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. ` ( v_2 - v_2)/( k- 1) `
C. ` ( v_2 - v_1)/( k- 1) `
D. ` ( v_2 - v_1)/( k- 1) `

Answer

B

Explanation

` hv_1 = hv_0 + 1/2 mu_1^2` …(1)
` hv_2 = hv_0 + 1/2 mu_1&2` …(2)
` :. 1/2 mi_1^2 = 1/k = 1/k ( 1/2 mu_2^2)`
`:.` From eq.(i)
` hv_1 = hv_0 + 1/( 2k) mu_2^2` …(iii)
or ` 1/2 mu_2^2 = khv_1 - khv_0` ..(iv)
By eqws . (ii) and (iv)
` hv_2 = hv_0 - khv_0 + khv_1`
or ` v_0 ( 1 - k ) = v_2 - kv_1`
or ` v_0 = ( kv_1 - v_2)/( k -1)`.