AQA A-Level Physics Paper 1, June 2022: Question 17

1 mark · Medium difficulty · Multiple Choice

Identify which particle has the smallest de Broglie wavelength among four given options with different particle types and speeds.

Practise this question

Question

Multiple choice question 17 asking which particle has the smallest de Broglie wavelength, worth 1 mark. The options are A: an electron moving at 4 × 10^3 m s^-1, B: a proton moving at 4 × 10^3 m s^-1, C: an electron moving at 8 × 10^5 m s^-1, and D: a proton moving at 8 × 10^5 m s^-1, each with a corresponding selection box.
Question text

17 Which particle has the smallest de Broglie wavelength?

[1 mark]

A an electron moving at 4 × 103 m s−1

B a proton moving at 4 × 103 m s−1

C an electron moving at 8 × 105 m s−1

D a proton moving at 8 × 105 m s−1

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 17 is D (AO2), which corresponds to a proton moving at 8 × 10^5 m s^-1.

17 D (AO2) a proton moving at 8 × 105 m s−1

How to answer it

De Broglie Wavelength Comparison

Question 17 • Multiple Choice • 1 Mark

What this question tests

This question assesses your understanding of particle-wave duality, specifically the de Broglie wavelength formula ( lambda = h / mv ). You must demonstrate how changes in mass ( m ) and velocity ( v ) inversely affect the de Broglie wavelength without necessarily needing to perform full calculations for every option.

Question 17: Choosing the Smallest Wavelength

✅ Correct Answer

D (a proton moving at 8 × 10⁵ m s⁻¹)

Mark Awarded: 1 / 1 (AO2 - Application)

💡 Key Knowledge

  • De Broglie Equation: lambda = h / p = h / (mv)
  • Mass Relationship: A proton is much more massive than an electron ( m_proton >> m_electron ).
  • Proportionality: Wavelength is inversely proportional to both mass and velocity ( lambda ∝ 1 / (mv) ).

🧠 Exam Technique

To find the smallest wavelength, you need the largest denominator ( mv or momentum) in the fraction h / mv . Look for the combination of the heaviest particle (proton) and the highest velocity (8 × 10⁵ m s⁻¹).

❌ Common Errors

  • Confusing inverse proportionality and picking the lightest particle moving slowest.
  • Wasting time calculating exact numerical values for all four options when a proportional reasoning approach takes seconds.

📐 Step-by-Step Breakdown (Why D is correct)

  1. Identify the constants and variables: Planck's constant ( h ) is constant. Therefore, lambda depends entirely on momentum ( p = mv ).
  2. Compare masses: Protons have a significantly larger mass than electrons (approx. 1836 times larger). Option B and D involve protons; A and C involve electrons. Automatically, options A and B can be ruled out because they have lower velocities and smaller masses compared to options C and D.
  3. Compare high-velocity options (C vs D): Option C is an electron at 8 × 10⁵ m s⁻¹. Option D is a proton at 8 × 10⁵ m s⁻¹. Since the proton has a much larger mass and the same high velocity, its momentum ( mv ) is the largest among all choices.
  4. Conclusion: Maximum momentum results in the minimum (smallest) de Broglie wavelength, making D the correct choice.

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.