When the frequency of light incident an a metallic plate is doubled , the KE of the emitted photoelectrons will be :

Options

A. doubled
B. halved
C. more than doubled
D. increases but less than doubled.

Answer

C

Explanation

`KE_(1) = hv - W_(0), KE_(2) = 2 hv - W_(0)`
If `KE_(2)` were equal to `2 hv - 2 W_(0), KE_(2)` would have been `2 xx KE_(1)`.
As `(2 hv - W_(0)) gt (2 hv - 2 W_(0))`, hence,
`KE_(2) gt 2KE_(1)`