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

Answer

2

Explanation

`(1)/(k)=(hv_(1)-varphi)/(hv_(2)-varphi)`
`hv_(2) - varphi = khv_(1) - k varphi`
`varphi = (h(kv_(1)-v_(2)))/(k-1)`