Photoelectric work function of a metal is $1\mathrm{eV}.$ Light of wavelength $\lambda =3000\text{Å}$ falls on it. The photo electrons come out with velocity

$\mathrm{hv}=\mathrm{W}+\frac{1}{2}{\mathrm{mv}}^2\text{ or }\frac{\mathrm{hc}}{\lambda }=\mathrm{W}+\frac{1}{2}{\mathrm{mv}}^2$

Here $\lambda =3000\text{Å}=3000\times {10}^{-10}\mathrm{m}$

and $\mathrm{W}=1\mathrm{eV}=1.6\times {10}^{-19}$ joule

$\begin{array}{l}\frac{\left(6.6\times {10}^{34}\right)\left(3\times {10}^8\right)}{3000\times {10}^{-10}}\\ =(1.6\times {10}^{-19})+\frac{1}{2}\times (9.1\times {10}^{-31}){\mathrm{v}}^2\end{array}$

Solving we get $\mathrm{v}\cong {10}^6\mathrm{m}/\mathrm{s}$