Photoelectric emission is observed from a metallic surface for frequencies `v_(1)` and `v_(2)` of the incident light rays `(v_(1) gt v_(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. `(Kv_(1)-v_(2))/(K-1)`
B. `(Kv_(2)-v_(1))/(K-1)`
C. `(v_(2)-v_(1))/(K)`
D. `(v_(1)-v_(2))/(K-1)`

Answer

A

Explanation

By Einstein's photoelectric equation
`hv=hv_(0)+1/2 mv_("max")^(2)` or `hv=hv_(0)=K_("max")`
In this case, `h(v_(1)-v_(0))=K_(1)` and `h(v_(2)-v_(0))=K_(2)`
`:.(v_(1)-v_(0))/(v_(2)-v_(0))=(K_(1))/(K_(2))=1/K` (Given)
`:.Kv_(1)-Kv_(0)=v_(2)-v_(0)`
`:.Kv_(1)-v_(2)=v_(0)(K-1)`
`:.` The threshold frequency , `v_(0)=(Kv_(1)-v_(2))/(K-1)`