If the frequency of incident light falling on a photosensitive metal is doubled, the kinetic energy of the emitted photoelectron is

Question

If the frequency of incident light falling on a photosensitive metal is doubled, the kinetic energy of the emitted photoelectron is (A) unchanged (B) halved (C) doubled (D) more than twice its initial value

Tardigrade Admin · Accepted Answer

By Einstein's photoelectric equation, The kinetic energy of photoelectrons

K E = h v - ϕ

where

v =

frequency of incidence light
and

K E^{^{'}} = 2 h v - ϕ

So, the kinetic energy is more than twice its initial value.

Q. If the frequency of incident light falling on a photosensitive metal is doubled, the kinetic energy of the emitted photoelectron is

unchanged

halved

doubled

more than twice its initial value

reduced to $\frac{1}{4}$ th

By Einstein's photoelectric equation, The kinetic energy of photoelectrons
$K E = h v - ϕ$
where $v =$ frequency of incidence light
and $K E^{^{'}} = 2 h v - ϕ$
So, the kinetic energy is more than twice its initial value.