The kinetic energies of the photoelectrons are `E_(1)` and `E_(2)` with wavelength of incident light `lamda_(1)` and `lamda_(2)`. The work function of the metal is :

Options

A. `(E_(1)lamda_(1)-E_(2)lamda_(2))/(lamda_(2)-lamda_(1))`
B. `(E_(1)E_(2))/(lamda_(1)-lamda_(2))`
C. `((E_(1)-E_(2))lamda_(1)lamda_(2))/(lamda_(1)-lamda_(2))`
D. `(lamda_(1)lamda_(2)E_(1))/((lamda_(1)-lamda_(2))E_(2))`

Answer

A

Explanation

`E=(hc)/(lambda_(1))-w and E_(2)=(hc)/(lambda_(2))-w`
`:.(E_(1)+w)/(E_(2))+w=(lambda_(2))/(lambda_(1)) or w=(E_(1) lambda_(1)-E_(2)lambda_(2))/(lambda_(2)-lambda_(1))`