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

Options

A. `V_(0)/2`
B. `2V_(0)`
C. `V_(0)+(hc)/(2elambda_(0))`
D. `V_(0)-(hc)/(2elambda_(0))`

Answer

D

Explanation

From Einstein.s photoelectric equation
`" "eV_(0)=(hc)/lambda_(0)=W_(0)`
`" "eV.=(hc)/(2lambda_(0))-W_(0)`
Subtracting
`" "e(V_(0)-V.)=(hc)/(lambda_(0))[1-1/2]=(hc)/(2lambda_(0))`
or `" "V.=V_(0)-(hc)/(2elambda_(0))`