Physics - Modern Physics Question with Solution | TestHub

Work function,${\mathrm{\varphi }}_0=6.2\mathrm{ }\mathrm{eV}$

Stopping potential${\mathrm{V}}_0=5\mathrm{V}$

As${\mathrm{eV}}_0=\mathrm{hv}-{\mathrm{\varphi }}_0\mathrm{ }$

or${\mathrm{eV}}_0+{\mathrm{\varphi }}_0=\frac{\mathrm{hc}}{\mathrm{\lambda }}$

or$\mathrm{\lambda }=\frac{\mathrm{hc}}{{\mathrm{eV}}_0+{\mathrm{\varphi }}_0}$

$\mathrm{\lambda }=\frac{6.62\times {10}^{-34}\mathrm{ }\mathrm{J} \mathrm{s}\times 3\times {10}^8\mathrm{ }\mathrm{m} {\mathrm{s}}^{-1}}{1.6\times {10}^{-19}\times 5 \mathrm{J}+6.2\times 1.6\times {10}^{-19}\mathrm{ }\mathrm{J}}$

$=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}(5+6.2)} \mathrm{m}$

$=\frac{19.86\times {10}^{-26}}{17.92\times {10}^{-19}}=1.10\times {10}^{-7}\mathrm{ }\mathrm{m}$.

Thus the wavelength of the incident radiation lies in the ultraviolet region.