AQA AS Level Physics Paper 1, June 2025: Question 7

8 marks · Medium difficulty · Extended Answer

Define threshold frequency, calculate stopping potential and work function for magnesium, and sketch and explain graphs of maximum kinetic energy against frequency for metals with different work functions.

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Question

Exam question with four parts, numbered 07.1 to 07.4. Part 07.1 asks to state the meaning of threshold frequency (2 marks). Part 07.2 gives the maximum kinetic energy of photoelectrons from magnesium as 5.47 × 10⁻¹⁹ J and asks to calculate the stopping potential Vs (1 mark). Part 07.3 gives an incident frequency of 1710 THz and asks to calculate the work function of magnesium (2 marks). Part 07.4 shows Figure 11, a graph of maximum kinetic energy Ek(max) versus frequency f with a straight-line graph for magnesium having positive gradient and negative y-intercept. Students are asked to sketch and label lines for metal X and metal Y, where work function of X > magnesium > Y, and explain their reasoning (3 marks).
Question text

07.1 State what is meant by the threshold frequency of radiation in the

photoelectric effect.

[2 marks]

When magnesium is illuminated with monochromatic electromagnetic radiation,

photoelectrons are emitted.

For this radiation, the maximum kinetic energy of the photoelectrons is 5.47 × 10−19 J

and the stopping potential is Vs.

07.2 Calculate Vs.

[1 mark]

Vs = V

07.3 The radiation incident on the magnesium has a frequency of 1710 THz.

Calculate the work function of the magnesium.

[2 marks]

work function = J

07.4 The frequency f of the electromagnetic radiation is varied.

Figure 11 shows the variation of Ek(max) with f for magnesium.

*23* Figure 11

Metal X and metal Y are illuminated by radiation with the same range of frequencies

as used for the magnesium.

Sketch, on Figure 11, graphs to show how Ek(max) varies with f for X and for Y.

Label your graphs X and Y.

Go on to explain your reasoning.

work function of X > work function of magnesium > work function of Y

[3 marks]

Mark scheme

Show the mark scheme Mark scheme table for question 07: 07.1 awards 2 marks for stating minimum photon frequency needed for photoelectrons to be emitted. 07.2 awards 1 mark for stopping potential of 3.4(2) V. 07.3 awards 1 mark for converting 1710 THz to 1.71 × 10¹⁵ Hz or using hf = phi + Ek(max), and 1 mark for 5.9 × 10⁻¹⁹ J. 07.4 awards 3 marks: 1 mark for parallel lines with identical gradient, 1 mark for correct frequency intercepts (X to the right, Y to the left of magnesium), and 1 mark for explanation that the gradient equals Planck's constant h or intercept depends on work function / threshold frequency.

Question Answers Additional Comments/Guidance Mark AO

07.1 Idea that it is the 2 AO1

minimum (photon) frequency needed

for (photo)electron(s) to be emitted

07.2 3.4(2) (V) 1 AO1

07.3 Any one from 2 AO2

1st bullet point

• converts 1710 THz to 1.71 × 1015 (Hz)

Accept any number (without prefix) equivalent

to 1.71 × 1015 Hz

• use of h × 𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑓𝑓 = ∅ + Ek(max)

Accept answer in range 5.8 × 10−19 to

5.9 × 10−19 (J)

5.9 × 10−19

Mp3 is for correct reasoning consistent with

07.4 Gradients of the lines are the same as gradient of 3 AO2

correct MP1.

magnesium’s line 29

Condone MP3 where an attempt made to

X line intersects with the frequency axis to the right of the parallel lines for X and Y

magnesium line’s intersection point AND Y line intersects with OR

the frequency axis to the left of the magnesium line’s

intersection point (and still negative y-intercept) MP3 is for correct reasoning consistent with

correct MP2

Because:

gradient is equal to Planck constant, h, (which has a constant

value) In support of MP1

OR

intercept on Ek(max) is (–) the work function

In support of MP2

OR

lower work function has a lower threshold frequency

In support of MP2

OR

Y’s photoelectrons will have a greater Ek(max) for any given

frequency as it has a lower work function. In support of MP2

If no other mark given,

Max 1 for intercepts correct on Ek max axis

and correct support seen in MP3.

Total 8

How to answer it

Photoelectric Effect: Work Function, Stopping Potential & Graphs

Topic Summary

What this question tests:

  • Precise Definitions: Threshold frequency and photoelectric emission conditions.
  • Stopping Potential: The relationship between stopping potential (Vs) and maximum photoelectron kinetic energy (Ek(max)).
  • Einstein's Photoelectric Equation: Converting SI prefixes ( THz ) and solving for the work function (Φ).
  • Graphical Analysis: The meaning of gradient (Planck constant, h ), x-intercept (threshold frequency, f₀ ), and y-intercept (−Φ) on an Ek(max) vs. f graph.

Question 07.1

Definition of Threshold Frequency (2 Marks)

✅ Correct Answer

The minimum frequency of incident radiation/photons [1 Mark]
needed to emit (photo)electrons from the surface of a metal [1 Mark].

Mark Scheme Breakdown:
• MP1: Minimum (photon) frequency needed.
• MP2: For (photo)electron(s) to be emitted / released.

🧠 Exam Technique

Always state two key parts for threshold frequency:

  1. It is an extreme value: say minimum (never "the frequency needed").
  2. The consequence: to release/emit electrons from the metal surface.

❌ Common Errors

  • Forgetting the word "minimum" (costs Mark 1 instantly).
  • Saying "emit photons" instead of electrons.
  • Confusing frequency with energy (giving the definition of work function instead).

💡 Key Knowledge

Because energy is delivered in discrete quanta (E = hf), an electron can only absorb energy from a single photon. If f < f₀ , the photon energy is less than the work function Φ, so no electrons can escape regardless of beam intensity.

Question 07.2

Calculating Stopping Potential Vs (1 Mark)

📐 Step-by-Step Calculation

  1. State the formula:
    Ek(max) = e × Vs
  2. Rearrange for stopping potential:
    Vs = Ek(max) / e
  3. Substitute values:
    Vs = (5.47 × 10⁻¹⁹ J) / (1.60 × 10⁻¹⁹ C)
  4. Calculate final answer:
    Vs = 3.42 V (accepts 3.4 V or 3.42 V)

✅ Correct Answer

Vs = 3.42 V (or 3.4 V)

Mark Scheme: 3.4(2) (V) [1 Mark, AO1]

❌ Common Errors

  • Multiplying by the elementary charge e instead of dividing.
  • Incorrect value for e (always check the data booklet: 1.60 × 10⁻¹⁹ C ).

Question 07.3

Calculating the Work Function of Magnesium (2 Marks)

📐 Step-by-Step Calculation

  1. Convert prefix THz to Hz:
    1 THz = 10¹² Hz
    f = 1710 THz = 1710 × 10¹² Hz = 1.71 × 10¹⁵ Hz [1 Mark]
  2. State Einstein's photoelectric equation:
    h f = Φ + Ek(max)
  3. Rearrange for work function (Φ):
    Φ = h f − Ek(max)
  4. Substitute constants and values:
    h = 6.63 × 10⁻³⁴ J s
    h f = (6.63 × 10⁻³⁴)(1.71 × 10¹⁵) = 1.134 × 10⁻¹⁸ J
    Φ = 1.134 × 10⁻¹⁸ − 5.47 × 10⁻¹⁹
  5. Evaluate:
    Φ = 5.87 × 10⁻¹⁹ J (or 5.9 × 10⁻¹⁹ J) [1 Mark]

✅ Correct Answer

work function = 5.9 × 10⁻¹⁹ J
(Acceptable range: 5.8 × 10⁻¹⁹ J to 5.9 × 10⁻¹⁹ J)

Mark Scheme Breakdown:
• MP1: Frequency conversion to 1.71 × 10¹⁵ Hz OR correct use of h × f = Φ + Ek(max) .
• MP2: Final answer in range 5.8 × 10⁻¹⁹ to 5.9 × 10⁻¹⁹ J .

❌ Common Calculation Traps

  • Tera (T) prefix error: Confusing tera (10¹²) with giga (10⁹) or mega (10⁶).
  • Sign error: Adding Ek(max) to hf rather than subtracting it. Remember: the photon energy is shared between escaping (Φ) and moving ( Ek(max) ).

Question 07.4

Graph Sketching and Scientific Reasoning (3 Marks)

Context: ΦX > ΦMg > ΦY

✅ Correct Answer & Requirements

  • Line Gradients: Both lines X and Y must be straight and parallel to the magnesium line (same gradient) [1 Mark].
  • Axis Intercepts:
    • Line X intersects the frequency axis to the right of magnesium.
    • Line Y intersects the frequency axis to the left of magnesium (and still has a negative y-intercept) [1 Mark].
  • Scientific Reasoning: [1 Mark for any valid point consistent with sketch]:
    • The gradient equals Planck's constant ( h ), which is constant for all metals.
    • f₀ = Φ / h , so a higher work function means a higher threshold frequency.
    • The y-intercept equals −Φ.

🧠 Diagram Description for Revision

What your sketch must look like on Figure 11:
1. Draw line X completely parallel to the magnesium line, shifted to the right. Its x-intercept ( f₀,X ) is higher than magnesium's, and its y-intercept is more negative.
2. Draw line Y completely parallel to the magnesium line, shifted to the left (between 0 and magnesium's intercept). Its x-intercept ( f₀,Y ) is lower than magnesium's.
Examiner Guidance: If lines are not parallel, MP3 can still be awarded if the explanation correctly references why they should be parallel or why intercepts differ.

💡 Linking Equation to Graph: y = mx + c

Rearrange Einstein's equation to match a straight line:

Ek(max) = hf − Φ  ↔  y = mx + c

  • y = Ek(max)
  • gradient (m) = h (universal constant, lines MUST be parallel)
  • x = f
  • y-intercept (c) = −Φ (negative work function)
  • x-intercept (y = 0) = f₀ = Φ / h (threshold frequency)

❌ Common Errors

  • Drawing lines that converge or cross the magnesium line (different gradients).
  • Swapping X and Y: remember, larger work function (ΦX) requires a higher threshold frequency, shifting the line to the right.
  • Failing to provide reasoning in the written lines below the graph.

Topics

Physics · 3.1 Measurements and their errors · 3.2 Particles and radiation

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