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)/(2elamda_(0))`
D. `V_(0)-(hc)/(2elamda_(0))`

Answer

D

Explanation

`eV_(0)=(hc)/(lamda_(0))-W_(0) and eV'=(hc)/(2lamda_(0))-W_(0)`
Subtracting them, we have
`e(V_(0)-V')=(hc)/(lamda_(0))[1-(1)/(2)]=(hc)/(2lamda_(0)) implies V'=V_(0)-(hc)/(2elamda_(0))`