AS Daily A Level Physics question
A photoelectric cell has a clean metal photocathode with work function 2.2 eV. Monochromatic light of wavelength 310 nm shines on it. The collector electrode is biased relative to the photocathode until the photocurrent just falls to zero; define this required collector potential as (collector minus photocathode). The light intensity is then doubled at the same wavelength. You then consider replacing the source with 620 nm light at the original intensity. Which statement must be true? Use hc ≈ 1240 eV·nm.
Answer
The correct answer is D.
Correct: D — The required collector potential is about −1.8 V and is unchanged by doubling intensity; with 620 nm light no photoelectrons are emitted so zero volts already stops the current. Reason: photon energy at 310 nm is 1240/310 = 4.0 eV, so the maximum electron kinetic energy is 4.0 − 2.2 = 1.8 eV; to stop these electrons the collector must be negative by about 1.8 V relative to the photocathode, and changing intensity does not change electron energy. At 620 nm the photon energy is 1240/620 = 2.0 eV, below the 2.2 eV work function, so no electrons are emitted and the current is already zero at 0 V. A is wrong because the sign is incorrect (the collector must be negative to repel electrons), intensity does not change the stopping potential, and 620 nm is below threshold so no stopping potential is needed at all. B is wrong because −0.9 V is not supported by the numbers (the correct magnitude is about 1.8 V), and 620 nm would produce no photoelectrons, not a weak current. C is wrong because the stopping potential is not positive in this definition (collector minus photocathode must be negative to stop electrons), and 620 nm does not produce any photoelectrons so there is no "less negative" value to apply.