Question

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

Calculate:

(i) The energy of the photon (eV)
(ii) The kinetic energy of the emission
(iii) The velocity of the photoelectron.

$[1eV=1.6020\times {10}^{-19}J]$.

Answer 1

(i)The energy of the photon is

$E=\frac{hc}{\lambda }$

$=\frac{6.626\times {10}^{-34}Js\times 3\times {10}^{8}m/s}{4.0\times {10}^{-7}m}$

$=4.97\times {10}^{-19}J$.

Convert the unit into electron volts.

$E=\frac{4.97\times {10}^{-19}}{1.602\times {10}^{-19}}=3.10eV$

(ii) The kinetic energy of emission is

$K.E=h\nu -h{\nu }_o=3.10eV-2.13eV=0.97eV$.

(iii) The expression for the velocity of emitted photoelectron is

$v=\sqrt{\frac{2K.E}{m}}$

$=\sqrt{\frac{2\times 0.97\times 1.602\times {10}^{-19}}{9.11\times {10}^{-31}}}=5.84\times {10}^{5}m/s$.