Question

A photon of wavelength $4\times {10}^{-7}$ m strikes on metal surface, the work function of the metal being $2.13eV$.
Calculate

(i) the energy of the phpton (eV),

(ii) the kinetic energy of the emission,
(iii) the velocity of the photoelectron. $(1eV=1.6020\times {10}^{-19}J).$

Answer 1

i)

$E=h\nu =\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{4\times {10}^{-7}}=4.97\times {10}^{-19}J$

$=\frac{4.97\times {10}^{-19}}{1.6\times {10}^{-19}}=3.1eV$

ii)

kinetic energy of emission,

$=h\nu -h{\nu }_0=3.1-2.13=0.97eV$

iii)

$\frac{1}{2}m{v}^2=0.97eV=0.97\times 1.6\times {10}^{-19}$

$\frac{1}{2}(9.11\times {10}^{-31})\times v^2=0.97\times 1.6\times {10}^{-19}$

$v^2=34.1\times {10}^{10}$

$v=5.84\times {10}^{5}m{s}^{-1}$