A $40W$ ultraviolet light source of wavelength 2480 Å illuminates a photosensitive metal surface placed $2m$ away. Calculate the number of photons emitted from the source per second and the number incident on unit area of the metal surface per second. The photo-electric work function for the metal is $3.7eV$. Calculate the kinetic energy of the fastest electrons ejected from the surface. Also calculate the maximum wavelength of light that can produce the photoelectric effect from the given metal surface

The energy of incident photon in $eV$ is given by

$E=\frac{12400}{2480}eV=5eV$

$\Rightarrow E=\left(5\right)\left(1.6\times {10}^{-19}\right)J=8\times {10}^{-19}J$

The number of photons emitted per second by the source is

$n=\frac{P_{\text{source}}}{E_{\text{each photon}}}=\frac{40}{8\times {10}^{-19}}=5\times {10}^{19}\text{ photons/s }$

The number of photons incident on the metal surface per second per unit area is

${\phi }_N=\frac{n}{A}=\frac{n}{4\pi r^2}=\frac{5\times {10}^{19}}{4\times 3.14\times (2{)}^2}=0.995\times {10}^{18}\approx {10}^{18} {\mathrm{s}}^{-1}{\mathrm{m}}^{-2}$

The kinetic energy of the fastest electron is given by

$K_{\max }=E-{\varphi }_0$$=5-3.7=1.3 \mathrm{eV}$

The threshold wavelength for the given metal surface for a work function of $3.7eV$ is given by

${\lambda }_0=\frac{12400}{3.7}=3351Å$