The work function of a certain metal is $3.31 \times {10}^{–19 }\mathrm{J}$. Then the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000$ Å is (given $\mathrm{h}=6.62\times {10}^{–34} \mathrm{Js},$$\mathrm{C}=3\times {10}^8{\mathrm{ms}}^{-1},$$\mathrm{e}=1.6\times {10}^{–19} \mathrm{coulomb}$).

Given data:

The work function of a certain metal is $3.31 \times {10}^{–19 }J.$

Formula used:

$\mathrm{E}={\mathrm{W}}_0+\mathrm{KE}$

Detailed solution:

Work function  ${\mathrm{W}}_0=3.31\times {10}^{-19} \mathrm{J}$
Wavelength of incident radiation
$$\begin{gathered} \lambda =5000\times {10}^{-10} \mathrm{m} \\ \mathrm{E}={\mathrm{W}}_0+\mathrm{KE} \end{gathered}$$
(According to Einstein equation)
$\frac{\mathrm{hc}}{\lambda }=3.31\times {10}^{-19}+\mathrm{KE}$

$\mathrm{KE}=-3.31\times {10}^{-19}+\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{5000\times {10}^{-10}}$

$=-3.31\times {10}^{-19}+\frac{6.62\times 3}{5}\times {10}^{-19}$

$=(-3.31\times 1.324\times 3)\times {10}^{-19}$

$=(3.972-3.31)\times {10}^{-19}=0.662\times {10}^{-19} \mathrm{J}$

$\Rightarrow \mathrm{E}=\frac{0.662\times {10}^{-19}}{1.6\times {10}^{-19}}=0.41\mathrm{eV}$