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. `(upsilon_2 - upsilon_1)/(k-1)`
B. `(Kupsilon_1 - upsilon_2)/(K-1)`
C. `(Kupsilon_2 - upsilon_1)/(K-1)`
D. `(upsilon_2 - upsilon_1)/(K)`

Answer

A, C, D

Explanation

`KE_1 , KE_2 = 1 :K, hupsilon_1 - hupsilon_0 = hupsilon_2 -hupsilon_0 = 1 :K`
`(hupsilon_1 - hupsilon_0)/(h upsilon_2 - hupsilon_0) = 1/K implies K (upsilon_1 - upsilon_2) = upsilon_2 - upsilon_0`
`implies K upsilon_1 - K upsilon_0 = upsilon_2 - upsilon_0`
`implies upsilon_0 (1-K) = upsilon_2 - K upsilon_1 implies upsilon_0 = (upsilon_2 - K upsilon_1)/(1-K)`