WORKBOOK ANSWERS
AQA Physics
Sections 1–3 Measurements and
their errors  Particles and radiation 
Waves
This Answers document provides suggestions for some of the possible answers that might be
given for the questions asked in the workbook. They are not exhaustive and other answers
may be acceptable, but they are intended as a guide to give teachers and students feedback.
Section 1
Measurements and their errors
Topic 1 Use of SI units and their prefixes
1
a
7 × 10–11 m; 70 pm (1)
b
6.328 × 10–7 m; 632.8 nm (1)
c
1.41 × 108 kg; 141 Gg (1)
Topic 2 Limitations of physical
measurements
2 Number of measurements = 19 (1)
mean average =
((3.1´1) + (3.4 ´1) + (3.5 ´ 3) + (3.6 ´ 7) + (3.7 ´ 5) + (3.8 ´ 2)) mm
19
= 3.6 mm
(1)
There do not appear to be any outliers.
Maximum value = 3.8
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1
Minimum value = 3.3
error in measurement =
(3.8 - 3.3) mm
2
= 0.25 mm or 0.3 mm (2 s.f.) (1)
so mean average of measurements = 3.6  0.3 mm
Topic 3 Estimation of physical quantities
3 All approximate values:
a
2 m × 10 m × 5 m = 100 m3 (1)
b
0.5 kg (1)
c
10 m in about 10 seconds = 1 m s–1 (1)
d
Yellow flame about 1000°C (or 1 × 103°C) (1)
e
about 1 N (1)
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Section 2
Particles and radiation
Topic 1 Particles
1 C (1)
2
a
specific charge =
charge
mass
=
+1.264 ´10
((79 ´ 1.672 ´10
b
specific charge = 0; atoms are neutral (1)
a
B (1)
b
C (1)
c
A (1)
d
B (1)
-27
-17
C
) +(118 ´1.675 ´10
(1) = 3.8 ´10 C kg
7
-27
-1
)) kg
3
4 Similarities: same number of protons (Z = 1) (1); same number of electrons (e = 1) (1)
Differences: different number of neutrons (0, 1 and 2) (1)
5
a
210
84
Po ® 206
Pb + 42 He
82
(3: 1 mark for each nuclide)
b
90
38
Sr ® 90
Y + -10e + ne
39
(3: 1 mark for each decay particle)
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(1)
6
Correct shape (1)
Correctly labelled axes (1)
Minimum at about 0.8 fm separation (1)
Values of separation on x-axis (1)
Repulsive / Attractive force labels (1)
7
B (1)
8
E = 6 eV = (6 × 1.6 × 10–19) J = 9.6 × 10–19 J (1)
E = hf =
hc
l
Þl =
hc 6.6 ´ 10-34 J s ´ 3.0 ´ 108 m s-1
=
= 2.1´ 10-7 m (1) = 210 nm
E
9.6 ´ 10-19 J
This is an ultraviolet photon. (1)
9 D (1)
10 Required energy of photon, E = 2 × 105.6 MeV (1) = 211.2 MeV = (211.2 × 106 × 1.6 × 10–
19) J = 3.4 × 10–11 J (1)
l=
hc 6.6 ´ 10-34 J s ´ 3.0 ´ 108 m s-1
=
(1) = 5.82 ´ 10-15 m (1)
E
3.4 ´ 10-11 J
11 D (1)
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12
(1 mark for each correct particle, electron and electron antineutrino can be either way around.)
13 A (1)
14 Production: Sbefore = 0 + 0 = 0; Safter = 0 + 0 + 1 – 1 = 0, so conserved. (1)
Decay: Sbefore = –1; Safter = 0 + 0, so not conserved. (1)
This is possible under the rules of weak interaction decay, so these equations are
possible. (1)
15 C (1)
16
1 mark for each correct particle. Positron and electron-neutrino can be either way around.
17
a
p + e+ → e– + 0 + K+
Conservation
quantity
Before interaction
After interaction
Quantity
conserved?
p
e+
Total
e–
0
K+
Total
Q
+1
+1
+2
–1
0
+1
0

B
+1
0
+1
0
+1
0
+1

L
0
–1
–1
+1
0
0
+1

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S
0
0
0
0
+1
-1
0

Interaction not possible.
(1 mark for each correct row and correct conclusion)
b
p + ne ® e+ + n
Conservation
quantity
Before interaction
After interaction
Quantity
conserved?
p
𝜈̅𝑒
Total
e+
n
Total
Q
+1
0
+1
+1
0
+1

B
+1
0
+1
0
+1
+1

L
0
–1
–1
–1
0
–1

S
0
0
0
0
0
0

Interaction possible
(1 mark for each correct row and correct conclusion)
Exam-style questions
1 a
Subatomic
particle
Quark
structure
Baryon
or
meson
or lepton
Relative
charge
Baryon
number
Lepton
number
Strangeness
K- meson
𝑠𝑢̅
meson
-1
0
0
+1
proton
uud
Baryon
+1
+1
0
0
positron
n/a
Lepton
+1
0
-1
0
(1 for each correct row)
b
For example: neutron (1) (udd) (1) and anti-neutron (1) (𝑢̅𝑑̅ 𝑑̅ ) (1)
c
Similar: mass-energy; baryon number; strangeness (any 1)
Differences: charge; lepton number (Any 1)
d
p + p ® 2g (1)
Energy of  = rest energy of proton = 938.3 MeV = 1.5 × 10–10 J (1)
l=
ei
11
6
hc 6.6 ´ 10-34 J s ´ 3.0 ´ 108 m s-1
=
= 1.32 ´ 10-15 m (1)
E
1.5 ´ 10-10 J
C ® 115 B + +10e + ne
(1 for each correct decay particle)
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e ii
(1 mark for each correct particle)
e iii
specific charge =
charge
5 ´ 1.6 ´ 10-19 C (1)
=
= 4.3 ´ 107 C kg-1(1)
mass ((5 ´ 1.673 ´ 10-27 ) (1) + (6 ´ 1.675 ´ 10-27 ) (1)) kg
f
Before interaction
After interaction
K-
Total
–
𝜈̅𝜇
Total
Q
–1
–1
–1
0
–1

B
0
0
0
0
0

L
0
0
+1
−1
0

S
−1
−1
0
0
0

Conservation
quantity
Quantity conserved?
This decay could happen as strangeness is not necessarily conserved in weak
interactions.
(1 for each correct row plus conclusion)
2
a
Your writing should be legible and the spelling, punctuation and grammar should be sufficiently
accurate for the meaning to be clear. Your answer needs to be marked as a whole and is
assigned to one of three levels according to the following criteria:
High level (good to excellent): 5 or 6 marks. Your answer is clearly organised, logical and
coherent, using the appropriate specialist vocabulary correctly. You have given correct
examples of hadrons and leptons and identified the differences between hadrons and leptons
(hadrons affected by strong nuclear reaction and are made of quarks). Leptons are fundamental
and do not experience the strong nuclear reaction. Hadrons are divided into baryons and
mesons. Baryons three quarks, mesons quark anti-quark pair. Similarities between groups: all
experience weak interaction and if charged the electromagnetic interaction. All have rest mass.
Intermediate level (modest to adequate): 3 or 4 marks. Your answer may be less well organised
and not fully coherent. You have used fewer examples of specialist vocabulary, or the specialist
vocabulary is used incorrectly. You give correct examples of hadrons and leptons and identified
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one difference between hadrons and leptons (e.g. hadrons affected by strong nuclear reaction or
are made of quarks). Leptons are fundamental. Hadrons are divided into baryons and mesons.
Low level (poor to limited): 1 or 2 marks. Your answer is poorly organised and may not be
relevant or coherent. You have used only a little correct specialist vocabulary. You have only
identified two correct properties of hadrons and leptons.
A 6 mark answer may include most of the following points:

(correct) example of a hadron and a lepton

mention of strong and weak interactions

mention of quark structure of hadrons

leptons are fundamental particles

identification of baryons and mesons

correct quark structure of baryons and mesons

similarities between leptons and hadrons e.g. they all have rest mass; all are affected by
weak interaction; if charged both experience electromagnetic interaction
b
Particle
Antiparticle
Lepton or Hadron
Electron-neutrino
Electron antineutrino
Lepton
Electron
Positron
Lepton
Neutron
Antineutron
Hadron
(1 for each correct row)
ci
Exchange particles (1)
c ii
Particle
Particle interaction involved
Virtual photon
Electromagnetic
W
Weak
(1 for each row)
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Topic 2 Electromagnetic radiation and
quantum phenomena
1 C (1)
2 Maximum kinetic energy of emitted photoelectrons = eVs = 3.02 eV (1)
Energy of photons = work function + max. kinetic energy = (2.28 + 3.02) = 5.30 eV (1)
l=
hc 6.6 ´ 10-34 J s ´ 3.0 ´ 108 m s-1
=
= 2.33 ´ 10-7 m (1)
E
(5.30 ´ 1.6 ´ 10-19 ) J
3 A (1)
4 E = hf = 6.6 × 10–34 J s × 3.6 × 1015 Hz = 2.376 × 10–18 J = 14.9 eV
(1)
Ionisation energy of hydrogen = 13.6 eV (1)
Max. ke of ionised electron = 14.9 – 13.6 = 1.3 eV (1) = 2.1 × 10–19 J (1)
5 C (1)
6 Highest energy photon: transition from n = 4 to n =1 (–0.85 + 13.6) eV = 12.75 eV (1)
Lowest energy photon: transition from n = 4 to n = 3 (–0.85 + 1.51) = 0.66 eV (1)
lhighest energy =
lhighest energy =
hc
Ehighest
hc
Ehighest
=
6.6 ´ 10-34 J s ´ 3.0 ´ 108 m s-1
= 9.7 ´ 10-8 m (1)
(12.75 ´ 1.6 ´ 10-19 ) J
=
6.6 ´ 10-34 J s ´ 3.0 ´ 108 m s-1
= 1.9 ´ 10-6 m (1)
(0.66 ´ 1.6 ´ 10-19 ) J
7 B (1)
8
l=
h
h
6.6 ´ 10-34 J s
Þv =
(1) =
(1) = 1.3 ´ 106 m s-1 (1)
mv
ml
(9.11´ 10-31 kg ´ 564 ´ 10-12 m)
Exam-style questions
1
ai
Energy (1) required to remove an electron from the surface of a metal via the
photoelectric effect. (1)
a ii
The energy of incident photon is constant (1). Energy is needed for an electron to
leave the surface of the metal, overcoming the metal’s work function. (1) The
maximum kinetic energy Ekmax of the emitted photoelectron is the energy of photon
minus the work function, . (1) Electrons that are deeper in the surface require
energy to get to the surface (1)
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a iii
hc
l
= Ekmax + f (1) Þ l =
6.6 ´ 10-34 J s ´ 3.0 ´ 108 m s-1
= 2.89 ´ 10-7 m (1 + 1 s.f.)
(3.51´ 10-20 + (4.07 ´ 1.6 ´ 10-19 ) (1)) J
b
New observation unable to be explained using existing theory (1). New theory
developed to explain all observations including new observation. (1)
c
Ek = 21 mv 2 Þ 2mEk = p (1)
l=
h
h
=
p
2mEk
(1) =
6.6 ´ 10-34 J s
2 ´ 9.11´ 10
-31
kg ´ 3.51´ 10
-20
= 2.61´ 10-9 m (1)
J
di
The energy required to remove an electron in the ground state (1) from an atom (1).
d ii
Excitation involves electrons moving from a lower energy level to a higher energy
level. (1)
d iii
∆E = (–8.99 + 16.91) eV = 7.92 eV (1) = (7.92 × 1.6 × 10–19) J = 1.27 × 10–18 J (1)
l=
hc 6.6 ´ 10-34 J s ´ 3.0 ´ 108 m s-1
=
= 1.56 ´ 10-7 m (1)
-18
E
1.27 ´ 10 J
2
Your answer should be legible and the spelling, punctuation and grammar should be sufficiently
accurate for the meaning to be clear. Your answer should be looked at as a whole.
High level (good to excellent answer): 5 or 6 marks. Your answer is clearly organised, logical and
coherent, using correct specialist vocabulary. It provides a comprehensive description which
includes a clear explanation of constant energy level differences and how electrons can be
excited by electron collisions. Your answer clearly links the energy of a photon to its frequency
and includes a clear explanation of the reason why atoms of a certain element emit photons of a
characteristic frequency. Your answer should relate the energy difference between levels to the
energy of the emitted photons, and states that this energy difference is constant.
Intermediate level (modest to adequate): 3 or 4 marks. Your answer is less well organised and
not fully coherent and your use of specialist vocabulary may not be correct. Your answer
provides an explanation of energy levels and how excitation takes place by electron collision
with orbital electrons and you have explained how an orbital electron loses energy by emitting a
photon.
Low level (poor to limited): 1 or 2 marks. Your answer is poorly organised and may not be
relevant or coherent. There is little correct use of specialist vocabulary. You have included some
mention of energy levels and the idea of excitation of electrons.
A 6-mark answer might include the following points:

electrons bombard the atoms of the gas and give energy to electrons in the atoms

the electrons move to higher energy levels (and are excited).

excited electrons move down to lower energy levels by losing energy via the emission
of photons
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
photons have energy hf

photons of characteristic frequencies emitted from atoms of a particular element; this is
because atoms have discrete energy levels
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Section 3
Waves
Topic 1 Progressive and stationary
waves
1 B (1)
2
3
4
´ 2p =
3p
p
(1); A–C = 0 (1); B–C = (one whole cycle plus)
(1)
2
2
a
A–B =
b
A and C are in phase (1)
c
2 complete waves pass from A to C in 5.2 s, therefore wave period = 2.6 s so
frequency = 1/(2.6 s) = 0.38 Hz (1);  = c/f = (0.68 m s–1)/(0.38 Hz) = 1.6 m (1)
3 A (1)
4
a
Differences: (any 1) axis of vibration in different orientations relative to direction of
travel; compressions/rarefactions v waves/troughs
Similarities: (any 1) wavelength, wave speed, frequency, amplitude, period
bi
b ii
Graph is at a maximum at 0o, 180o and 360o (1) and at zero intensity at 90o and
270o. (1)
c
Polarising filters reduce glare (1) off water surfaces.
5 A (1)
6
f=
1 T
1
78 N
=
(1) ´
(1) = 137 Hz (1) (140 Hz to 1 s.f.)
2l m 2 ´ 0.6 m
(2.3 ´ 10-3 kg/0.8 m)
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Exam-style questions
1
ai
/2 or 90o (1)
a ii
3/2 or 270o (1)
a iii
 or 180o (1)
b
The rock particle first moves upwards to its maximum upwards displacement (1)
before then moving downwards to its maximum downwards displacement and then
moves back upwards to its equilibrium.(1)
c
Wave must be transverse. (1) Only transverse waves can be polarised. (1)
di
Number of waves passing a point per second. (1)
d ii
l=
e
The rock particle will displace from equilibrium to a maximum amplitude in the
direction of propagation of the wave (1) and then to a maximum amplitude in the
opposite direction of propagation of the wave (1) before returning to equilibrium.
f
vp =
c 4.8 ´ 103 m s-1
=
(1) = 873 m (1) (870 m to 2 s.f.)
f
5.5 Hz
M
r
=
67.5 ´ 109 Pa
(1) = 5000 m s-1 (1)
3
-3
2.75 ´ 10 kg m
(1)
2
Your answer should be legible and the spelling, punctuation and grammar should be sufficiently
accurate for the meaning to be clear. Your answer should be looked at as a whole.
High level (good to excellent) 5 or 6 marks. Your answer is clearly organised, logical and
coherent, using correct specialist vocabulary. It fully describes waves travelling in opposite
directions AND superimposing on each other forming a series of nodes and antinodes with
stated features.
Intermediate level (modest to adequate) 3 or 4 marks. Your answer is less well organised and not
fully coherent and your use of specialist vocabulary may not be correct. It describes waves
travelling in opposite directions OR the superposition of waves and some of the features of the
standing wave.
Low level (poor to limited) 1 or 2 marks. Your answer is poorly organised and may not be
relevant or coherent. There is little correct use of specialist vocabulary. One correct key feature
or one relevant remark regarding formation given.
A 6-mark answer might include the following points

4 nodes of zero amplitude

3 antinodes of maximum amplitude

the wavelength is 0.80 m

the end antinodes are in phase and the middle and end antinodes are in antiphase

between a node and an antinode, the amplitude of oscillation increases
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
the waves reflect off the clamp (and the metal bar)

the waves travelling in opposite directions superpose on each other

the waves have the same wavelength and frequency (and similar amplitude)

destructive interference occurs at nodes and constructive interference occurs at
antinodes

energy is not transferred along the string
Topic 2 Refraction, diffraction and
interference
1 C (1)
2
ai
Two wave sources are coherent if they have the same frequency and emit waves in
phase (1) or with a constant phase difference.
a ii
A source is monochromatic if it emits waves with the same (constant) frequency
and wavelength. (1)
b
Wear laser safety spectacles to reduce the intensity of the laser beam. Put a ‘laser
safety’ sign on the door to avoid unauthorised access to the laboratory. (1 mark for
description, 1 mark for explanation)
c
w=
lD
s
=
632.8 ´ 10-9 m ´ 4.6 m (1)
= 0.015 m (1)
0.20 ´ 10-3 m (1)
3 A (1)
1.3 ´ 10-3 m
= 3.03 ´ 10-6 m (1)
4
330
nl = d sin q
d=
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nl 632.8 ´ 10 -9 m
=
= 0.21
d
3 ´ 10-6 m
sin-1 0.21= 12.05 Þ AB = 5.4 m ´ tan12.05 = 1.15 m (1)
sin q1 =
nl 2 ´ 632.8 ´ 10-9 m
=
= 0.42
d
3 ´ 10-6 m
sin-1 0.42 = 24.69 Þ AC = 5.4 m ´ tan24.69 = 2.48 m (1)
sin q 2 =
Distance between n = 1 and n = 2 is (2.48 – 1.15) m = 1.33 m (1) (1.3 m 2 s.f.)
5 A (1)
6
a
Reflected at P (1); Refracted at back face of cube (1) at same angles (1)
b
n1 sin q1 = n2 sin q 2 Þ n2 =
n1 sin q1
sin q 2
(1) =
1´ sin 41°
= 1.496 (1) » 1.49
sin 26°
Exam-style questions
1
a
c decreases (1); f stays constant (1);  decreases (1)
b
Total internal reflection; beam reflects off core/cladding boundary (1) obeying law of
reflection (1)
ci
n=
c1 3.00 ´ 108 m s-1
=
(1) = 1.49 (1)
cs 2.01´ 108 m s-1
c ii
sin qc =
d
n2
æ 1.44 ö
(1) Þ qc = sin-1 ç
= 75.1° (1)
n1
è 1.49 ÷ø
Angle of refraction at air/core boundary = (90 – 75.1)o = 14.9o (1)
æ 1.49 ´ sin 14.9° ö
n1 sin q1 = n2 sin q 2 (1) Þ q1 = sin-1 ç
÷ø = 22.5° (1)
1.00
è
e
Absorption of the signal by the material of the fibre. (1)
Modal dispersion — rays inside the fibre taking slightly different paths. (1)
2
a
Your answer should be legible and the spelling, punctuation and grammar should be sufficiently
accurate for the meaning to be clear. Your answer should be looked at as a whole.
High level (good to excellent) 5 or 6 marks
Your answer is clearly organised, logical and coherent, using correct specialist vocabulary.
Your diagram and text fully describe the experiment, the measurements that need to be taken
© Jeremy Pollard 2015
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and the mathematical calculation of the wavelength of the laser light. Your answer contains two
safety precautions and at least two ways of improving accuracy.
Intermediate level (modest to adequate) 3 or 4 marks
Your answer is less well organised and not fully coherent and your use of specialist vocabulary
may not be correct. It describes the experiment although it may be missing some details. Your
diagram may not fully describe the experiment and you may have omitted key measurements OR
how to use them to calculate the wavelength of the laser light. You may have at least one safety
precaution and some reference to improving accuracy.
Low level (poor to limited) 1 or 2 marks
Your answer is poorly organised and may not be relevant or coherent. There is little correct use
of specialist vocabulary. You may have a simple diagram and an incomplete method, but you do
not have the key measurements or details of how to calculate the wavelength of the laser light.
You may have a safety precaution OR a way of improving accuracy.
A 6-mark answer might include the following points

a fully labelled diagram

a comprehensive method

the key measurements required (order number and spacing; grating spacing and
grating-screen distance).

reference to the grating equation: n = d sin 

two safety precautions: wearing laser safety goggles and displaying a laser safety sign.

ways of improving accuracy such as: precision of measurements (suitable measuring
instruments); repeating measurements; consideration of the width of order spots
bi
Order spacing is not regular for a grating (fringe spacing is regular for a double slit)
(1)
b ii
w=
b iii
Young’s double slit experiment shows light behaving as a wave (1). Two previous
models (Huygens — wave model; and Newton — particle model) did not
conclusively prove that light was wave or particle in nature (1). Young’s experiment
confirmed Huygens wave model (1).
lD
ws
1.8 ´ 10-2 m ´ 0.15 ´ 10-3 m
Þl =
(1) =
(1) = 6.4 ´ 10-7 m (1)
s
D
4.2 m
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