Download QAT Trial Answers 2015

2015 TRIAL HIGHER SCHOOL CERTIFICATE
EXAMINATION
PHYSICS – MAPPING GRID
Core plus Options
Question
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Answer
D
A
A
D
B
C
D
B
C
D
D
D
B
C
A
A
A
B
B
C
Disclaimer
Every effort has been made to prepare this Examination in accordance with the Board of Studies documents. No guarantee or warranty is
made or implied that the Examination paper mirrors in every respect the actual HSC Examination question paper in this course. This paper
does not constitute ‘advice’ nor can it be construed as an authoritative interpretation of Board of Studies intentions. No liability for any
reliance, use or purpose related to this paper is taken. Advice on HSC examination issues is only to be obtained from the NSW Board of
Studies. The publisher does not accept any responsibility for accuracy of papers which have been modified.
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
Page 1
Section I:
Part B:
Extended
Response
21
22
23
24
25
26
27
28
29
30
31
4
6
6
2
7
4
7
5
4
5
5
H2; H6
H5; H6; H11; H14
H2; H12; H14; H15
H6
H3; H7; H9
H3; H4; H14
H3; H9; H13
H2; H13; H14
H1; H7 H10
H3; H10; H13
H1; H2; H10; H14
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
9.2.2
9.2.1; 9.2.3
9.2.1
9.2.4
9.3.1
9.3.2; 9.3.4
9.3.2; 9.3.5
9.4.4
9.4.2
9.4.3
9.4.1
1–4
2–6
1–6
2–4
1–5
2–6
1–5
1–6
1–5
1–6
1–6
Page 2
2015 TRIAL HIGHER SCHOOL CERTIFICATE
EXAMINATION
PHYSICS – MARKING GUIDELINES
Section I
Part A – 20 marks
Questions 1-20 (1 mark each)
Question
Correct Response
Outcomes Assessed
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
D
A
A
D
B
C
D
B
C
D
D
D
B
C
A
A
A
B
B
C
H1; H2; H3
H3; H5; H6; H9
H2; H9; H14
H1; H9; H13
H5; H9
H3; H7
H1; H3
H9; H10; H11
H1; H10; H14
H2; H7; H9
H2; H14
H9
H2; H3; H11; H14
H1; H3; H7
H1; H3; H8; H10
H1; H3; H4; H16
H1; H6; H7
H2; H7; H9
H2; H6; H11
H2; H9; H13
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
Targeted
Performance Bands
1-2
1-2
1-2
1-2
2-3
2-3
2-3
2-3
2-3
2-3
3-4
3-4
3-4
3-4
3-4
3-4
3-4
4-5
4-5
5-6
Page 3
Section I
Part B – 55 marks
Question 21 (4 marks)
21 (a) (2 marks)


Criteria
All three forces correctly identified (direction arrows not required)
Two of the forces correctly identified
Marks
2
1
Sample answer
weight, m g ↓; normal reaction FN ; tension T 
21 (b) (1 mark)
Criteria


Correct identification of T and
The direction must be clearly identified (resultant vector)
Mark
1
Sample answer
FR = T 
21 (c) (1 mark)
Criteria

Identification of circular motion
Mark
1
Sample answer
The ball will undergo uniform circular motion.
Question 22 (6 marks)
22 (a) (2 marks)




Criteria
Sun in the middle of the twin orbit
Earth and Vulcan must be on the same orbit (same radius), on opposite sides
Earth, Vulcan and the Sun must be labelled appropriately
One of the above features correctly shown
Marks
2
1
Sample answer
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
Page 4
22 (b) (2 marks)



Criteria
Correctly stating that Vulcan could have twice Earth’s mass in the same orbit
Providing a correct principle of Physics to support the statement
Providing one of the above outcomes
Marks
2
1
Sample answer
It would certainly be possible for Vulcan to have twice the mass of Earth, yet still follow the
same orbit. The orbital period of a satellite depends upon the mass of the central object, here
the Sun; it does not depend on the mass of the satellite.
22 (c) (2 marks)



Criteria
Shows that weight is the same as the gravitational attraction of the Earth
Finds Spock’s mass, and substitutes into the correct formula yielding an answer
Determines Spock’s mass on the ground on Earth
Marks
2
1
Sample answer
Mr Spock’s weight on Earth is 1200 newtons, so his mass is 1200/9.8 = 122.45 kg
In any location the gravitational attraction of Earth for Mr Spock is the same as his weight:
24
m E mS
 122.45
11 6.0  10
Wt'  G
 6.67  10
 490 N
2
2
rorbit
10 000  10 3




Question 23 (6 marks)






Criteria
An appropriate neat, labelled diagram
An accurate description of how the investigation proceeded
An assessment of both the validity and reliability of the procedure
Attempting all the above outcomes, but one or two weak responses
Two of the above outcomes attempted
One of the above outcomes attempted
Marks
5-6
3-4
2
1
Sample answer
A group of three students carried out this experiment, each with a separate task to perform.
A strip of transparent plastic tape 20 cm long, marked with an accurate scale in cm, was
attached to a metal bob at the bottom. The tape was located between the sides of a light-gate,
and pulled gently upwards until the bob touched the top of the light-gate. This meant the zero
mark of the tape was precisely aligned with the light-gate. After the device was switched on
the tape was released. Almost immediately the timer measured the interval for the 20-cm tape
to drop through, and the computer determined its acceleration using s = ½ g t2 [g = 0.40/t2]
The experiment was valid because the initial velocity of the bob was zero until dropped, and
the friction of the tape through the light-gate was negligible. Reliability was maximised by
repeating the trial ten times, producing the identical result almost every time.
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
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Question 24 (2 marks)



Criteria
Identifies that the width of the comet remains unchanged
Substitutes correctly into the length contraction formula, yielding an answer
Makes a correct statement about either of these
Marks
2
1
Sample answer
The observed width of the comet remains unchanged at 36 km since its relative velocity in
that direction is zero. However, its length would be observed to be shorter, according to the
formula:
Question 25 (7 marks)
25 (a) (1 mark)

Criteria
A correct definition of the Motor Effect
Mark
1
Sample answer
A current-carrying conductor within an external magnetic field experiences a force.
25 (b) (2 marks)



Criteria
A correct identification of the magnetic pole at one end of the coil
Correctly relating this to the movement of the coil
A correct identification of one of these processes
Marks
2
1
Sample answer
Since the DC current is flowing as shown in the diagram the right-hand end of the speaker
coil would become a north pole. This would be repelled by the adjacent central North Pole, so
the coil would move to the left.
25 (c) (4 marks)






Criteria
States that the incoming audio signal is AC, varying in frequency and intensity
Describes how the coil moves back and forth in response
Identifies the speaker cone attached to the coil
The speaker cone produces sound waves
Unsuccessful attempt to relate one or two of the above outcomes
A correct identification of one of these processes
Marks
4
2-3
1
Sample answer
If current flows in one direction through the speaker coil it is repelled by the central north
pole of the permanent magnet – if it flows in the opposite direction it is attracted. But the
signal received by the speaker coil is alternating current, varying both in frequency and
intensity, causing the speaker coil to oscillate back and forth at a changing rate and intensity.
The coil is attached to a cone, often made of stiff paper. The moving coil causes the cone to
oscillate back and forth as well, creating a correspondingly varying sequence of compressions
and rarefactions in the air – sound waves corresponding to the input electric signal.
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
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Question 26 (4 marks)





Criteria
Draws one circuit with a 1.5 volt battery and a switch
Draws a second circuit with an output of 18000 V
Links the two with a transformer (induction coil) and finds the ratio of turns
Omits one of the above outcomes
Successfully completes on or two of the outcomes
Marks
4
3
1-2
Sample answer
Question 27 (7 marks)
27 (a) (2 marks)



Criteria
A correct explanation of back-emf (switching off is not necessary)
Lenz’s law must be stated
One of the above outcomes
Marks
2
1
Sample answer
As electric current starts to flow through a coil each of its loops forms a magnetic field which,
in accordance with Lenz’s law, opposes the change in flux, so the operating voltage takes time
to reach its maximum value, delayed by the ‘back-emf’. (When the current is switched off the
voltage drop to zero is similarly delayed.)
27 (b) (3 marks)





Criteria
Describes the change in current direction through the coils of a DC motor
Relates this to reduced operating voltage or current due to back-emf
Relates the slowing-down of the motor to increased current or torque
Successfully displays understanding of two of the above outcomes
Makes a relevant statement about reversal of current in the DC motor coils
Marks
3
2
1
Sample answer
As a DC motor applies torque onto a load it is forced to slow down. Of course, DC motors are
fitted with commutator and brushes to allow the coil to continue to turn in the same direction.
The reversal of direction of current through the coil causes back-emf every time, meaning that
the operating voltage is always less than the applied voltage. When a load forces the motor to
slow down the reversal of current direction through the coils is less frequent, hence the actual
voltage – and current – is greater, hence the torque it applies increases:  = B A n I cos .
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
Page 7
27 (c) (2 marks)



Criteria
Identifies the high rate of current reversal when AC is applied
Relates this to a comparatively minor variation when the motor runs slower
One of these outcomes is displayed
Marks
2
1
Sample answer
When a universal motor is operating with an AC supply the current through its coils reverses
100 times every second in addition to the rate at which its coils are rotating. As a result the
slower rate of the coils when a load is applied makes very little difference to the back-emf
experienced in the coils.
Question 28 (5 marks)


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




Criteria
An appropriate labelled diagram associated directly with BCS theory
A correct argument in support of a proposal of BCS theory
A correct argument opposing a proposal of BCS theory
A general summary of the discussion
Omitting a summary of the discussion
Not including the required appropriate labelled diagram or
Not providing both an argument for and against BCS theory
Stating one correct statement concerning BCS theory
Mark
5
4
2-3
1
Sample answer
BCS theory proposes that the electrons in superconductors at temperatures below their critical
temperatures form pairs in special electron-lattice-electron interactions to form what is known
as ‘Cooper-pairs’. There is experimental evidence that supports this proposition.
The theory also proposes that as the first electron moves through the metal crystal it attracts
the adjacent positive ions inwards, thus creating a slightly positive region which thus attracts
the second electron. However, if the region does attract the second electron it would also keep
the initial electron from leaving. Moreover, the positive ions supposedly drawn inwards by
that first electron are simultaneously being attracted by billions of other electrons in the metal
crystal, so there would seem no reason for this proposition to occur. Neither is BCS theory
able to account for ‘high-temperature’ ceramic superconductors at all.
In actual fact, advanced quantum theory is required to adequately assess BCS theory, and
even then it has shown to be seriously flawed.
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
Page 8
Question 29 (4 marks)
29 (a) (1 mark)

Criteria
Substituting correctly into the correct formula yielding a result
Mark
1
Sample answer
According to Einstein’s explanation of the photoelectric effect, Ψ = h fTH
Hence 5.32  10-19 = 6.63  10-34 fTH
 fTH = 8.02  1014 hertz.
29 (b) (2 marks)
Criteria
 Correct substitution into the correct formula, yielding a wavelength
 Correctly matches this wavelength with visible light to identify type
 One of the above outcomes successfully achieved
Marks
2
1
Sample answer
Since v = f   3.0  108 = 8.02  1014  TH = 3.74  10-7 m [i.e. 374 nm]
The wavelength range of visible light is from 390 – 690 nm. The wavelength of this radiation
is shorter than that of visible light, so it is ultra-violet radiation.
29 (c) (1 mark)

Criteria
Correctly substitutes into the correct relationship to get an answer
Mark
1
Sample answer
Since 1 eV = 1.60 × 10-19 joules,  3.57 eV = 5.71 × 10-19 J
Max KE of the photoelectron, EK = h f – Ψ = 5.71 × 10-19 – 5.32 × 10-19 = 3.9 × 10-20 J.
Question 30 (5 marks)
30 (a) (3 marks)





Criteria
Correct identification of all three band-theory diagrams
Justification of all three identifications
Either all three band-theory diagrams correctly identified or
One diagram correctly identified with correct justification
One diagram correctly identified
Marks
3
2
1
Sample answer
The diagram on the left shows the semiconductor doped with a valence-3 impurity, hence is a
p-type, because the acceptor level permits a nearby electron to jump into a ‘hole’ created by
the impurity, locating the electron within the ‘forbidden gap’, closer to the conduction band.
The second diagram shows the undoped semiconductor; the ‘forbidden gap’ between its
valence band and conduction band has no available levels.
The diagram on the right shows the semiconductor doped with an n-type valence-5 dopant,
because the donor level close to the conduction band indicated that an electron can easily gain
sufficient thermal energy to jump the gap and thus move freely through the crystal.
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
Page 9
30 (b) (2 marks)
Criteria
 Correctly identifies the order of electrical conductivity
 Justifies at least two of these
 Correctly identifies the order of electrical conductivity
Marks
2
1
Sample answer
The undoped semiconductor has the least electrical conductivity of the three. Because any
valence electrons must gain sufficient thermal energy to jump right across the ‘forbidden gap’
to reach the conduction band, less are able to do so at any instant than if it has been doped.
The semiconductor doped with n-type impurity has the most electrical conductivity (provided
the proportions of dopant are equal), because a number of electrons are already located close
to the conduction band, therefore requiring little energy to be able to drift through the crystal.
The conductivity of the semiconductor doped with p-type impurity is between these two.
Question 31 (5 marks)





Criteria
At least one appropriate labelled diagram
Descriptions of two phenomena known to be properties of light, not of particles
Identifies two correct properties but does not contrast them with particles or
The diagram provided is unlabelled, or does not display e-m properties
Correctly identifies 1 or 2 wavelike properties of cathode-rays
Sample answer
Marks
4-5
3
1-2
Green glow
Shadow
The German view of the nature of cathode-rays was that they were a form of electromagnetic
radiation different from visible light or radio waves, and therefore possessing neither electric
charge nor mass.
The ‘Maltese Cross’ cathode-ray tube showed that cathode-rays produce (green) light when
they strike glass. U-V light was already known to make certain crystals fluoresce, whereas it
is not a known property of particles to cause materials to fluoresce.
This tube also showed that cathode-rays form a shadow behind a solid target. While particles
also form a type of shadow behind an impenetrable barrier, it is not usual for such a shadow to
be as clearly defined as the ‘Maltese Cross’ demonstrated.
Cathode-rays were also found to expose photographic film – certainly a common property of
electromagnetic radiation, unknown to be a property of particles.
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
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Section II
Question 32 – Medical Physics (25 marks)
32 (a) (i) (1 mark)
Criteria
 Identifies the correct principle
Mark
1
Sample answer
The physical principle governing the use of endoscopes is total internal reflection.
32 (a) (ii) (2 marks)
Outcomes Assessed: H1, H3, H4
Targeted Performance Bands: 3-5




Criteria
Correctly describes the function of the incoherent bundle
Correctly describes the function of the coherent bundle
Reversed the two functions or
Provided a correct statement about either
Marks
2
1
Sample answer
The incoherent bundle of fibres transmits the bright light from outside down to where the
body part is being observed to illuminate it. This light then reflects off the surfaces of the
internal structures. It is then collected by the coherent bundle and transmitted as a coherent
arrangement from which a clear image can be formed.
32 (b) (i) (1 mark)
Criteria
 Correct substitution into the correct formula yielding an answer
Mark
1
Sample answer
Z =    Z  1.07  10 3  1590  1.70  10 6 rayls


PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
Page 11
32 (b) (ii) (3 marks)






Criteria
Correct substitution into the correct formula yielding an answer
Appropriate analysis of the interaction – a weak reflection of the signal
Correct substitution into the correct formula yielding an answer
Incorrect, or no analysis of the interaction
Incorrect substitution into the correct formula or
Attempted analysis of the interaction without use of the formula
Sample answer
I r Z 2  Z 1 

I o Z 2  Z 1  2


Marks
3
2
1


Z  Z M   1.61  10 6  1.70  10 6  7.4  10 4
I
 r  B
2
I o Z B  Z M  2
1.61  10 6  1.70  10 6
Reflection of the sound will take place at the boundary because the two materials have
different acoustic impedances. The main factor affecting the proportion of ultrasound
reflected is the difference between the two values; here they are very similar, resulting in a
weak, but detectable, reflection.
2
2
2
32 (b) (iii) (2 marks)
Criteria
 Identifies that the bones would cause too much unwanted reflection
 Describes a method to avoid the influence of the bones on the examination
 Correct statement about one of these
Marks
2
1
Sample answer
Bones have a much greater influence on the reflection of ultrasound than do the other body
tissues, due to both their greater density and the velocity of ultrasound passing through them.
Consequently, the two bones of the lower leg would produce reflections strong enough to
drown out the reflection from the muscle/blood interface. To avoid this it is necessary to site
the ultrasound source and detector such that unwanted reflections can be avoided.
32 (c) (i) (4 marks)





Criteria
Identifies 2 advantages of CAT-scans over X-rays (clarity, multi-images)
Gives reasons for their importance in this instance
Relates these to the choices the surgeon needs to make
Successfully provides responses to two of the above outcomes
States 1 or 2 correct comments about CAT-scans or this situation
Marks
4
3
1-2
Sample answer
Prior to operating, modern-day surgeons have the advantage of CAT-scans to assist them in
making important decisions. An X-ray provides a 2-D image on a photographic plate, whilst
CAT-scans allow a 3-D image to be built up by isolating ‘slices’ of the body as the machine
rotates around the body. A CAT-scan will display the type of break in the bone, whether there
are few or several fragments (and their accurate location), and whether the jagged bone is
likely to cause significant damage to surrounding tissues (muscles, nerves or blood vessels).
The main sections of the bone must be twisted so they fit exactly, with any fragments as well.
In this case the surgeon must judge whether the prosthetic hip may have been damaged, or if
it has been displaced from within the femur. If so, the operation will be lengthy and difficult,
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
Page 12
because some of the bone will be attached to it will need to be removed, and the metal hip
will need to be re-joined to the femur once the break has been stabilised.
Because of the resolution provided by a CAT-scan, as well as the distinct views it provides,
such choices become more valid and reliable.
32 (c) (ii) (1 mark)
Criteria
 Correct comparison between a single burst of X-rays and ‘bombardment’
Mark
1
Sample answer
The process of CAT (computed axial tomography) necessitates the body to be bombarded
with high-energy X-rays, whereas a single burst is required for an effective X-ray. Inevitably
this means that a CAT-scan is more hazardous than an X-ray.
32 (c) (iii) (2 marks)
Criteria
 Identifies that an X-ray provides enough clarity to see operation results
 Identifies another advantage of using a simple X-ray (other than hazard)
 Correct identification of one of these
Marks
2
1
Sample answer
The X-ray is able to show the essential details of the surgery, the plate, the screws through the
bone and wires wrapped tightly around it clearly enough to indicate a successful procedure
without needing the finer details. CAT-scans are far more expensive that X-rays, and after the
operation such expense is unwarranted.
32 (d) (2 marks)
Criteria
 Correctly states one advantage of MRI scans over CAT-scans
 Correctly states one disadvantage of MRI scans over CAT-scans
 Correct identification of one of these
Marks
2
1
Sample answer
Although both CAT-scans and MRI scans are useful for investigating the brain for a possible
tumour, CAT-scans are superior for analysis of hard and soft tissues whilst MRI is better at
comparing differences in soft tissue, such as most brain cancers. However, MRI scans are far
more expensive than CAT-scans, and are not nearly as available because the machines are
also very expensive.
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
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32 (e) (7 marks)

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


Criteria
Thorough analysis; properties of the materials used, why they are used, the
processes they undergo
How the material enters the body, is detected and forms images
Substantial analysis, covering most of the above
Good analysis of at least three of the required outcomes above
One or two correct statements concerning PET scans
Marks
6-7
4-5
3
1-2
Sample answer
PET scans are distinct from all the other types because they can reveal how a certain part of a
living body is functioning, and not just how it looks.
The scan is initiated when the radiologist selects a specific radioisotope that will be carried
through the patient’s bloodstream and located principally in the organ to be investigated. The
properties of the radioisotope are that it is radioactive, it releases positrons as it decays, the
daughter nucleus it forms is a safe, stable element, and its half-life is short – in the range of
several hours to several days. The radioisotope must release positrons because once emitted a
positron rapidly annihilates with an electron, creating two -rays emitted in exactly opposite
0 
directions. A common example is gallium-67: 67
 67
31 Ga  1 
30 Zn
The -rays can be detected by special scanners, from which an image is formed. Isotopes that
decay by releasing positrons do not naturally exist on Earth. They have an excess number of
protons for stability; there are too few neutrons in their nuclei to maintain stability.
Their half-life has to be short because sufficient positrons must be released to form a clear,
useful image, yet the radioisotope should not stay in the body of the patient for long; after all,
it is radioactive, and therefore potentially harmful. Because their half-life is so short, these
isotopes are almost always made-to-order using a cyclotron, ready to use as soon as possible.
Different bodily organs often accumulate special elements, such as iodine in the thyroid
gland, and as such the isotope of a suitable radioisotope can usually be chosen.
Once the appropriate isotope has been selected it is combined with a blood-soluble compound
and injected into the patient. Sufficient time is allowed for the isotope to have accumulated in
the organ to be scanned. This usually takes 30 to 60 minutes. Then -ray scanners are located
to capture the images. The greater the intensity of -rays the darker the image produced.
Following the procedure the patient is advised to drink copious quantities of water to flush the
remaining radioisotopes out of the body as quickly as possible.
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
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Question 33 – Astrophysics (25 marks)
33 (a) (i) (4 marks)
Criteria
 Descriptions of 2 advantages of ground-based over space-based telescopes
 One description and one statement
 One or two stated advantages of ground-based over space-based telescopes
Marks
4
3
1-2
Sample answer
Ground-based telescopes can be fitted with much larger mirrors than can be mounted on space
telescopes, potentially giving them superior sensitivity and resolution. One example is the
Keck telescope in Hawaii, a pair of separate building each housing a 10-metre diameter mirror
at a height more than 4 km above sea-level, and well above most clouds and pollution. It is
currently the largest optical and infra-red telescope in the world. Its two mirrors are linked by
computer, so its resolution is increased by interferometry, another advantage of ground-based
telescopes not (currently) available in space. The KLA array, currently under construction in
Australia, New Zealand and South Africa, provides another example of interferometry that is
unlikely ever to be matched by technology in space.
Another serious disadvantage of space-based telescopes is the cost of launching them, placing
them in precise orbit, and maintaining them. When the HST was initially put into orbit its
performance was well below expectations, and a further US$1.2 billion upgrade was needed
to fit a device that improved its usefulness enormously.
33 (a) (ii) (3 marks)





Criteria
Describes atmospheric problems reducing sensitivity and/or resolution
Describes the objective mirror made of parts on a flexible base
Describes how the computer-controlled adjustment of the sectors occurs
Coherently relates two of the above outcomes
One correct statement related to atmospheric effects and telescopes
Marks
3
2
1
Sample answer
Ground-based telescopes, especially those not far above sea-level, perform well below their
theoretical capacity in terms of sensitivity and resolution due to atmospheric effects. These
include ‘seeing’, a general name given to turbulence in the atmosphere, mainly due to winds
or temperature differences. Even our eyes note colour and brightness variations when looking
at celestial objects; this effect is magnified when viewing them through powerful telescopes.
They seem to dance about, changing colour and brightness continually. Adaptive optics is a
method used to minimise the problem. The objective mirror is composed of many interlacing
sectors on rubber bases so they are flexible. Light from nearby ‘standard’ bright stars is taken
into the CCD of the telescope and analysed by a computer hundreds, even thousands of times
per second. As variations from the standard intensities and colours are detected the computer
adjusts the sectors to counteract the differences, allowing the observed celestial object to be
observed and measured far more accurately.
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33 (a) (iii) (1 mark)
Criteria
 Identifies that a lens cannot be made of rubber-backed sectors
Mark
1
Sample answer
The objective of a refracting telescope is a lens. As such it cannot be made of rubber-backed
sectors because light could not pass through it.
33 (b) (i) (2 marks)
Outcomes Assessed: H3, H12, H14
Targeted Performance Bands: 2-4




Criteria
Identifies that most stars in the sky are small or cool
Describes that blue filters reduce the amount of yellow/red passing through
Reversed definitions of sensitivity and resolution or
Correct statement about either
Marks
2
1
Sample answer
The colour of stars ranges from blue (largest and hottest) through white and yellow to red
(smallest and coolest). The majority of the stars in the Universe are main sequence stars, and
the majority of these are small. When observed through a blue filter the smaller, cooler stars
appear smaller than when observed through a visual filter because, like blackbodies, they emit
less violet and blue radiation than yellow and red, and blue filters reduce the amount of longwavelength radiation passing through.
33 (b) (ii) (1 mark)
Criteria
 States that some stars emit equal amounts of violet/blue as red/yellow or
 States that the star must be white
Mark
1
Sample answer
One star keeps the same apparent size (i.e. brightness) when viewed through a blue or a visual
filter because it is a white A-type star. These emit approximately equal amounts of violet and
blue light as red and yellow light.
33 (b) (iii) (1 mark)
Criteria
 Identifies that the star is blue (or any of the alternatives)
Mark
1
Sample answer
Since C.I. = B – V, a star having a colour index of –0.2 indicates to astronomers that it is a
blue O-type main sequence star, with high luminosity, mass over 8 times that of the Sun, and
with a short life-span, probably only a few million years.
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33 (c) (i) (1 mark)
Criteria
 Identifies that the period of the binary is (around) 80 years
Mark
1
Sample answer
The light curve shows that the period of this eclipsing binary is about 80 years.
33 (c) (ii) (2 marks)
Criteria
 Identifies star Z as the Red Giant.
 Relates smaller orbital path to greater mass to faster evolution
 Identifies star Z as the Red Giant
Marks
2
1
Sample answer
From the second diagram showing the orbits, star Z has the greater mass since its orbital path
is smaller. It is most likely to be the Red Giant, since stars having greater mass evolve more
quickly, and a Red Giant is in a later evolutionary stage than a Main Sequence star.
33 (c) (iii) (3 marks)






Criteria
Determines the mean orbital radius to be 7.0  1011 m
Determines the orbital period of the binary to be (80  365  3600) s
Substitutes correctly into the correct formula yielding an answer
Successfully manages two of the above outcomes
Substitutes incorrectly into the correct formula or
Successfully finds either the orbital period or orbital radius in correct units
Marks
3
2
1
Sample answer
The mean orbital radius of a binary is taken to be the semi-major axis of the two stars; in this
case it is ½  1.4  1010  103 (in metres) = 7.0  1012 m.
Its orbital period must also be changed into seconds, = 80  365  24  3600 = 2.52  109 s
4 2 r 3
Subbing these into the appropriate formula: m1  m2 
gives m1 + m2 = 3.19  1031 kg
GT 2
33 (d) (7 marks)





Criteria
Extensive response including pre-star, protostar, MS, RG, SG, supernova
and corpse, including the balance between gravitation and radiation
Strong response including description of most of the above stages
Limited, but correct descriptions of MS, RG, SG supernova and corpse
Descriptions of Main Sequence, Red Giant and supernova
One or two descriptions of stellar processes
Marks
7
5-6
4
3
1-2
Sample answer
The Crab Nebula shows a modern image of a supernova explosion of a massive star, the light
of which was observed by astronomers in 1054. A supernova ends the actual life of a massive
star, which began as part of a gaseous nebula. Millions of years later the material collected
together under gravitational forces and began to rotate, as part of a massive accretion disc.
Material collided, broke apart, and collided again; most falling inwards to the central mass,
which eventually gained sufficient mass to become a protostar. The energy lost by all the inPHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
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falling material changed to heat, and along with the increasing pressure the hydrogen trapped
in the core began to fuse into helium. This energy radiated slowly out from the core through
the dense material, during which time further material agglomerated to become part of the star
so by the time the light reached its surface, and radiated outwards, the star was massive, as
much as ten times the mass of our Sun. Once the energy was emitted from the new, hot, blue
main sequence star, it drove away the remaining gas and dust from which it had formed, aided
by the ‘solar wind’ of protons, electrons and helium ions also expelled from its surface. The
core reaction of the fusion was probably the CNO-cycle, and because of the enormous inward
gravitational pressure of the massive star, the fusion reaction had to be very rapid in order to
counterbalance it through the radiation pressure outwards.
A hot star like this will have a ‘life’ on the Main Sequence, fusing hydrogen in its core into
helium, for several million years – far less than the 9-10 billion years of our Sun. Eventually
the hydrogen in the core of the star was exhausted, a catastrophe. With no radiation pressure
to prevent it, gravitation caused the star to collapse inwards. Gravitational energy changed to
heat, whilst the pressure on the stellar core increased. This initially created the conditions for
a new shell of hydrogen surrounding the central core to begin to fuse into helium. In addition,
the huge increase in temperature and pressure in the central core reached a point where the
helium ‘ash’ there reached a point that caused it to spontaneously start to fuse as well, the
helium fusing into carbon using what is called the ‘triple-alpha reaction’. This point is called
the ‘helium flash’.
The vast radiation pressure from the combined fusion sources overcame the gravitational
pressure, forcing the star to expand, forming a Red Giant. This stage of the star’s evolution
was much shorter than that on the Main Sequence. Once the helium fuel within the central
core was also exhausted, again the radiation pressure dropped below the gravitational pressure
causing the star to collapse inwards once more. The star was massive enough for the
collapsing material to raise the core temperature again, allowing a new shell of hydrogen to
commence fusion, whilst the carbon now in the central core also started to fuse – the ‘carbon
flash’. The radiation pressure was yet greater than before, so the star became a Supergiant.
Depending solely upon the mass of the star these collapses and expansions continued, though
for a shorter time in each case.
Eventually the material in the central core did not reach ‘flash point’ despite the huge loss of
gravitational potential energy. The increasing pressure on the material in the core was enough
to collapse its atoms inwards. Electrons were crushed into the protons forcing them to become
neutrons, with another huge loss in volume, loss in gravitational potential energy, and massive
temperature rise, which caused the Supernova that was observed almost 1000 years ago. Most
of the atmosphere was blasted away to become part of other nebulae. The rest of the star has
condensed into a massive core about 20 km in diameter, a neutron star, a stellar corpse.
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Question 34 – From Quanta to Quarks (25 marks)
34 (a) (i) (2 marks)
Criteria
 Response mentions the wavelengths of the visible waves
 Identifies an (apparent) geometrical series
 One of the above correctly identified
Marks
2
1
Sample answer
When studying the visible spectral lines of excited hydrogen gas, Balmer, a mathematics
teacher, noticed that the wavelengths of the spectral lines displayed an apparent geometrical
series relationship.
34 (a) (ii) (2 marks)
Criteria
 Response includes the wavelength formula of at least one other scientist
 Identifies Rydberg’s success in developing a single formula to predict 
 One of the above outcomes correctly supplied
Marks
2
1
Sample answer
Following Balmer’s discovery, Lyman noticed a similar relationship between the wavelengths
of the U-V spectral lines of hydrogen. He developed a formula similar to that of Balmer to
‘predict’ these wavelengths. Paschen and others followed this lead with respect to the I-R
lines of excited hydrogen. Rydberg developed a single formula incorporating all these series
 1
1
1 
into a single formula,  R  2  2 

ni 
n f
34 (a) (iii) (2 marks)
Criteria
 Identifies that nf = 2
 Correct substitution into the correct formula to yield an answer
 One of the above outcomes correctly supplied
Sample answer
Since this is the Balmer (visible) series, nf
 1
1
1 
1
 R 2  2  
 1.097 × 10 7
n



ni 
 f

PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
Marks
2
1
= 2

1
1
 2   2.304  10 6
2
5 
2
   4.34  10 7 m
Page 19
34 (a) (iv) (2 marks)
Criteria
 Shows (diagram or words) that a single spectral line is split into 3 (or more)
 States that a (strong) magnetic field causes this
 One of the above outcomes correctly supplied
Marks
2
1
Sample answer
The Zeeman effect: a single spectral line is observed to split into several lines when the light
source is located within a strong magnetic field.
34 (b) (i) (3 marks)





Criteria
Description of the initial Davisson-Germer experiment
Description of the accident, and the resulting procedure
Description of what was observed afterwards, and its consequence
Statements of two of the above outcomes
States one of the above outcomes
Marks
3
2
1
Sample answer
Davisson and Germer were experimenting reflecting electrons from the surface of nickel in a
vacuum. To avoid producing X-rays they were using low-energy electrons. By accident some
air entered the vacuum, oxidising the nickel surface. To remove it they heated the nickel
strongly, with a vacuum pump to remove the air. This did remove the oxygen, but the heating
made the nickel crystals larger. When their experiment with the slow electrons recommenced
they discovered that the reflected electrons now produced patterns similar to those of X-rays,
i.e. displaying wave-like nature, confirming de Broglie’s assumption of ‘particle-waves’.
34 (b) (ii) (1 mark)
Criteria
 Correct substitution into the correct formula yielding an answer
Sample answer
6.626  10 34
h

 
mv
9.109  10 31  3.0  10 6

 
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES

Mark
1
 2.42  10 10 m
Page 20
34 (c) (i) (1 mark)
Criteria
 A correct definition of half-life
Mark
1
Sample answer
The half-life of a radioactive isotope is defined to be the time required for exactly half the
nuclei initially present in the sample to remain unchanged.
34 (c) (ii) (2 marks)
Criteria
 Correctly explaining the number of α –particles through change in mass
 Correctly explaining the number of -particles through variation in charge
 One of the above numbers correctly identified
Marks
2
1
Sample answer
Thorium-232 contains 90 protons and 142 neutrons in its nucleus.
Radium-224 contains 88 protons and 136 neutrons in its nucleus.
The mass of an α-particle is 4 amu (the mass of a -particle is effectively zero). Since the loss
in nuclear mass from Th-232 to Ra-224 is 8, 2 α-particles have been emitted in this series.
The loss of protons via 2 α-particles is 4, but the difference in atomic number is only 2. This
means that 2 neutrons have been converted to protons by the emission of -particles, so 2
-particles were also emitted in this breakdown series.
34 (d) (3 marks)
Criteria
 Clear description (labelled diagram ideal) of the cloud chamber
 Clear outline of how the ‘lines’ are produced
 Comparison between (at least) α-particle and -particle tracks
 Statements of two of the above outcomes
 States one of the above outcomes
Marks
3–4
2
1
Sample answer
The main part of the Cloud Chamber is a cylindrical enclosure about 2 cm high and 10 cm in
diameter, with transparent walls and top. A source of bright light is attached, and a small hole
on the side allows different radioactive sources to be tested in the chamber. A metal ring on
top of the cylinder was charged positively and the base, covered in paper soaked in alcohol,
was negatively charged. The base extended into a flask full of alcohol and dry ice, lowering
the temperature to around -30 C. The evaporated alcohol was below its dew-point, so formed
droplets around any charged particles inside the cylinder. When an α-particle source is located
a large number of straight lines is seen spreading out rapidly being the newly-formed droplets.
With a -particle source the pattern is similar except that the lines are much thinner, and not
as clearly defined as with α-particles. When replaced by a -ray source little response is noted,
apart from some -particles that are also released.
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34 (e) (7 marks)










Criteria
Describes Fermi’s initial results
Describes of at least two further events until Fermi’s initial nuclear reactor
Describes at least three requirements for a nuclear reactor to function
Writes a correct nuclear equation for a fission reaction
Attempts to use that equation to assess the energy release
All required key outcomes attempted, but 1 or 2 sub-descriptions missing
One or two of the key required outcomes omitted or incorrect and
Reasonable detail included in discussions of other outcomes
Two of the key required outcomes reasonably attempted
One of the key outcomes attempted
Marks
7
5-6
3-4
2
1
Sample answer
With the intent of transmuting nuclei to form a new element, Fermi bombarded a sample of
uranium with neutrons. Uranium was at that time the known element with the highest atomic
number, and neutrons had been shown to induce transmutation. Fermi discovered that this
procedure greatly increased the radioactivity of the sample, and in addition Fermi found that a
number of different products had been formed. His experiment was repeated by teams of
researchers such as Joliot-Curie and Savich, then Hahn and Strassmann, who discovered that
the element lanthanum, atomic number 57, was one product. Scientists quickly realised that
the atom had been ‘split’, and the importance of Einstein’s prediction was clear, especially as
Hahn and Strassmann were Germans, and World War 2 was imminent. Meitner and Frisch
soon mathematically explained how fission works, and experiments were carried out. Neils
Bohr visited the USA to announce the findings, and Americans were soon involved. Szilard
and Zinn established the basic mechanics of fission, distinguishing fissile and fertile isotopes,
and identifying the release of several neutrons in each fission. This gave rise to the possibility
of self-sustaining fission ‘chain-reactions’. Fermi, who had moved to the USA, modelled such
a chain-reaction using mouse-traps loaded with ping-pong balls to demonstrate the factors that
could affect the likelihood of such an event. The outbreak of war caused many researchers to
secret work making a bomb – the Manhattan Project – and in 1942 Fermi constructed the first
nuclear reactor as part of this. The reactor demonstrated a controlled nuclear reaction, a chainreaction that could be accelerated if it began to subside, or decelerated when it started reacting
too quickly.
A nuclear fission reactor requires fuel, sufficient to maintain ‘critical mass’, the exact amount
such that of the average 2.5 neutrons released with each fission just one causes another fission
(with the others escaping). Control rods quickly absorb neutrons; they are essential to slow a
chain-reaction if it proceeds to rapidly. A coolant is essential both to stop the rods melting and
to transfer heat out of the reactor for use, usually by boiling water into steam used to generate
electricity. Moderators improve the efficiency of the fuel by slowing the neutrons down; slow
neutrons are far more efficient fission-inducers than fast ones, which often bounce off nuclei.
Although Fermi’s initial reactor had no radiation shielding, enormous numbers of neutrons
and -radiation are emitted from functioning reactors, both of which are lethal to humans.
One example of a fission reaction is:
Using the Binding energy/nucleon Vs number of nucleons graph provided the energy released
can be estimated. (235  7.6) + 0  (142  8.3) + (90  8.8) + 0
Binding energy of products = 1970 MeV; binding energy of initial nucleus = 1786 MeV.
Thus, approximately 200 MeV of energy is released by a single fission, as Einstein predicted!
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Section I
Part A
Question Marks
1
1
2
1
3
1
4
1
5
1
6
1
7
1
8
1
9
1
10
1
11
1
12
1
13
1
14
1
15
1
16
1
17
1
18
1
19
1
20
1
Content
9.1; 9.2.2
9.2.4
9.2.1; 9.2.2
9.2.1
9.2.2
9.2.4
9.2.3
9.3.2; 9.3.5
9.3.2; 9.4.1
9.3.1; 9.3.4
9.3.1
9.3.3; 9.4.1
9.3.3
9.3.4
9.4.1
9.4.2
9.4.4
9.4.3
9.4.4
9.4.3
Syllabus/Course Outcomes
H2; H6; H11
H1; H3; H4; H16
H3; H5; H6; H9
H5; H9
H1; H6; H7
H2; H14
H1; H3; H7
H2; H9; H14
H9; H10; H11
H9
H2; H9; H13
H2; H7; H9
H4; H7; H16
H3; H7
H2; H3; H11; H14
H1; H10; H14
H1; H3; H8; H10
H1; H3
H1; H9; H13
H1; H3; H9
Section I
Part B
Question Marks
21
4
22
6
23
6
24
2
25
7
26
4
27
7
28
5
29
4
30
5
31
5
Content
9.2.2
9.2.1; 9.2.3
9.2.1
9.2.4
9.3.1
9.3.2; 9.3.4
9.3.2; 9.3.5
9.4.4
9.4.2
9.4.3
9.4.1
Syllabus/Course Outcomes
H2; H6
H5; H6; H11; H14
H2; H12; H14; H15
H6
H7; H9
H3; H4; H14
H1; H9; H13
H2; H13; H14
H1; H10
H3; H10; H13
H1; H2; H10; H14
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
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Section II
Question Marks
32
25
33
25
34
25
Content
Medical Physics
Astrophysics
From Quanta to
Quarks
PHTRY15_2_CORE PLUS OPTIONS_GUIDELINES
Syllabus/Course Outcomes
H1; H3; H4; H8
H1; H3; H12; H14
H1; H8; H9; H11; H14; H15
Page 24