The photoelectric threshold wavelength for silver is `lamda_(0)`. The energy of the electron ejected from the surface of silver by an incident wavelength `lamda(lamdaltlamda_(0))` will be

Options

A. `hc(lamda_(0)-lamda)`
B. `(hc)/(lamda_(0)-lamda)`
C. `(h)/(c)/((lamda_(0)-lamda)/(lamdalamda_(0)))`
D. `hc((lamda_(0)-lamda)/(lamdalamda_(0)))`

Answer

D

Explanation

(d) : According to Einstein's photoelectric equation
`K=(hc)/(lamda)-(hc)/(lamda_(0))=hc((1)/(lamda)-(1)/(lamda_(0)))=((lamda_(0)-lamda)/(lamdalamda_(0)))`