The photoelectric threshold wavelength of silver is $3250\times {10}^{-10} \mathrm{m}$. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536\times {10}^{-10} \mathrm{m}\text{ is }$

$\left[\text{ Given }h=4.14\times {10}^{-15}\mathrm{eVs}\text{ and }c=3\times {10}^8 \mathrm{m}{ \mathrm{s}}^{-1}\right]$

$$\begin{gathered} K_{\max }=hv-{\varphi }_0=hv-h{v}_o=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0} \\ \text{ where }{\lambda }_0=\text{ threshold wavelength } \\ \text{ or }\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0} \\ \begin{array}{l}\text{ Here, }h=4.14\times {10}^{-15}\mathrm{eVs},c=3\times {10}^8 \mathrm{m}{ \mathrm{s}}^{-1}\\ {\lambda }_o=3250\times {10}^{-10} \mathrm{m}=3250\stackrel{0}{A}\\ \lambda =2536\times {10}^{-10} \mathrm{m}=2536\stackrel{0}{A},m=9.1\times {10}^{-31} \mathrm{kg}\\ hc=4.14\times {10}^{-15}\mathrm{eVs}\times 3\times {10}^8 \mathrm{m}{ \mathrm{s}}^{-1}=12420\mathrm{eV}\mathrm{AA}\\ \therefore \frac{1}{2}m{v}^2=12420\left[\frac{1}{2536}-\frac{1}{3250}\right]\mathrm{eV}=1.076\mathrm{eV}\\ v^2=\frac{2.152\mathrm{eV}}{m}=\frac{2.152\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}\\ \therefore v\approx 6\times {10}^5 \mathrm{m}{ \mathrm{s}}^{-1}=0.6\times {10}^6 \mathrm{m}{ \mathrm{s}}^{-1}\end{array} \end{gathered}$$