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_(1))/(K-1)`
D. `(v_(2)-v_(1))/K`

Answer

B

Explanation

`hv_(1)=hv_(@)-1/2"mu"_(1)^(2)`………..(1)
`hv_(2)=hv_(@)-1/2"mu"_(2)^(2)`………(2)
`1/2 "mu"_(1)^(2)=1/K(1/2"mu"_(2)^(2))`
From equation(1)
`hv_(1)=hv_(@)-1/(2k)"mu"_(2)^(2)` ………(3)
`1/2"mu"_(2)^(2)=Khv_(@)-Khv_(1)`…..(4)
From eqn (4) & (2)
`hv_(2)=hv_(@)-Khv_(@)+Khv_(1)`
`v_(@)=(Kv_(1)-v_(2))/((K-1))`