$\lambda =500 \mathrm{nm}\to \mathrm{v}=\frac{\mathrm{C}}{\lambda }=\frac{3\times {10}^8}{500\times {10}^{-9}}=6\times {10}^{14} \mathrm{Hz}$

$v_0=4.3\times {10}^{14} \mathrm{Hz}$

For photo electric effect, $\mathrm{hv}={\mathrm{hv}}_0+\mathrm{KE}$

$\mathrm{KE}=\mathrm{hv}-{\mathrm{hv}}_0=\mathrm{h}\left(\mathrm{v}-{\mathrm{v}}_0\right)=6.6\times {10}^{-34}\left(6\times {10}^{14}-4.3\times {10}^{14}\right)$

$=6.6\times 1.7\times {10}^{-20} \mathrm{J}$

$\mathrm{K}.\mathrm{E}. =\frac{1}{2}{\mathrm{mv}}^2;\mathrm{v}=\sqrt{\frac{2\times \mathrm{K}·\mathrm{E}}{\mathrm{m}}}$

$\mathrm{V}=\sqrt{\frac{2\times 6.6\times 1.7\times {10}^{-20}}{9.1\times {10}^{-31}}}=\sqrt{24.65\times {10}^{10}}=5\times {10}^5 \mathrm{m}/\mathrm{s}$