Q. A photo metal is illuminated by light of wavelengths λ1 and λ2 respectively. The maximum kinetic energies of electrons emitted in the two cases are E1 and E2 respectively. The work function of metal is
(a) (E2 λ1 − E1 λ2 ) /λ1
(b) (E1 λ1 + E2 λ2 ) /(λ1 + λ2)
(c) (E1 λ1 + E2 λ2 ) /(λ1 − λ2)
(d) (E2 λ2 − E1 λ1 ) /(λ1 − λ2)
Ans: d
Sol: $\large \frac{hc}{\lambda_1} = \phi + E_1 $ ….(i)
$\large \frac{hc}{\lambda_2} = \phi + E_2 $ ….(ii)
On dividing (i) by (ii)
$\large \frac{\lambda_2}{\lambda_1} = \frac{\phi + E_1}{\phi + E_2} $
$\large E_2 \lambda_2 + \lambda_2 \phi = E_1 \lambda_1 + \lambda_1 \phi $
Hence , $\large \phi = \frac{E_2 \lambda_2 – E_1 \lambda_1}{\lambda_1 -\lambda_2} $