AQA A-Level Physics Paper 1, June 2025: Question 12

1 mark · Medium difficulty · Multiple Choice

Identify which change results in a greater stopping potential in a photoelectric effect experiment.

Practise this question

Question

Question 12 asks: 'In a photoelectric effect experiment, monochromatic light is incident on a metal surface and the stopping potential is measured. Which change results in a greater stopping potential?' Options are: A: using a metal that has a greater work function, B: using a light source that emits more photons per second, C: using a light source that emits light of a shorter wavelength, D: using a metal surface that has a positive charge.
Question text

12 In a photoelectric effect experiment, monochromatic light is incident on a metal surface

and the stopping potential is measured.

Which change results in a greater stopping potential?

[1 mark]

A using a metal that has a greater work function

B using a light source that emits more photons per second

C using a light source that emits light of a shorter wavelength

D using a metal surface that has a positive charge

Mark scheme

Show the mark scheme Mark scheme for question 12 shows the correct answer is C, 'using a light source that emits light of a shorter wavelength', assessed under AO1.

12 C using a light source that emits light of a shorter wavelength AO1

How to answer it

Photoelectric Effect: Factors Affecting Stopping Potential

📋 What this question tests

This question assesses your understanding of Einstein's photoelectric equation and the definition of stopping potential (Vs). Specifically, it tests your ability to distinguish between variables that affect the energy of individual emitted electrons (frequency, wavelength, work function) versus variables that affect only the rate of emission (intensity, photon flux).

Question 12 Breakdown

Multiple Choice Question (1 Mark)

✅ Correct Answer

C: using a light source that emits light of a shorter wavelength

Mark Scheme: 1 mark (AO1)
Requires selection of option C only.

💡 Key Knowledge

  • Einstein's Photoelectric Equation:
    hf = Φ + Ek(max)
  • Photon Energy & Wavelength:
    Since c = fλ , photon energy is E = hc/λ .
  • Stopping Potential Relation:
    The stopping potential Vs halts the fastest photoelectrons:
    eVs = Ek(max) = (hc / λ) − Φ
  • Decreasing wavelength λ increases photon energy, which directly increases Ek(max) and therefore increases Vs .

📐 Step-by-Step Option Elimination

  1. Option A — Greater work function (Φ):
    Rearranging the equation: eVs = hf − Φ . If Φ increases while incident light remains unchanged, Ek(max) decreases. Therefore, stopping potential Vs decreases. (Incorrect)
  2. Option B — Emits more photons per second:
    Photons per second corresponds to the intensity of the beam. A higher intensity increases the number of electrons emitted per second (photocurrent), but does not change the kinetic energy of individual electrons. Stopping potential remains unchanged. (Incorrect)
  3. Option C — Shorter wavelength (λ):
    Expressing stopping potential in terms of wavelength:
    Vs = (hc / eλ) − (Φ / e)
    As λ decreases, hc / λ increases. Since work function Φ and fundamental constants h, c, e are fixed, Vs must increase. (Correct)
  4. Option D — Metal surface has a positive charge:
    Adding an initial static positive charge to the emitter surface exerts an electrostatic attractive force holding electrons in, effectively raising the barrier to remove electrons from the bulk metal and reducing their free kinetic energy. It does not yield a higher stopping potential in the measuring circuit. (Incorrect)

🧠 Exam Technique

  • Write the governing equation first: Whenever you see "stopping potential", immediately write eVs = hf − Φ or eVs = hc/λ − Φ in the margin.
  • Isolate the variable of interest: Rearrange for Vs to clearly see whether the parameter is inversely or directly proportional.
  • Decouple intensity and energy: In modern physics MCQs, always remind yourself: Frequency / Wavelength controls Energy; Intensity controls Count / Current.

❌ Common Examiner Traps

  • Confusing Intensity with Energy (Option B): The most common student error is selecting Option B. Remember, wave theory predicts intensity increases kinetic energy, but the photon model proves it only increases the emission rate.
  • Wavelength vs. Frequency confusion: Students often forget that wavelength is inversely proportional to frequency ( f = c/λ ). A shorter wavelength means a higher frequency and thus higher photon energy.
  • Misinterpreting Work Function (Option A): A higher work function requires more energy just to liberate the electron, leaving less surplus energy for Ek(max) .

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.