If `lamda_o and lamda` be the threshold wavelength and wavelength of incident light , the velocity of photoelectron ejected from the metal surface is :

Options

A. `sqrt((2h)/(m)(lamda_(0)-lamda)`
B. `sqrt((2hc)/(m)(lamda_(0)-lamda))`
C. `sqrt((2hc)/(m)((lamda_(0)-lamda)/(lamdalamda_(0))))`
D. `sqrt((2h)/(m)((1)/(lamda_(0))-(1)/(lamda)))`

Answer

C

Explanation

Absorbed energy=Threshold energy+kinetic energy of photoelectrons
`(hc)/(lamda)=(hc)/(lamda_(0))+(1)/(2)mv^(2)`
`v=sqrt((2hc)/(m)((lamda_(0)-lamda))/(lamdalamda_(0)))`.