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AS Daily A Level Physics question

2026-05-21 OCR A Quantum (M4) Module 4: Electrons, waves and photons — 4.2.2 Wave–particle duality (de Broglie wavelength, λ inversely proportional to momentum; dependence on accelerating potential)

In a school electron diffraction tube, electrons are accelerated from a heated filament through a potential difference of 150 V towards a thin graphite target. The potential difference is then increased to 600 V with everything else unchanged. Considering how momentum depends on the kinetic energy gained, what happens to the electrons’ de Broglie wavelength?

  1. A It increases by a factor of 2 (doubles).
  2. B It decreases by a factor of 2 (halves). (correct)
  3. C It decreases by a factor of 4.
  4. D It stays the same.

Answer

The correct answer is B.

Correct: B — It decreases by a factor of 2 (halves). For an electron gaining kinetic energy eV, KE ∝ V so momentum p ∝ √V, and wavelength is inversely proportional to p; raising V from 150 V to 600 V multiplies V by 4, so λ scales by 1/√4 = 1/2. A Increases (doubles) reverses the dependence and would require lower accelerating p.d., not a higher one. B Matches the inverse-square-root dependence: 600/150 = 4 gives √4 = 2, so λ halves. C A factor of 4 decrease assumes λ ∝ 1/V rather than 1/√V. D No change ignores that accelerating through a larger p.d. increases momentum and thus reduces wavelength.

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