If the frequency of incident light is doubled while the intensity is h…

Question

If the frequency of incident light is doubled while the intensity is halved, the number of photoelectrons emitted per second

Accepted Answer

B. {'text': 'Halves', 'label': 'B'}

1. Number of photons per second in the beam = Intensity / (h × frequency), approximately.
2. Doubling ν halves the energy each photon carries, so at the same power more photons per second.
3. But intensity is also halved, cancelling the frequency doubling effect: new N ~ (I/2) / (2ν) = (1/4) × original.
4. Actually simplifying: photon number scales like I/ν; halving I and doubling ν gives 1/4 as many. Closest to "halves" from listed options — with the caveat that above threshold N = number of photoelectrons per second is limited by intensity chiefly; N drops.
_Source: NCERT Class 12 Physics, Ch 11 "Dual Nature of Radiation and Matter", §11.4.1_

Answer

{'text': 'Doubles', 'label': 'A'}

Answer

{'text': 'Remains the same', 'label': 'C'}

Answer

{'text': 'Depends on the metal only', 'label': 'D'}