AQA A-Level Physics Paper 1, November 2020: Question 2

✅ Correct Answer

• Photon energy is directly proportional to frequency (E = hf).
• A minimum photon energy (the work function) is required to release a photoelectron; thus a minimum (threshold) frequency is needed.

💡 Key Knowledge

• MP1: Links photon energy to frequency ( E = hf ).
• MP2: Explains the concept of work function and minimum energy removal.

❌ Common Errors

• Forgetting to mention photons (mentioning general "waves" or "energy levels" instead loses marks).
• Stating "threshold frequency" without explaining why it is required (must link to minimum energy/work function).

✅ Correct Answer

1.9 × 10¹⁴ (electrons per second)

📐 Step-by-Step Calculation

1. Identify Maximum Current: From Figure 2, the saturation current (maximum current) I = 30 μA = 30 × 10⁻⁶ A .
2. Recall Current Definition: Current is charge per unit time ( I = Q / t = n e / t ), where e = 1.6 × 10⁻¹⁹ C .
3. Rearrange for Rate: Number of electrons per second ( n / t ) = I / e .
4. Evaluate: (30 × 10⁻⁶) / (1.6 × 10⁻¹⁹) = 1.875 × 10¹⁴ ≈ 1.9 × 10¹⁴ s⁻¹ .

❌ Common Errors

• Forgetting the micro prefix ( 10⁻⁶ ) for the current value from the graph.
• Failing to use correct significant figures (the graph scale supports 2 sig figs).

✅ Correct Answer

• The number of photoelectrons released per second depends entirely on the intensity of the em radiation (number of incident photons), not the potential difference.
• A constant (saturation) current is reached when all photoelectrons released each second are successfully collected at the anode due to the positive potential.

🧠 Exam Technique

• Explicitly contrast light intensity with potential difference to show you understand that increasing voltage past a certain point cannot pull out more electrons than the light source is liberating.

✅ Correct Answer

• Photoelectrons are emitted with a range of kinetic energies from zero up to a maximum value.
• When the potential difference V is negative, photoelectrons lose kinetic energy / do work against the electric field as they travel to the anode.
• As V becomes increasingly negative, fewer photoelectrons possess enough initial kinetic energy to overcome the retarding potential, reducing the detected current.

💡 Key Knowledge

• MP1: Mention range of KE.
• MP2: Explain energy conversion (KE to electrical potential energy) under negative V .
• MP3: Link fewer electrons having sufficient threshold energy to cross over.

✅ Correct Answer

• Stopping potential is directly related to the maximum kinetic energy of photoelectrons ( KE_max = e V_s ).
• Since KE_max = hf - work function , reducing the work function while keeping incident frequency constant increases the maximum kinetic energy.
• A higher maximum kinetic energy means a larger negative retarding potential is required to stop the fastest electrons, so the stopping potential increases (becomes more negative).

🧠 Exam Technique

• Quote the relationship clearly using Einstein's equation rearranged for stopping potential: V_s = (hf - φ) / e . Show step-by-step how decreasing φ affects V_s .

❌ Common Errors

• Omitting the word "maximum" when discussing kinetic energy (marks are routinely withheld if max KE is not specified, as electrons have a range of energies).