AQA AS Level Physics Paper 1, June 2022: Question 4
9 marks · Medium difficulty · Extended Answer
Discuss how the rate of loss of charge from an isolated negatively-charged metal plate depends on the frequency and intensity of incident electromagnetic radiation, and calculate the work function given the maximum kinetic energy and frequency.
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Question text
04 An isolated metal plate is given a negative charge. Electromagnetic radiation is
incident on the plate. The plate loses its charge due to the photoelectric effect.
04.1 Discuss how the rate of loss of charge from the plate depends on the frequency and
intensity of the incident radiation.
In your answer you should explain why:
• the plate loses its charge
• the photoelectric effect occurs only for frequencies greater than a particular value
• the rate of loss of charge increases with intensity for radiation above that particular
value of frequency.
[6 marks]
04.2 Charged particles are emitted from the metal plate with a maximum kinetic energy
of 1.1 eV when radiation of frequency 1.2 × 1015 Hz is incident on the plate.
Calculate, in eV, the work function of the metal.
[3 marks]
work function = eV
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
04.1 The mark scheme gives some guidance as to what statements The following statements are likely to be 6 4 x AO1
are expected to be seen in a 1- or 2-mark (L1), 3- or 4-mark (L2) present.
and 5- or 6-mark (L3) answer. Guidance provided in section 3.10 2 x AO2
Area A Loses its charge:
of the ‘Mark Scheme Instructions’ document should be used to
assist marking this question. • Emission of electrons from the surface (when
electromagnetic radiation is incident on plate)
(A)
Mark Criteria • Number of surplus electrons remaining on
plate decreases with time / (photo)electrons
All three areas (as outlined alongside) covered with at least carry away negative charge(B)
two aspects covered in some detail.
6 marks can be awarded even if there is an error and/or parts Area B Frequency:
of one aspect missing. • Minimum energy required /work function (C)
A fair attempt to analyse all three areas. If there are several • A photon must supply this energy in one
errors or missing parts then 5 marks should be awarded. interaction. (D)
Two areas successfully discussed, or one discussed and two • The energy of a photon is directly proportional
4 others covered partially. Whilst there will be gaps, there should to its frequency / E =hf (E)
only be an occasional error. • Minimum frequency is the threshold frequency
One area discussed and one discussed partially, or all three (F)
3 covered partially. There are likely to be several errors and Area C Intensity:
omissions in the discussion.
• Increased intensity (at same frequency)
Only one area discussed or makes a partial attempt at two results in more photons per second incident
areas. on plate. (G)
1 One of the three areas covered without significant error. • Must increase the number of photons per
second even if frequency increases. (H)
0 No relevant analysis.
• More electrons released from plate every
second so loses charge more rapidly. (I)
04.2 Use of E = hf or converts their photon energy in J to eV / For use of E = hf: 3 1 x AO1
converts 1.1 (eV) to 1.76 × 10−19 (J) -34 15 −19
6.63 x 10 x 1.2 x 10 / 7.956 × 10 (J) / 4.97 eV seen 2 x AO2
MP2:
rearrangement of terms is insufficient.
Correct substitution in eV or J with or without
rearrangement (condone one consistent POT error)
Use of hf = Φ + Ek(max)
Expect to see
(Φ =) 4.97 – 1.1 / Φ + 1.1= 4.97 /
(Φ =) 7.956 × 10−19 − 1.76 × 10−19 / (Φ =) 6.196 × 10−19
/ Φ + 1.76 × 10−19 = 7.956 × 10−19
Condone one error in either hf or Ek(max) or signs but must
be rearranged where Φ would be subject.
Common error seen in E ) = 6.875 x 1018
k(max
Examples:
(Φ =) 7.956 × 10−19− 1.1 (=– 1.1) /
(Φ =) 6.63 x 10-34 x 1.2 x 1015 − 1.1 /
(Φ =) 4.97 − 1.76× 10−19 (=4.97)
Condone error in
Accept a correctly rounded answer to 2 or more
significant figure.
Condone answer (with working seen) = 6.1 or 6.07
for 2 marks.
Φ = 3.9 eV (Calculator displays 3.8725)
( )
Total 9
How to answer it
Photoelectric Effect & Work Function Analysis
What this question tests
This question assesses your deep conceptual understanding of the photoelectric effect (threshold frequency, photon interaction, work function, and intensity) alongside quantitative application of the photoelectric equation hf = Φ + Ek(max) using energy unit conversions between Joules and electron-volts (eV).
Discuss how rate of loss of charge depends on frequency and intensity
💡 Key Knowledge (The 3 Core Areas)
- Area A (Charge Loss): Incident EM radiation knocks electrons out of the metal surface. Photoelectrons carry away negative charge, reducing the plate's surplus negative charge over time.
- Area B (Frequency): A minimum energy (work function) is required to release an electron. Each photon transfers its energy E = hf to a single electron in a one-to-one interaction. Hence, frequency must exceed a threshold frequency ( f₀ ).
- Area C (Intensity): Increasing intensity at a constant frequency increases the number of photons arriving per second, resulting in more electrons emitted per second, speeding up the rate of charge loss.
🧠 Exam Technique & Level Marking
This is a 6-mark extended-prose Level of Response question (LUR). Examiners look for structured coverage across all three bullet points:
- Level 3 (5-6 marks): All 3 areas covered with detail and clarity.
- Level 2 (3-4 marks): Two areas successfully discussed, or minor omissions.
- Level 1 (1-2 marks): Only one area covered or fragmented points.
❌ Common Errors
- Confusing intensity with frequency (e.g., claiming high intensity can cause emission when frequency is below the threshold).
- Failing to explicitly link the emission of negatively charged photoelectrons to the reduction of the plate's negative charge.
Calculate the Work Function in eV
Charged particles are emitted with a maximum kinetic energy of 1.1 eV when radiation of frequency 1.2 × 10¹⁵ Hz is incident on the plate. Calculate, in eV, the work function of the metal.
✅ Correct Answer
Work function (Φ) = 3.9 eV (Accept 3.87 to 3.9 eV)
📐 Step-by-Step Calculation
- Calculate incoming photon energy in Joules:
E = hf = (6.63 × 10⁻³⁴ J s) × (1.2 × 10¹⁵ Hz) = 7.956 × 10⁻¹⁹ J - Convert photon energy from Joules to eV:
E = (7.956 × 10⁻¹⁹ J) / (1.60 × 10⁻¹⁹ J eV⁻¹) = 4.9725 eV - Rearrange photoelectric equation for work function (Φ):
Φ = hf - Ek(max)
Φ = 4.9725 eV - 1.1 eV = 3.8725 eV - Round appropriately:
Φ = 3.9 eV (to 2 significant figures)
❌ Common Calculation Traps
- Unit mismatch: Subtracting Joules directly from electron-volts without converting both quantities to the same unit first.
- Power of 10 errors: Incorrectly inputting standard form numbers into calculators when finding hf .
Mark 2: Correct application/rearrangement of hf = Φ + Ek(max) .
Mark 3: Final correct answer of 3.9 eV (or 3.87 eV with working shown).
Topics
Physics · 3.2 Particles and radiation
Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.