Question

Photoelectric work function of a metal is $1eV$. Light of wavelength $\lambda =3000\stackrel{o}{A}$ falls on it. The photoelectrons come out with velocity :

Answer 1

The Einstein's equation for photoelectric effect is given by , $K_{max}=h\nu -W$

or $\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-W$

or $v=\sqrt{\frac{2(hc/\lambda -W)}{m}}=\sqrt{\frac{2\left[\right(6.62\times {10}^{-34}\times 3\times {10}^8\left)/\right(3000\times {10}^{-10})-(1\times 1.6\times {10}^{-19}\left)\right]}{9.1\times {10}^{-31}}}$

$=1.04\times {10}^{6}m/s\sim {10}^{6}m/s$ (where mass of electron $m=9.1\times {10}^{-31}kg$)