Question

A photon of wavelength $4\times {10}^{-7}m$ strike on metal surface,the work function of the metal being $2.13eV$. $(1eV=1.6020\times {10}^{-19}J)$ calculate the velocity of the photoelectron in $n\times {10}^{5}m{s}^{-1}$.

Answer 1

(i) $E/Photon=\frac{hc}{\lambda }$
$=\frac{6.626\times {10}^{-34}\times 3.0\times {10}^8}{4\times {10}^{-7}}$
$=4.97\times {10}^{-19}J=3.10eV$
(ii) $E=\frac{1}{2}m{u}^2+w$
$4.97\times {10}^{-19}=\frac{1}{2}m{u}^2+2.13\times 1.602\times {10}^{-19}$
$\therefore KE=\frac{1}{2}m{u}^2=4.97\times {10}^{-19}-3.41\times {10}^{-19}$
$\therefore =1.56\times {10}^{-19}J=0.97eV$
(iii) $\frac{1}{2}m{u}^2=KE$
$\frac{1}{2}\times 9.108\times {10}^{-31}\times u^2=1.56\times {10}^{-19}$
$\therefore u=5.85\times {10}^{5}m{s}^{-1}$