A metal surface is illuminated by a 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 emitted photoelectrons will become.

Options

A. twice the original value
B. four times the original value
C. one fourth of the original value
D. unchanged

Answer

D

Explanation

The maxium kinetic energy of photoelectrons is given by `KE_(max) = h (v-v_(0))" "…..(i)`
Where h = Planck's constant
v = frequnecy of radiation .
and `v_(0) =` thereshold frequency.
It can be see from Eq. (i), that the maximun KE of emitted photoelectron is proportional to the frequency of radiation and is independent of the intensity of radiation, so it remains unchaged .