AS Daily A Level Physics question
In a photoelectric experiment on a clean metal, monochromatic light with frequency 1.5 f0 (f0 is the threshold frequency) produces a stopping potential and a steady photocurrent. At the same time, an electron diffraction tube produces electrons accelerated through a potential V, giving a de Broglie wavelength λ. The light intensity is then doubled, and the accelerating voltage is increased to 4V. Which statement must be true?
Answer
The correct answer is A.
Correct: A — The stopping potential is unchanged; the photocurrent doubles; the electron de Broglie wavelength halves. At fixed frequency above threshold, stopping potential depends on photon energy (frequency), not intensity; doubling intensity doubles the photon rate so the photocurrent doubles. Quadrupling the accelerating voltage increases electron speed by √4, so λ decreases by 2 (halves). A … B is wrong because it confuses intensity with photon energy (stopping potential does not double with intensity), and it reverses de Broglie dependence by claiming λ doubles rather than decreasing with higher voltage. C is wrong due to a threshold-frequency misunderstanding: emission already occurs at 1.5 f0 without needing higher intensity; and λ does change with accelerating voltage. D is wrong because stopping potential does not scale with intensity, photocurrent should increase (not halve) when intensity is doubled, and λ would be one quarter only if voltage increased by 16×, not 4×.