Photoelectric emission is observed from a surface for frequencies `v_(1) "and" v_(2)` of the incident radiation `(v_(1) gt v_(2))` . If maximum kinetic energies of the photo electrons in the two cases are in the ratio `1:K` , then the threshold frequency is given by:

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_(2))/(K)`

Answer

B

Explanation

`hv_(1) = hv_(0) +1 rArr h[v_(1) -v_(0)] = 1` ..........(1)
`hv_(2) = hv_(0) +k rArr h[v_(2)-v_(0)] = k` .........(2)
`((1))/((2)) =(1)/(K) = (v_(1)-v_(0))/(v_(2)-v_(0))`
`v_(2) - v_(0) = Kv_(1) - Kv_(0), Kv_(0) - v_(0) = Kv_(1) - v_(2)`
`v_(0) [K-1] = Kv_(1) - v_(2) v_(0) = (Kv_(1)-v_(2))/(K-1)`