Question

A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value then the maximum KE of the emitted photoelectrons would be ______.

Options

• twice the original value

• four times the original value

• one-fourth of the original value

• unchanged

Answer 1

A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value then the maximum KE of the emitted photoelectrons would be unchanged.

Explanation:

The maximum kinetic energy of photoelectrons is given by `"KE"_"max" = "h"("v" - "v"_0)`   ...(i)

where, h = Planck's constant,

v = frequency of radiation

and v₀ = threshold frequency.

It can be seen from Eq. (i), that the maximum KE of emitted photoelectron is proportional to the frequency of the radiation and is independent of the intensity of radiation, so it remains unchanged.