AQA AS Level Physics Paper 2, June 2022: Question 12

1 mark · Medium difficulty · Multiple Choice

Calculate the energy difference between two higher-energy states X and Y of an atom given the wavelengths of photons produced by transitions from X and Y to the ground state.

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Question

An energy level diagram showing a ground state at the bottom and two higher-energy states labeled X and Y above it, with an arrow on the right pointing upwards and labeled energy. Text states that a transition from X to the ground state produces a photon of wavelength 147 nm, and a transition from Y to the ground state produces a photon of wavelength 160 nm. Below, multiple-choice options A, B, C, and D give different energy values in joules.
Question text

12 The diagram shows the ground state and two higher-energy states X and Y of an atom.

A transition from X to the ground state produces a photon of wavelength 147 nm.

A transition from Y to the ground state produces a photon of wavelength 160 nm.

What is the energy difference between X and Y?

[1 mark]

A 1.5 × 10−17 J

B 1.4 × 10−18 J

C 1.2 × 10−18 J

D 1.1 × 10−19 J

Mark scheme

Show the mark scheme A table showing the correct answer for question 12 is D (AO1), corresponding to 1.1 x 10^-19 J.

12 D (AO1) 1.1 × 10−19 J

How to answer it

Energy Level Transitions & Photon Wavelengths

AQA AS Level Physics • Quantum Physics

What this question tests

This question assesses your understanding of discrete energy levels in atoms, the relationship between photon energy and wavelength (hc / lambda), and how to apply the principle of conservation of energy to find energy level differences.

Question 12: Multiple Choice Solution

Detailed Breakdown and Step-by-Step Analysis

✅ Correct Answer

D ( 1.1 × 10⁻¹⁹ J )

Marks available: 1 mark (AO1)

💡 Key Knowledge

  • The energy of a emitted photon equals the difference in energy between two levels: ΔE = hc / λ
  • Planck constant ( h ) = 6.63 × 10⁻³⁴ J s
  • Speed of light ( c ) = 3.00 × 10⁸ m s⁻¹
  • Nanometres must be converted to metres using ×10⁻⁹ .

🧠 Exam Technique

Instead of calculating the absolute energy of state X and state Y separately (which introduces rounding errors and wastes valuable time), factor out hc and calculate the difference directly using: ΔE = hc ( (1 / λ_Y) - (1 / λ_X) ) or find both energies first to 3+ significant figures.

❌ Common Errors

  • Unit Neglect: Forgetting to convert nm to m (omitting 10⁻⁹ ).
  • Subtraction Order: Subtracting wavelengths directly ( 160 - 147 ) instead of calculating energies and finding the difference. Wavelengths are inversely proportional to energy!

📐 Step-by-Step Calculation

Step 1: Find the energy for the transition from state X to the ground state ( λ = 147 nm )

E_X = hc / λ_X = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (147 × 10⁻⁹) = 1.353 × 10⁻¹⁸ J

Step 2: Find the energy for the transition from state Y to the ground state ( λ = 160 nm )

E_Y = hc / λ_Y = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (160 × 10⁻⁹) = 1.243 × 10⁻¹⁸ J

Step 3: Calculate the energy difference between X and Y

ΔE = E_X - E_Y = (1.353 × 10⁻¹⁸) - (1.243 × 10⁻¹⁸) = 1.10 × 10⁻¹⁹ J

This matches option D.

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.