In a photoelectric effect measurement, the stoppingg potential for a given metal is found to be `V_(0)` volt, when radiation of wavelength `lamda_(0)` is used. If radiation of wavelength `2lamda_(0)` is used with the same metal, then the stopping potential (in V) will be

Options

A. `(V_0)/(2) `
B. `2V_0`
C. `V_0 + (hc)/(2e lamda_0)`
D. `- (hc)/(2e lamda_0)`

Answer

D

Explanation

`eV_0 = (hc)/(lamda_0) - W_0 " and " eV. = (hc)/(2lamda_0) - W_0`
substracting them we have
`e(V_0 - V.) = (hc)/(lamda_0) [1-1/2] = (hc)/(2lamda_0) " or " V. = V_0 - (hc)/(2e lamda_0)`