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A2 Daily A Level Physics question

2026-01-31 OCR A Quantum I: intensity vs photon rate OCR-A 4.5 Quantum physics: photon model of light OCR-A 4.5.1 Photons: energy of a photon (E ∝ f), intensity and photon number

A lab source can emit monochromatic red light of wavelength 660 nm or blue light of 440 nm. The output is adjusted so the intensity at a photon-counting detector is identical in both cases. Which statement must be true when switching from the red to the blue setting?

  1. A The photon count rate increases by about 1.5× because the shorter wavelength provides more photons at the same intensity.
  2. B The photon count rate stays the same because the intensity at the detector is unchanged.
  3. C The photon count rate decreases to about 2/3 of the red rate because each blue photon carries about 1.5× more energy. (correct)
  4. D The photon count rate decreases to about 1/2 of the red rate because the wavelength is much smaller.

Answer

The correct answer is C.

Correct: C — The photon count rate decreases to about 2/3 of the red rate because each blue photon carries about 1.5× more energy. A Higher photon energy at fixed intensity means fewer photons per second, not more, so the rate does not increase by 1.5×. B Equal intensity does not imply equal photon rate when photon energy changes; with higher-energy photons, the count must fall. C With the same intensity, photon rate is inversely proportional to energy per photon; since 660/440 ≈ 1.5, the blue count is about 1/1.5 ≈ 2/3 of the red. D A drop to 1/2 would require the photon energy to double (i.e., wavelength to halve), but 440 nm is 2/3 of 660 nm, giving a factor of about 2/3, not 1/2.