AQA AS Level Physics Paper 2, June 2025: Question 10

1 mark · Medium difficulty · Multiple Choice

Calculate the frequency of a photon emitted during an electron transition to the ground state from an excited energy level.

Practise this question

Question

Energy level diagram showing four levels labelled 0.0 eV, -1.6 eV, -3.7 eV, and -5.5 eV, with a note that the ground state is not shown. The question states that a transition from -1.6 eV to the ground state emits a photon of frequency 2.1 × 10^15 Hz, and asks for the frequency emitted when an electron falls from -3.7 eV to the ground state. Four options are given: A: 4.3 × 10^14 Hz, B: 8.9 × 10^14 Hz, C: 1.2 × 10^15 Hz, D: 1.6 × 10^15 Hz.

Mark scheme

Show the mark scheme Mark scheme table indicating for question 10 the correct option is D with value 1.6 × 10^15 Hz, mapped to assessment objective AO2.

How to answer it

Atomic Energy Levels & Photon Emission Frequency

📋 What this question tests

This question assesses your understanding of atomic excitation and de-excitation, calculating the energy of emitted photons using ΔE = hf , converting between electron-volts (eV) and Joules (J), and deducing energy values when an unlabelled reference level (the ground state) is omitted from the diagram.

Question 10 Breakdown

Multiple Choice: Finding the transition frequency to an unshown ground state

✅ Correct Answer

D: 1.6 × 10¹⁵ Hz

When an electron drops from -3.7 eV to the ground state, the emitted photon has an energy 2.1 eV lower than the drop from -1.6 eV. This corresponds to a reduction in frequency of 0.51 × 10¹⁵ Hz, leaving 1.59 × 10¹⁵ Hz ≈ 1.6 × 10¹⁵ Hz.

💡 Key Knowledge

  • Photon energy equation: ΔE = E₂ - E₁ = hf
  • Planck's constant: h = 6.63 × 10⁻³⁴ J s
  • Unit conversion: 1 eV = 1.60 × 10⁻¹⁹ J
  • Conservation of energy: The frequency emitted depends purely on the magnitude of the difference between the initial excited level and the final level (ground state).

📐 Step-by-Step Calculation

Method 1: Direct Energy Level Deduction (Most common approach)

  1. Calculate the energy of the first emitted photon (in Joules):
    E₁ = hf₁ = (6.63 × 10⁻³⁴ J s) × (2.1 × 10¹⁵ Hz) = 1.392 × 10⁻¹⁸ J
  2. Convert this photon energy into electron-volts (eV):
    E₁ = (1.392 × 10⁻¹⁸ J) / (1.60 × 10⁻¹⁹ J eV⁻¹) = 8.70 eV
  3. Determine the ground state energy level (E_ground):
    Since the electron falls from -1.6 eV down to E_ground :
    ΔE₁ = -1.6 eV - E_ground = 8.70 eV
    E_ground = -1.6 eV - 8.70 eV = -10.30 eV
  4. Find the energy transition from -3.7 eV to the ground state:
    ΔE₂ = -3.7 eV - (-10.30 eV) = 6.60 eV
  5. Convert ΔE₂ to Joules and calculate frequency f₂:
    ΔE₂ = 6.60 × 1.60 × 10⁻¹⁹ J = 1.056 × 10⁻¹⁸ J
    f₂ = ΔE₂ / h = (1.056 × 10⁻¹⁸ J) / (6.63 × 10⁻³⁴ J s) = 1.59 × 10¹⁵ Hz ≈ 1.6 × 10¹⁵ Hz

Method 2: Fast Difference Method (Examiner shortcut)

Notice that the -3.7 eV level is |-1.6 - (-3.7)| = 2.1 eV lower than the -1.6 eV level. Therefore, the photon emitted has 2.1 eV less energy:

  • Frequency difference: Δf = ΔE / h = (2.1 × 1.60 × 10⁻¹⁹ J) / (6.63 × 10⁻³⁴ J s) = 5.07 × 10¹⁴ Hz = 0.51 × 10¹⁵ Hz
  • f₂ = f₁ - Δf = (2.10 - 0.51) × 10¹⁵ Hz = 1.59 × 10¹⁵ Hz ≈ 1.6 × 10¹⁵ Hz

❌ Common Traps & Wrong Distractors

  • Trap Option A (4.3 × 10¹⁴ Hz): Students assume the lowest level shown on the diagram (-5.5 eV) is the ground state! Falling from -3.7 eV to -5.5 eV gives ΔE = 1.8 eV , which equals 4.34 × 10¹⁴ Hz . The question explicitly states "The ground state is not shown."
  • Trap Option B (8.9 × 10¹⁴ Hz): Forgetting negative signs or using 3.7 eV directly as the transition energy ( 3.7 eV / h ≈ 8.9 × 10¹⁴ Hz ), assuming the ground state is 0.0 eV.
  • Unit conversion error: Failing to convert eV to Joules when dividing by Planck's constant h .

🧠 Exam Technique & Insight

  • Read the diagram notes carefully: Examiners frequently state when energy levels are missing. Underline words like "not shown" immediately.
  • Sanity check your frequency: The fall from -3.7 eV to ground is a smaller drop than from -1.6 eV to ground. Therefore, f₂ must be less than 2.1 × 10¹⁵ Hz . (All choices are less, but this confirms the logic).
  • Use powers of 10 consistently: Express frequencies in × 10¹⁵ Hz to make comparisons and mental approximations fast during a timed paper.
Mark Scheme Reference: Question 10 | Answer: D (1.6 × 10¹⁵ Hz) | Assessment Objective: AO2 (Application of knowledge in a theoretical context) [1 mark]

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.