Photoelectric work function of a metal is 1 eV. If a light of wavelength of 500 $\stackrel{\mathrm{o}}{\mathrm{A}}$ falls on the metal, the speed of the photoelectron emitted approximately will be: ( h = 6.625
x 10^-34 J.s)

$hv=w+\mathrm{KE}=w+\frac{1}{2}{m}_e{v}^2$

$\begin{array}{l}\text{ Or }h\frac{c}{\lambda }=1+\frac{1}{2}{m}_e{v}^2\\ \text{ Or }\frac{6.625\times {10}^{-34}\times 3\times {10}^8}{500\times {10}^{-10}}\end{array}$

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=1.6\times {10}^{-19}+\frac{1}{2}\times 9.1\times {10}^{-31}\times v^2\\ \text{ Or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{1}{2}\times 9.1\times {10}^{-31}\times {\mathrm{v}}^2\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=\frac{6.625\times 3}{5}\times {10}^{-18}-1.6\times {10}^{-19}=38.15\times {10}^{-19}\\ \text{ Or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{v}^2=\frac{38.15\times 2\times {10}^{12}}{9.1}\\ \text{ Or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{v}^2=8.385\times {10}^{12}\\ \text{ Or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}v=2.89\times {10}^6\mathrm{m}/\mathrm{s}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{v}{c}=\frac{3\times {10}^6}{3\times {10}^8}=\frac{1}{100}\end{array}$