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

Options

A. `v_(1) = v_(2)`
B. `(v_(1) - v_(2))/(h)`
C. `2 v_(1) - v_(2)`
D. `2v_(2) - v_(1)`

Answer

D

Explanation

`hv_(1) = hv_(0) + E` (K.E)..(i)
`hv_(2) = hv_(0) + (1)/(2) E or 2 hv_(2) = 2 hv_(0) + E`...(ii)
Eqn. (i) - Eqn (ii) gives `hv_(1) - 2hv_(2) = hv_(0) - 2 hv_(0)`
or `v_(1) - 2 v_(2) = - v_(0) or v_(0) = 2 v_(2) - v_(1)`