AQA A-Level Physics Paper 1, June 2024: Question 11

1 mark · Medium difficulty · Multiple Choice

Calculate the ratio of wavelengths of photons emitted from two different energy level transitions in an atom.

Practise this question

Question

Multiple choice question showing a diagram of three energy levels at -1.0 eV, -6.0 eV, and -21.0 eV. Energy change E1 is between -1.0 eV and -6.0 eV, emitting wavelength lambda 1, and energy change E2 is between -6.0 eV and -21.0 eV, emitting wavelength lambda 2. Four options A, B, C, and D are given for the ratio lambda 1 over lambda 2.
Question text

11 Three energy levels for an atom are shown.

Energy change E1 leads to the emission of a photon of wavelength λ1.

Energy change E2 leads to the emission of a photon of wavelength λ2.

λ1

What is ?

λ2

[1 mark]

A

B

C 3

D 4

Mark scheme

Show the mark scheme The mark scheme table indicates the correct answer is C, with a value of 3.

11 C 3 AO1

How to answer it

Energy Levels and Photon Wavelengths

What this question tests

This question assesses your understanding of atomic energy transitions, the relationship between energy change and photon frequency/wavelength (delta E = hf = hc / lambda), and proportional reasoning under timed multiple-choice conditions.

Question 1.1

Multiple Choice Answer Analysis

✅ Correct Answer: C (3)

Option C is correct because wavelength is inversely proportional to energy change. Transition 2 has three times the energy change of transition 1, meaning its emitted photon has one-third of the wavelength, resulting in a ratio lambda_1 / lambda_2 equal to 3.

💡 Key Knowledge

  • Energy transition equation: delta E = E_initial - E_final = hf = hc / lambda.
  • Inverses: Because E is inversely proportional to lambda (lambda = hc / delta E), a larger energy change yields a shorter wavelength.
  • Unit consistency: Since we are finding a ratio, energy units (eV) do not even need to be converted to joules, saving crucial exam time!

🧠 Exam Technique

  • Do not waste time converting electron-volts (eV) into joules (J) when calculating a simple algebraic ratio. The conversion factors (charge of an electron, Planck's constant, speed of light) will completely cancel out.
  • Always double-check whether the question asks for lambda_1 / lambda_2 or lambda_2 / lambda_1 to avoid inverting your final fraction by accident.

❌ Common Errors

  • Inverting the ratio: Many students correctly find the energy values but mistakenly calculate lambda_2 / lambda_1, selecting Option B (1/3) instead of C.
  • Arithmetic slips with negative numbers: Subtracting negative energy levels incorrectly (e.g., mixing up -6.0 eV and -21.0 eV).

📐 Step-by-Step Calculation

  1. Find energy change E_1:
    delta E_1 = |-1.0 eV - (-6.0 eV)| = 5.0 eV
  2. Find energy change E_2:
    delta E_2 = |-6.0 eV - (-21.0 eV)| = 15.0 eV
  3. Set up the relationship with wavelength:
    Since lambda = hc / delta E, wavelength is inversely proportional to delta E. Therefore:
    lambda_1 / lambda_2 = delta E_2 / delta E_1
  4. Calculate the final ratio:
    lambda_1 / lambda_2 = 15.0 eV / 5.0 eV = 3
Mark Allocation: [1 mark] awarded for selecting option C. Tested under AO1 (Demonstrate knowledge and understanding of scientific ideas).

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.