AS Daily A Level Physics question
In a school photoelectric-effect experiment with a clean photocathode, monochromatic light of wavelength 400 nm gives a stopping potential of 0.80 V at a certain intensity. The light is then changed to 360 nm while the intensity is halved. With all other conditions the same, which statement must be true about the new stopping potential? (Useful data: (hc)/e ≈ 1.24 × 10^-6 V m.)
Answer
The correct answer is C.
Correct: C — It increases to about 1.1 V; intensity affects current but not the stopping potential. Decreasing λ from 400 nm to 360 nm increases photon energy, raising the maximum electron kinetic energy and thus the stopping potential; ΔV ≈ (hc/e)(1/360 nm − 1/400 nm) ≈ 1.24×10^-6×(2.78×10^6 − 2.50×10^6) ≈ 0.34 V, so V_s ≈ 0.80 + 0.34 ≈ 1.14 V. A Intensity only changes the emission rate (photocurrent), not the maximum kinetic energy, so the stopping potential does not halve. B The effects do not cancel: shorter wavelength (higher photon energy) increases V_s; intensity does not reduce V_s. C This correctly identifies that V_s rises by about 0.34 V while intensity has no effect on V_s. D Although the work function sets the intercept, the change in stopping potential here depends only on the change in 1/λ, so it can be calculated without φ.