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

Options

A. `(v_(1)-v_(2))/((n-1)`
B. `(nv_(1)-v_(2))/((n-1)`
C. `(nv_(2)-v_(1))/((n-1)`
D. `(v_(1)-v_(2))/((n-1)`

Answer

B

Explanation

`hv-hv_(0)=k_max`
`h(v_(1)-v_(0))=k_(1)`
`h(v_(2)-v_(0))=K_(2)`
Dividing equation (1)by(2)
`implies(v_(1)-v_(0))/(v_(2)-v_(0))=(k_(1))/k_(2)=(1)/n`
`impliesnv_(1)-nv_(0)=v_(2)-v_(0)therefore" "v_(0)=(-nv-v_(2))/(n-1)`