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A2 Daily A Level Physics question

2026-03-22 OCR A Quantum II: photoelectric effect; stopping potential; work function; de Broglie wavelength (intro) OCR A Module 4.5.1 Photoelectric effect: threshold frequency, stopping potential, work function

In a vacuum photocell, a clean sodium cathode (work function 2.3 eV) is illuminated with monochromatic 300 nm light, giving a stopping potential about 1.8 V. The lamp is replaced with a 600 nm source of the same intensity. What is the new stopping potential? Support your answer with a short numerical estimate.

  1. A 0 V, because the photon energy at 600 nm (~2.1 eV) is below the 2.3 eV work function so no electrons are emitted. (correct)
  2. B About 0.9 V, roughly half, since doubling wavelength halves frequency and thus halves the stopping potential.
  3. C About 1.8 V, unchanged, because the intensity stayed the same so the electron energy is unchanged.
  4. D A small negative value (around −0.9 V), because longer-wavelength light gives slower electrons that need reverse bias to be collected.

Answer

The correct answer is A.

Correct: A — 0 V, because the photon energy at 600 nm (~2.1 eV) is below the 2.3 eV work function so no electrons are emitted. A states there are no photoelectrons, so the stopping potential is zero; 1240/600 ≈ 2.07 eV < 2.3 eV confirms this. B wrongly assumes stopping potential scales directly with frequency from zero, ignoring the threshold; below threshold there is no emission, not a halved value. C confuses intensity with photon energy; intensity affects photocurrent, not the maximum kinetic energy or stopping potential. D misunderstands sign and collection; with no photoelectrons there is no current to stop, so no finite or negative stopping potential applies.