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

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

TestHub · Accepted Answer

The maximum kinetic energy of photoelectrons is given by $K E_{m a x} = h (v - v_{0})$ …. (i)
Where h = Planck’s constant,
v= frequency of radiation
and $v_{0}$ = threshold frequency.
It can be seen from Eq. (i), that the maximum KE of emitted photoelectrons is proportional to the frequency of radiation and is independent of the intensity of radiation, so it remains unchanged.

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