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

As `eV_(0)=(hc)/(lamda_(0))=W_(0)`……..1
and `eV.=(hc)/(2lamda_(0))-W_(0)`……….2
Subtracting eq 1 from eq. 2
`e(V_(0)-V.)=(hc)/(lamda_(2))(1-1/2)`
`=(hc)/(2lamda_(0))`
`V.=V_(0)-(hc)/(2elamda_(0))`