Photoelectric emission is observed from a metallic surface for frequencies `v_(1)` and `v_(2)` of the incident light rays `(v_(1) gt v_(2))`. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of `1 : k` , then the threshold frequency of the metallic surface is

Options

A. `(v_(1)-v_(2))/(n-1)`
B. `(nv_(1)-v_(2))/(n-1)`
C. `(nv_(2)-v_(1))/(n-1)`
D. `(v_(1)-v_(2))/(n)`

Answer

B

Explanation

`E_(1) = h (v_(1)v_(0))`
`E_(2) = h(v_(2)-v_(0))`
`(E_(1))/(E_(2)) = (v_(2)-v_(0))/(v_(1)-v_(0)) rArr n = (v_(2)-v_(0))/(v_(1)-v_(0))`