The stopping potential for photoelectrons from a metal surface is `V_(1)` when monochromatic light of frequency `upsilon_(1)` is incident on it. The stopping potential becomes `V_(2)` when monochromatic light of another frequency is incident on the same metal surface. If h be the Planck’s constant and e be the charge of an electron, then the frequency of light in the second case is:

Options

A. `v_1-e/h(V_2+V_1)`
B. `v_1+e/h(V_2+V_1)`
C. `v_1-e/h(V_2-V_1)`
D. `v_1+e/h(V_2-V_1)`

Answer

D

Explanation

`hv_1=eV_1+phi_0`
`hv_2=eV_2+phi_0`
`h(v_2-v_1) =e(V_2-V_1)`
`implies v_2=e/h(V_2-V_1) +v_1`