A2 Daily A Level Physics question
A vacuum photocell with a clean metal cathode has a threshold frequency f0. With light of frequency 1.5 f0 at some intensity, a potential difference of 0.80 V (with an appropriate polarity) just reduces the photocurrent to zero. The light frequency is then increased by 20% while the intensity is halved. Which statement about the new stopping condition is correct?
Answer
The correct answer is A.
Correct: A — 1.28 V with the anode negative relative to the photocathode. A Photon energy is proportional to frequency; raising f from 1.5 f0 to 1.8 f0 increases the excess above the work function from 0.5 to 0.8 (in units of the threshold energy), so the stopping potential scales by 0.8/0.5 = 1.6, giving 1.28 V, and the anode must be negative to repel the electrons. B Uses the correct magnitude but the wrong polarity: a positive anode would attract electrons and would not stop the current. C Incorrectly assumes halving intensity halves the stopping potential; intensity changes the emission rate, not the maximum electron energy, so the magnitude is wrong (the polarity named here is the correct way round). D Assumes a simple 20% rise from 0.80 V to 0.96 V, ignoring the fixed work-function offset, and also has the polarity reversed.