INTERNATIONAL AS

PHYSICS
PH02
Unit 2 Electricity, waves and particles

Mark scheme
January 2020
Version: 1.0 Final

*201XPH02/MS*
 MARK SCHEME – INTERNATIONAL AS PHYSICS – PH02 – JANUARY 2020

Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant
questions, by a panel of subject teachers. This mark scheme includes any amendments made at the
standardisation events which all associates participate in and is the scheme which was used by them in
this examination. The standardisation process ensures that the mark scheme covers the students’
responses to questions and that every associate understands and applies it in the same correct way.
As preparation for standardisation each associate analyses a number of students’ scripts. Alternative
answers not already covered by the mark scheme are discussed and legislated for. If, after the
standardisation process, associates encounter unusual answers which have not been raised they are
required to refer these to the Lead Examiner.

It must be stressed that a mark scheme is a working document, in many cases further developed and
expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark
schemes on the basis of one year’s document should be avoided; whilst the guiding principles of
assessment remain constant, details will change, depending on the content of a particular examination
paper.

Further copies of this mark scheme are available from oxfordaqaexams.org.uk

Copyright information

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their own internal use, with the following important exception: OxfordAQA cannot give permission to schools/colleges to photocopy any material that is
acknowledged to a third party even for internal use within the centre.

Copyright © 2020 Oxford International AQA Examinations and its licensors. All rights reserved.

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 MARK SCHEME – INTERNATIONAL AS PHYSICS – PH02 – JANUARY 2020

Level of response marking instructions

Level of response mark schemes are broken down into levels, each of which has a descriptor. The
descriptor for the level shows the average performance for the level. There are marks in each level.

Before you apply the mark scheme to a student’s answer read through the answer and annotate it (as
instructed) to show the qualities that are being looked for. You can then apply the mark scheme.

Step 1 Determine a level

Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the
descriptor for that level. The descriptor for the level indicates the different qualities that might be seen in
the student’s answer for that level. If it meets the lowest level then go to the next one and decide if it
meets this level, and so on, until you have a match between the level descriptor and the answer. With
practice and familiarity you will find that for better answers you will be able to quickly skip through the
lower levels of the mark scheme.

When assigning a level you should look at the overall quality of the answer and not look to pick holes in
small and specific parts of the answer where the student has not performed quite as well as the rest. If
the answer covers different aspects of different levels of the mark scheme you should use a best fit
approach for defining the level and then use the variability of the response to help decide the mark within
the level, ie if the response is predominantly level 3 with a small amount of level 4 material it would be
placed in level 3 but be awarded a mark near the top of the level because of the level 4 content.

Step 2 Determine a mark

Once you have assigned a level you need to decide on the mark. The descriptors on how to allocate
marks can help with this. The exemplar materials used during standardisation will help. There will be an
answer in the standardising materials which will correspond with each level of the mark scheme. This
answer will have been awarded a mark by the Lead Examiner. You can compare the student’s answer
with the example to determine if it is the same standard, better or worse than the example. You can then
use this to allocate a mark for the answer based on the Lead Examiner’s mark on the example.

You may well need to read back through the answer as you apply the mark scheme to clarify points and
assure yourself that the level and the mark are appropriate.

Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be
exhaustive and you must credit other valid points. Students do not have to cover all of the points
mentioned in the Indicative content to reach the highest level of the mark scheme.

An answer which contains nothing of relevance to the question must be awarded no marks.

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 MARK SCHEME – INTERNATIONAL AS PHYSICS – PH02 – JANUARY 2020

Question Marking guidance Additional comments/guidelines Mark

01.1 The photon (energy) is absorbed OR the atom/electron is excited ✓

(Atom moves to) −1.57 eV level ✓ Evidence for this can come from the diagram i.e. 2
upward arrow to the correct level

01.2 6✓ 1

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 MARK SCHEME – INTERNATIONAL AS PHYSICS – PH02 – JANUARY 2020

Question Marking guidance Additional comments/guidelines Mark

02.1 The maximum angle of incidence Accept “incidence” explained.

at which refraction occurs ✓ Condone minimum angle of incidence at which total
internal reflection occurs
1
Accept the angle of incidence that produces an
°
angle of refraction of 90 / along the boundary

n
02.2 Use of sin𝜃𝜃 = n2 to find n2 (= 1.3) ✓
1
𝑐𝑐
Use of 𝑛𝑛 = ✓ For max 1 mark, condone 1.52 =
3.0 ×108
3
𝑣𝑣
𝑣𝑣
2.3 × 108 (m s−1) ✓

03 any four:
For V1: Same current through R and var R, ✓
V1 decreases ✓
V = IR therefore larger R gets more V ✓
because voltage is shared in the ratio of the resistances ✓
and the variable resistor receives a larger (share of the) voltage ✓ OR: the current (in that branch) decreases ✓
4
Since R is fixed, and V=IR, V decreases✓
V2 stays the same ✓
because the parallel branches are independent or V2 is still emf ✓

Incorrect conclusion negates explanation marks

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 MARK SCHEME – INTERNATIONAL AS PHYSICS – PH02 – JANUARY 2020
04.1 Any two: For mp1 allow a description of diffraction
Diffraction at (each) slit ✓ For MP2 accept discussion of phase difference at
2
Interference (between light from different slits) ✓ maxima / minima
Idea that different wavelengths have maxima at different angles ✓

04.2 correct read-off of λ from the diagram = 619 (nm) ✓ Ignore POT error for mp1
hc
use of Eγ = 𝜆𝜆
✓ 3
3.2 × 10–19 (J)✓ Allow 2 marks max if 618 or 620nm are used.

Question Marking guidance Additional comments/guidelines Mark

05.1 clear rearrangement including:

squaring f correctly ✓ 2
substituting m/l for μ ✓

2 1 For MP1, evidence of gradient calculation must be

05.2 gradient determined correctly = 17.6 to 18.2 (Hz N– ) ✓
seen OR if a single point rather than a gradient is
correct substitution and rearrangement of 1/(4lm) ✓ used, evidence of extending the line to the origin
3
must be seen.

m = 5.3x10-3 (kg) to 5.5 x 10-3 (kg) ✓ must be at least 2 sf in final answer

05.3 smaller uncertainty for the tape measure ✓ MP1 condone smaller error or more precise for
smaller uncertainty 2
because the metre rule must be used multiple times ✓

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 MARK SCHEME – INTERNATIONAL AS PHYSICS – PH02 – JANUARY 2020

Question Marking guidance Additional comments/guidelines Mark

06.1 Use of W = QV ✓
2
9400 V ✓

06.2 1
Use of Ek = 2 mv2 

Use of p = mv  3

5.2 × 10−23 (kg m s–1) to at least 2 sf 

06.3 use of λ = h/p ✓ condone incorrect n for MP3

If 5 × 10−23 kg m s–1 used, d = 1.5 × 10−10(m)
n = 2 seen or implied ✓
4
use of (n) λ = d sin θ ✓

d = 1.5 × 10-10 (m) cao✓

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 MARK SCHEME – INTERNATIONAL AS PHYSICS – PH02 – JANUARY 2020

Question Marking guidance Additional comments/guidelines Mark

07.1 Minimum photon frequency required ✓ Must have reference to light / radiation etc.
2
for a (photo)electron to be emitted (from a metal surface) ✓

–19
07.2 φ = Eγ – KE (= 3.1 × 10 J) ✓
2
conversion of their φ to eV (= 2.0 eV) ✓ Condone 1sf for this question

07.3 Evidence of
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
✓ For one mark maximum, condone work function
(𝑝𝑝ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜) 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
(ecf) or KE used instead of photon energy if no other 2
16 –1 error made.
N = 3.5 × 10 (s ) ✓

07.4 any three: Accept speed / velocity for KE

(Photo)electrons have higher (maximum) kinetic energy ✓ For MP2 accept explanation using photoelectric
because each photon has more (double) energy ✓ effect equation.
3
half the number of photoelectrons emitted ✓
because half the number of photons incident on plate ✓ Award 1 mark for MP3 and MP4 if full qualitative
argument given

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 MARK SCHEME – INTERNATIONAL AS PHYSICS – PH02 – JANUARY 2020

Question Marking guidance Additional comments/guidelines Mark

08.1 Ratio of voltage to current ✓ accept simply V/I (as seen in spec) - with terms
1
defined

08.2 correct read-off (= 2.1 A)✓

2
3V/their current (= 1.4 Ω) ✓
2 2
08.3 Use of P = IV or P = I R or P = V /R ✓ ecf on 08.2 for MP1 and MP2
P = 6.3 ✓ If 1.4Ω used, P = 6.2 3
2 –3
kg m s ✓ MP3 is stand alone

08.4 less current passes through Y (than X) or Y receives less voltage ✓ Full marks for reverse argument
at lower currents/voltages, the resistance (ratio of voltage to current) is For MP2, accept the temperature is lower or there 3
less ✓ are fewer collisions between electrons and ions
so Y has a lower resistance ✓

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 MARK SCHEME – INTERNATIONAL AS PHYSICS – PH02 – JANUARY 2020

Question Marking guidance Additional comments/guidelines Mark

09.1 m
use of T = 2𝜋𝜋� ✓ must be at least 2 sf
k
2
0.19 (s) ✓

09.2 resonance ✓
the speed bumps provide a periodic driving force ✓ Ignore any reference to phase 3
which matches the natural frequency/time period of the bicycle's
suspension ✓

09.3 use of distance = speed × time and recognising that the time is the Answer is 1.2 m if 0.2 s is used from 09.1
same as in 09.1✓
If no other mark awarded, one mark can be given for
1.1 (m) ✓ calculation of frequency (5.2Hz) 2

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 MARK SCHEME – INTERNATIONAL AS PHYSICS – PH02 – JANUARY 2020

Question Marking guidance Additional comments/guidelines Mark

10.1 a length that is at least 5w must be measured ✓ MP1 evidence seen on diagram or through
calculation 2
w = 10.2 to 10.9 (mm) ✓
3 sf only

10.2 Absolute uncertainty is = ±1 mm ✓ expect eg 1 mm/105 mm

2
Divides by their original measured value x 100 ✓ Mp2: 1 or 2 sf only
MP1 and MP2 are separate marking points

10.3 λD ecf on their w

Use of w = ✓
s 2
520 (nm) – 535 (nm) ✓ ecf on 10.1

10.4 Attempt to add three % uncertainties ecf on 10.2 ✓ Expect to see 1.2% + 0.6% + their 10.2

Use of Absolute uncertainty = % uncertainty x λ (may be implied) ✓ 3

Final answer 2 sig fig only

Final answer 14 (nm) OR 15 (nm) cao ✓

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 MARK SCHEME – INTERNATIONAL AS PHYSICS – PH02 – JANUARY 2020

Question Key

11 B

12 D

13 B

14 A

15 C

16 D

17 C

18 A

19 C

20 B

21 B

22 C

23 A

24 C

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