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The number of photoelectrons emitted for light of a frequency $\nu$ (higher than the threshold frequency $\nu_0$) is proportional to
A
Frequency of light ($\nu$)
B
$\nu - \nu_0$
C
Threshold frequency ($\nu_0$)
D
Intensity of light
Detailed Solution
In the photoelectric effect one photon ejects at most one electron, provided $\nu > \nu_0$.
The number of photoelectrons emitted per second therefore depends on the number of photons falling on the surface per second.
For light of a given frequency, the number of photons per second is proportional to the intensity of the light.
The frequency (and $\nu - \nu_0$) decides only the maximum kinetic energy of the photoelectrons: $KE_{max} = h(\nu - \nu_0)$.
Hence the number of photoelectrons is proportional to the intensity of light.
The number of photoelectrons emitted per second therefore depends on the number of photons falling on the surface per second.
For light of a given frequency, the number of photons per second is proportional to the intensity of the light.
The frequency (and $\nu - \nu_0$) decides only the maximum kinetic energy of the photoelectrons: $KE_{max} = h(\nu - \nu_0)$.
Hence the number of photoelectrons is proportional to the intensity of light.