If a photocell is illuminated with a radiation of $1240 \stackrel{\circ }{\mathrm{A}}$, the stopping potential is found to be $8 \mathrm{V}$; then the work function of the emitter and the threshold wavelength are:

$\mathrm{W}=$ work function of metal, $\lambda =$ wavelength of incident light.

If $V_0$ be the stopping potential, then $K{E}_{\max }=e{V}_0$

So, $e{V}_0=\frac{hc}{\lambda }-\mathrm{W}$

$\mathrm{W}=\frac{hc}{\lambda }-e{V}_0=\frac{\left(6.62\times {10}^{-34}\right)\times \left(3\times {10}^8\right)}{1240\times {10}^{-10}}-\left(1.6\times {10}^{-19}\right)\left(8\right)$
$=3.2\times {10}^{-19}=2 \mathrm{eV}$ as $\left(1 \mathrm{eV}=1.6\times {10}^{-19}\right)$

If ${\lambda }_0$ be the threshold wavelength, $W=\frac{hc}{{\lambda }_0}$

${\lambda }_0=\frac{hc}{W}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{3.2\times {10}^{-19}}=6.2\times {10}^{-7} \mathrm{m}=6200 \mathrm{A}°$