The maximum kinetic energies of photoelectrons emitted from a metal are `k_(1)` and `k_(2)` when it is irradiated with light of wavelenght `lambda_(1)` and `lambda_(2)` respectively. Find work function of the metal.

Options

A. `(lambda_(1)K_(1)+lambda_(2)K_(2))/(lambda_(1)+lambda_(2))`
B. `(lambda_(1)K_(2)+lambda_(2)K_(1))/(lambda_(1)+lambda_(2))`
C. `(lambda_(1)K_(1)-lambda_(2)K_(2))/(lambda_(2)-lambda_(1))`
D. `(lambda_(1)K_(2)-lambda_(2)K_(1))/(lambda_(1)-lambda_(2))`

Answer

C

Explanation

In the first case, the K.E. , `K=hv-W_(0)=(hC)/(lambda)-W_(0)`
`:.K_(1)=(hC)/(lambda_(1))-W_(0)` or `(hC)/(lambda_(1))=K_(1)+W_(0)`
or `hC=lambda_(1)(K_(1)+W_(0))` ...(1)
Similarly, in the second case `hC=lambda_(2)(K_(2)+W_(0))`
`:.` From (1) and (2), `lambda_(1)(K_(1)+W_(0))=lambda_(2)(K_(2)+W_(0))`
`:.lambda_(1)K_(1)+lambda_(1)W_(0)=lambda_(2)K_(2)+lambda_(2)W_(0)`
`:.lambda_(1)K_(1)-lambda_(2)K_(2)=(lambda_(2)-lambda_(1))W_(0)`
`:.W_(0)=(lambda_(1)K_(1)-lambda_(2)K_(2))/(lambda_(2)-lambda_(1))`