At an incident radiation frequency of `v_(1)`, which is greater than the threshold frequency, the stopping potential for a certain metal is `V_(1)`. At frequency `2v_(1)`, the stopping potential is `3V_(1)`. If the stopping potential at frequency `4v_(1)` is `nV_(1)` then n is

Options

A. 2
B. 3
C. 6
D. 7

Answer

D

Explanation

We know,
`hv = phi + K.E.`
From the given data,
`hv1 = phi + eV_1" "…(1)`
`h ( 2 v_1 ) = phi + 3e V_1" "…(2)`
On dividing (1) by (2),
`(1)/(2) = ( phi + e V_1)/( phi + 3 e V_1)`
`rArr phi + 3 vV_1 = 2phi + 2 ev_1`
`therefore phi = e V_1`
From (1),
`hv_1 = eV_1 = eV_1`
`rArr 4 (2 eV_1) = eV_1 + n V_1`
`7 eV_1= n V_1`
`therefore n = 7`