.36 pc.
(2 marks)
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Page 19
Question 4 relates to the following information.
Arcturus is the third brightest star in the night sky. It is 36.7 light-years from Earth and has a measured light
intensity of 5.36 × 10-8 Wm-2.
Question 4
What is the luminosity of Arcturus?
A. 2.47 × 10–5 W
B. 9.07 × 10–4 W
C. 2.35 × 1011 W
D. 8.19 × 1028 W
(2 marks)
Question 5 relates to the following information.
The diagram in Figure 2 shows the absorption spectrum for our Sun.
700 nm
600 nm
500 nm
400 nm
Figure 2
Question 5
What would the absorption spectrum look like for a star that is travelling towards us and is of the same type as
our Sun?
A.
700 nm
600 nm
500 nm
400 nm
700 nm
600 nm
500 nm
400 nm
700 nm
600 nm
500 nm
400 nm
700 nm
600 nm
500 nm
400 nm
B.
C.
D.
(2 marks)
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Page 20
Question 6 relates to the following
information.
Astronomers have calculated that the spiral
galaxy NGC 634 shown in Figure 3 is
76.7 Mpc away from Earth.
Figure 3
Question 6
How far from the Earth is NGC 634 in light-years?
7
A. 2.35 × 10 l.y.
B. 2.50 × 108 l.y.
C. 23.5 l.y.
D. 250 l.y.
(2 marks)
Questions 7 to 10 relate to the Hertzsprung–Russell diagram shown in Figure 4 below.
Effective Temperature (K)
25000
10000
3000
6000
-10
Absolute Magnitude, M
-5
•
• • • • • •
•
• •
• •
D
• • • • • • •
•
• •
•
• • • • • •
• •••
•
•• ••
•
•
•
• • • •• •• • • •C
• • •
•
• • • • •
• • •• •
• ••
•
• •
•
•• •
•
•
•
•
• •
• •
• •A•
••
•
•
0
+5
•
• •
• •
• •
••
B
+10
+15
O
B
A
F
G
K
104
102
1
Luminosity (LStar/LSun )
•
10-2
10-4
M
Spectral Class
Figure 4
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Page 21
Question 7
The arrows labelled B point to the central wavy line. The stars that lie along this line are referred to as members
of:
A. the main sequence.
B. the dwarf sequence.
C. the giant sequence.
D. the luminosity sequence.
(2 marks)
Question 8
Referring to the Hertzsprung–Russell diagram shown in Figure 4 on the previous page, our Sun is currently
located at:
A. position A.
B. position B (top arrow).
C. position C.
D. position D.
(2 marks)
Question 9
Referring to the Hertzsprung–Russell diagram shown in Figure 4 on the previous page, our Sun will move to
what region after it runs out of hydrogen?
A. position A.
B. position B (top arrow).
C. position C.
D. position D.
(2 marks)
Question 10
Referring to the Hertzsprung–Russell diagram shown in Figure 4 on the previous page, our Sun will move to
what region after it becomes a white dwarf?
A. position A
B. position B (bottom arrow)
C. position C
D. position D
(2 marks)
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Page 22
Questions 11 and 12 relate to the following information.
In the late 1920s, Edwin Hubble used the red shift of the light emitted by distant galaxies to calculate the speed
at which they were moving away from Earth. He then plotted a graph of their speed against distance from Earth.
He discovered that there was a relationship between these two variables that could be expressed in the form:
v = H0d, where v = speed (km/s) and d = distance (Mpc). The currently accepted value for the constant Ho is
70 km/s/Mpc.
Question 11
The distant galaxy NGC 253 is moving away from Earth at a speed of 243 km/s. How far is NGC 253
from Earth?
A. 3.47 pc
B. 3.47 Mpc
C. 1.7 × 1044 pc
D. 11.7 × 1010 km
(2 marks)
Question 12
What is the estimated age of the universe if Hubble’s constant is taken to be 70 km/s/Mpc?
A. 1.40 × 1010 years
B. 1.40 × 1013 years
C. 6.88 × 1013 years
D. 4.43 × 1017 years
(2 marks)
End of Detailed Study – Astrophysics
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Page 23
Detailed Study – Energy from the nucleus
Question 1
How many protons, neutrons and nucleons does an atom of
142
56 Ba
have?
A. 56 protons, 86 neutrons and 142 nucleons
B. 86 protons, 56 neutrons and 142 nucleons
C. 86 protons, 142 neutrons and 56 nucleons
D. 56 protons, 142 neutrons and 86 nucleons
(2 marks)
Question 2
Which one of the statements A–D gives the best definition of nuclear binding energy?
A. Nuclear binding energy is the energy that joins protons and neutrons together to make a nucleus.
B. Nuclear binding energy is the energy needed to separate the protons and neutrons in the nucleus.
C. Nuclear binding energy is the energy needed to join an atom to its electrons.
D. Nuclear binding energy is the energy needed to separate all of the electrons from an atom.
(2 marks)
Question 3
Which one of the following equations is an example of nuclear fusion?
A.
238
92 U
→
234
90 Ba
+ 42 He + energy
B. 6CO2 + 6O2 → 12O2 + C6H12O6
C.
1
1 He
+ 21 H → 23 He +
D.
235
92 U
+ 01 n →
236
92 U
0
0γ
→
127
50 Sn
+
104
42 Ba
+ 5 01 n + energy
(2 marks)
Question 4
Which one of the statements A–D best describes the process of nuclear fission?
A. Nuclear fission is the process of splitting a suitable heavier element into two lighter nuclei.
B. Nuclear fission is the process in which two lighter nuclei join together to form a larger, more stable nucleus.
C. Nuclear fission is the process in which radioisotopes are created.
D. Nuclear fission is the process by which all radioisotopes undergo radioactive decay.
(2 marks)
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Page 24
Question 5
A controlled chain reaction is required to sustain nuclear fission inside a nuclear reactor. For this to occur, the
number of neutrons per fission needs to be:
A. equal to 0.
B. equal to 1.
C. greater than 1.
D. less than 1.
(2 marks)
Question 6
The uranium that is used as fuel in nuclear reactor has been enriched so that the percentage content of U-235
is about:
A. 99.3%
B. 97.0%
C. 3.0%
D. 0.7%
(2 marks)
Question 7
Cadmium is often used to control the rate of nuclear fission in a reactor because cadmium atoms:
A. deflect neutrons away from the fuel rods thus preventing nuclear fission.
B. cause the neutrons to slow down so that they are travelling too slow to cause nuclear fission.
C. cause the neutrons to speed up so that they are travelling too fast to cause nuclear fission.
D. absorb neutrons thus preventing them from causing nuclear fission.
(2 marks)
Question 8
The role of a moderator in a nuclear reactor is to:
A. absorb neutrons.
B. remove excess heat energy from the reactor.
C. slow down fast moving neutrons.
D. stop nuclear fission from occurring.
(2 marks)
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Page 25
Question 9
In the nuclear equation
235
92 U
+ 01 n →
236
92 U
→
91
36 Kr
+
142
86 Ba
+ X + energy, X is:
A. 3 01 n
B. 3 01 n
C.
0
3n
D.
3
0n
(2 marks)
Question 10
235
92 U
+ 01 n →
236
92 U
→
148
57 La
+
85
35 Br
+ 301 n + energy
If the mass defect as a result of this nuclear equation is 2.12 × 10–28 kg. How much energy would be released?
–11
eV
–20
eV
A. 1.91 × 10
B. 6.36 × 10
C. 0.40 eV
D. 120 MeV
(2 marks)
Question 11
Which of the statements A–D best explains why fast breeder reactors do not require a moderator?
A. For fission to occur, a collision with a slow-moving neutron is required.
B. The neutrons released during fission are slow moving.
C. The control rods slow the neutrons down so that fission can occur.
D. For fission to occur, a collision with a fast-moving neutron is required.
(2 marks)
Question 12
One of the fusion reactions taking place in the Sun involves two helium-3 nuclei fusing together to form
helium-4 and two protons. How does the mass of the reactants compare with that of the fusion products?
A. The mass of the products is the same as that of the reactants because mass cannot be created or destroyed.
B. The reactants have slightly more mass than the products because the binding energy for a helium-4 nucleus is
greater than that of two helium-3 nuclei.
C. The products have slightly more mass than the reactants because the binding energy for two helium-3 nuclei
is greater than the binding energy of a helium-4 nucleus.
D. The reactants have slightly more mass than the products because the total mass defect was transformed into
the kinetic energy of the emitted protons.
(2 marks)
End of Detailed Study – Energy from the nucleus
End of Section B
End of Exam
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5457 3
Page 26
Physics Practice Exam 1 – Formula Sheet
Nuclear physics and radioactivity
1
absorbed dose (Gy)
2
dose equivalent
3
activity level
4
nuclear transmutation
absorbed dose =
E
m
dose equivalent = absorbed dose × quality factor
A=
A0
tn
1 Becquerel (Bq) = 1 nuclear transmutation per second
Quality Factors
γ -ray
1
β -particle
1
slow neutron
fast neutron
α-particle
3
10
20
1 eV = 1.6 x10-19 J
Electricity
1.6 × 10-19 J
5
charge on an electron
6
electrical charge
7
electrical work
8
voltage
V = IR
9
power
P = VI
10
resistors in series
11
resistors in parallel
Q = It
W = QV
RTOTAL = R1 + R2 + … + Rn
1
RTOTAL
=
1
1
1
+
+ ... +
R1 R2
Rn
Astronomy
12
speed of light in a vacuum
13
speed
3.0 × 108 ms-1
v=
d
t
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5457 3
Page 27
14
M=
magnification
f0
fe
Astrophysics
15
light year
1 l.y. = 9.5 × 1015 m 16
parsec
1 pc = 3.1 × 1016 m
17
parallax
18
luminosity
19
Hubble’s Law
d pc =
1
p arc sec
L = b × 4πR2
v = H0 d
Energy from the nucleus
20
mass-energy equation
E = mc2
Prefixes/Units
p = pico = 10-12
n = nano = 10-9
µ = micro = 10-6
m = milli = 10-3
k = kilo = 103
M = mega = 106
G = giga = 109
t = tonne = 103 kg
End of Formula Sheet
Copyright © Pearson Australia 2012 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4425 5457 3
Page 28