When the frequency of the incident radiation on a metallic plate is doubled, KE of the photoelectrons will be

When the frequency of the incident radiation on a metallic plate is do

1.	When the frequency of the incident radiation on a metallic plate is doubled, KE of the photoelectrons will be
Answer» doubled halved more than doubled increases but less than doubled. Solution :`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)`