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

`h upsilon_(1) =phi + K` …(i)
`h upsilon_(2) =phi + kK` …(ii)
k(i) - (ii) `implies h(k upsilon_(1) - upsilon_(2)) =(k-1) phi = (k-1)h upsilon_(0)`
`upsilon_(0) =(k upsilon_(1) - upsilon_(2))/(k - 1)`