When light of frequency `v_(1)` is incident on a metal with work function W (where `hv_(1) gt W`) the photocurrent falls to zero at a stopping potential of `V_(1)`. If the frequency of light is increased to `v_(2)`, the stopping potential changes to `V_(2)`. Therefore the charge of an electron is given by

Options

A. `(W (v_(2) + v_(1)))/(v_(1) V_(2) + v_(2) V_(1))`
B. `(W (v_(2) + v_(1)))/(v_(1) V_(1) + v_(2) V_(2))`
C. `(W (v_(2) - v_(1)))/(v_(1) V_(2) - v_(2) V_(1))`
D. `(W (v_(2) - v_(1)))/(v_(2) V_(2) - v_(1) V_(1))`

Answer

C

Explanation

Acoording to Einstein's photoelectric equation,
`hv_(1) = W + eV_(1)`…(1)
and `hv_(2) = W + eV_(2)`…(2)
Solving equation (1) and (2) we get,
`e = (W (v_(2) - v_(1)))/(v_(1) V_(2) - v_(2) V_(1))`