Question

A 40 W ultraviolet light source of wavelength $2480\stackrel{o}{A}$ illuminates a magnesium (Mg) surface placed 2 m away. Determine the number of photons emitted from the surface per second and the number incident on unit area of Mg surface per second. The photoelectric work function for Mg is 3.68 eV. Calculate the kinetic energy of the fastest electrons ejected from the surface. Determine the maximum wavelength for which the photoelectric effect can be observed with Mg surface

Answer 1

Energy of each photon,
$E=\frac{hc}{\lambda }$
$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2480\times {10}^{-10}}J$
$=8.0\times {10}^{-19}J$
Number of photons emitted per second
$N=\frac{P}{E}$
$=\frac{40J{s}^{-1}}{8.0\times {10}^{-19}J}=5.0\times {10}^{19}{s}^{-1}$
These photons spread in all directions over surface area $4\pi r^2$, therefore the number of photons incident per unit area per second,
$N_i=\frac{N}{4\pi r^2}$
$=\frac{5.0\times {10}^{19}}{4\times 3.14\times (2{)}^2}$ (As $r=2m)$
$=9.95\times {10}^{13}{s}^{-1}$
From Einstein's photoelectric equation,
$E_k=hv-W$
$\Rightarrow E=hv=\frac{hc}{\lambda }$
$=8.0\times {10}^{-19}J$
$=\frac{8.0\times {10}^{-19}}{1.6\times {10}^{-19}}eV=5.0eV$
$\therefore E_k=5.0eV-3.68eV=1.32eV$
Threshold (or maximum) wavelength for photoelectrons emission,
${\lambda }_0=\frac{hc}{W}$
$=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{3.68\times 1.6\times {10}^{-19}}m$
$=3.373\times {10}^{-3}m$
$=3373\stackrel{o}{A}$