Edexcel IAL Physics: Waves and the Particle Nature of Light — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Waves and the Particle Nature of Light revision notes

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Section A

1. Distinguish between transverse and longitudinal waves, giving one example of each. [3]

2. State the wave equation and define each term. [2]

Section B

3. Light of wavelength 590 nm passes through a diffraction grating with 500 lines per mm.

(a) Calculate the grating spacing d. [2] (b) Calculate the angle of the first-order maximum. [3] (c) Determine the highest order that can be observed. [3]

4. In a double-slit experiment, explain why an interference pattern is only seen if the two sources are coherent, and define coherence. [3]

5. The work function of a metal is 2.30 eV.

(a) Calculate the threshold frequency. [3] (b) Light of frequency 8.0 × 10¹⁴ Hz is shone on the metal. Calculate the maximum kinetic energy of the emitted electrons in joules. [3] (c) Explain why increasing the intensity of light below the threshold frequency produces no emission, and why this supports the photon model. [4]

6. Calculate the de Broglie wavelength of an electron travelling at 2.0 × 10⁶ m s⁻¹. (m = 9.11 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s) [3]

7. State the condition for total internal reflection to occur, and calculate the critical angle for a medium of refractive index 1.50. [3]

8. Describe how a standing wave forms, and state the two equations for its speed on a stretched string and for radiation intensity. [3]

9. An electron transition between two atomic energy levels releases 3.0 × 10⁻¹⁹ J of energy. Calculate the frequency of the emitted radiation. (h = 6.63 × 10⁻³⁴ J s) [2]

10. Explain how Huygens’ construction is used to describe diffraction of a wave passing through a gap. [2]

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Answers

1. Transverse — the oscillations are perpendicular to the direction of energy transfer, e.g. light or water waves [1] [1]. Longitudinal — the oscillations are parallel to the direction of energy transfer, e.g. sound [1].

2. v = fλ [1], where v is the wave speed (m s⁻¹), f the frequency (Hz) and λ the wavelength (m) [1].

3. (a) d = 1 ÷ (500 × 10³) m [1] = 2.00 × 10⁻⁶ m [1]. (b) d sin θ = nλ [1]; sin θ = (1 × 590 × 10⁻⁹) ÷ 2.00 × 10⁻⁶ = 0.295 [1]; θ = 17.2° [1]. (c) Maximum when sin θ = 1 [1]; n = d ÷ λ = 2.00 × 10⁻⁶ ÷ 590 × 10⁻⁹ = 3.39 [1]; so the highest order is n = 3 [1].

4. Coherent sources have a constant phase difference [1] and the same frequency [1]. If the phase difference varied, the positions of constructive and destructive interference would change continuously and the pattern would not be stable enough to observe [1].

5. (a) φ = 2.30 × 1.60 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J [1]; f₀ = φ ÷ h = 3.68 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ [1] = 5.55 × 10¹⁴ Hz [1]. (b) E = hf = 6.63 × 10⁻³⁴ × 8.0 × 10¹⁴ = 5.30 × 10⁻¹⁹ J [1]; KE_max = E − φ = 5.30 × 10⁻¹⁹ − 3.68 × 10⁻¹⁹ [1] = 1.6 × 10⁻¹⁹ J [1]. (c) Each electron absorbs one photon only, and a photon’s energy depends solely on frequency [1]. Below the threshold, no single photon carries enough energy to overcome the work function [1], so increasing intensity merely delivers more photons, each still too weak [1]. A wave model would predict that energy accumulates until emission occurs, which is not observed — so light must be quantised into photons [1].

6. λ = h ÷ mv [1] = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 2.0 × 10⁶) [1] = 3.6 × 10⁻¹⁰ m [1].

7. Total internal reflection occurs when light travelling in the denser medium hits the boundary at an angle greater than the critical angle [1]. sinC = 1/n = 1/1.50 = 0.667 [1] → C = 41.8° [1].

8. A standing wave forms when two waves of the same frequency travelling in opposite directions superpose [1], producing fixed nodes (no displacement) and antinodes (maximum displacement) [1]. Speed on a string: v = √(T/µ), where T is tension and µ is mass per unit length; intensity of radiation: I = P/A, power per unit area [1].

9. f = E ÷ h = 3.0 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ [1] = 4.5 × 10¹⁴ Hz [1].

10. Every point on a wavefront is treated as a source of secondary wavelets [1]; at a narrow gap, the new wavefront formed from these secondary sources spreads out into the region beyond the gap, explaining the observed spreading of the wave [1].

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Where marks are usually lost

• Forgetting to convert lines per mm into a spacing in metres.
• Using eV instead of joules in the photoelectric equation.
• Saying higher intensity means higher energy photons.
• Rounding the maximum order up instead of down.
• Applying total internal reflection at the boundary from the less dense to the denser medium — it only occurs travelling from denser to less dense.
• Confusing a node (no displacement) with an antinode (maximum displacement) in a standing wave.