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When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectrons and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are respectively
A
N and 2T
B
2N and T
C
2N and 2T
D
N and T
Detailed Solution
Intensity of monochromatic light is proportional to the number of photons falling per second; each photon can eject one electron.
So the number of photoelectrons is proportional to the intensity: doubling I doubles the number to 2N.
Maximum kinetic energy: $KE_{max} = h\nu - \phi_0$, which depends only on the frequency of the light and the work function, not on the intensity.
The frequency is unchanged, so the maximum kinetic energy remains T.
Hence the values are 2N and T.
So the number of photoelectrons is proportional to the intensity: doubling I doubles the number to 2N.
Maximum kinetic energy: $KE_{max} = h\nu - \phi_0$, which depends only on the frequency of the light and the work function, not on the intensity.
The frequency is unchanged, so the maximum kinetic energy remains T.
Hence the values are 2N and T.