`K_(1)` be the maximum kinetic energy of photoelectrons emitted when metal surface is illuminated by light of wavelength `lamda_(1)`. Similarly `K_(2)` is corresponding to wavelength `lamda_(2)` for same metal. If `lamda_(1)=2lamda_(2)` then

Options

A. `2K_(1) = K_(2)`
B. `K_(1) = 2K_(2)`
C. `K_(1) lt (K_(2))/(2)`
D. `K_(1) gt 2K_(2)`

Answer

C

Explanation

`(K_(2))/(K_(1)) = ((hc)/(lambda_(2)) - phi)/((hc)/(lambda_(1)) - phi) = (2((hc)/(lambda_(2)) - phi))/(((hc)/(lambda_(2)) - 2phi)) rArr K_(1) lt (K_(2))/(2)`