Photoelectric emission is observed from a metallic surface for frequencies `v_1 and v_2` of the incident light rays `(v_1 gtv_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. `(f_1-f_2)/(k-1)`
B. `(kf_1-f_2)/(k-1)`
C. `(kf_2-f_1)/(k-1)`
D. `(f_2-f_1)/(k-1)`

Answer

B

Explanation

`K_("max")=hf-hf_0`
`impliesh(f_1-f_0)=K_1`
`h(f_2-f_0)=K_2`
`K_1/K_2=1/kimplies(f_1-f_0)/(f_2-f_0)=1/kimpliesf_0=(kf_1-f_2)/(k-1)`