Question

The work function of a metal is 2.5$\times {10}^{-19}$ J. (a) Find the threshold frequency for photoelectric emission. (b) If the metal is exposed to a light beam of frequency $6.0\times {10}^{14}$ Hz, what will be the stopping potential ?

Answer 1

Given $W_0=2.5\times {10}^{-19}J$

$\left(a\right)\text{We have}{W}_0=h{v}_0$

$\text{or}{V}_0=\frac{W_0}{hv0}$

$=\frac{2.5\times {10}^{-19}}{6.063\times {10}^{-34}}$

$=3.77\times {10}^{14}Hz$

$=3.8\times {10}^{14}Hz$

(b)$e{V}_0=hv-W_0$

$\text{or}{V}_0=\frac{hv-W_0}{e}$

$=\frac{6.63\times {10}^{-34}\times 6\times {10}^{10}-2.5\times {10}^{-19}}{1.6\times {10}^{-19}}$

$=\frac{6.97\times {10}^{-19}-2.5\times {10}^{-19}}{1.6\times {10}^{-19}}$

=$0.91V$