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))/(k - 1)`
B. `(kv_(1) - v_(2))/(k - 1)`
C. `(kv_(2) - v_(1))/(k - 1)`
D. `(v_(2) - v_(1))/(k)`

Answer

B

Explanation

By using `hv - hv_(0) = K_(max)`
`rArr h (v_(1) - v_(0)) = K_(1)` …..(i)
And `h(v_(2) - v_(0)) = K_(2)` …….(ii)
`rArr (v_(1) - v_(0))/(v_(2) - v_(0)) = (K_(1))/(K_(2)) = (1) /(K)` , Hence `v_(0) = (Kv_(1) - v_(2))/(K - 1)`.