Question

A beam of light has three wavelengths $4144\mathring{A}$, $4972\mathring{A}$ and $6216\mathring{A}$ with a total intensity of $3.6\times 10^{-3}W{m}^2$ equally distributed amongst the three wavelengths. The beam falls normally on an area $1.0c{m}^2$ of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds.

Answer 1

Energy of photon having wavelength $4144\mathring{A}$
$E_1=\frac{12375}{4144}eV=2.99eV$
Similarly, $E_2=\frac{12375}{4972}eV=2.49eV$ and
$E_3=\frac{12375}{6216}eV=1.99eV$
Since, only $E_1$ and $E_2$ are greater than the work function W=2.3 eV, only first two wavelengths are capable for ejecting photoelectrons. Given intensity is equally
distributed in all wavelengths. Therefore, intensity corresponding to each wavelength is
$\frac{3.6\times 10^{-3}}{3}=1.2\times 10^{-10}W/m^2$
Or energy incident per second in the given area $(A=1.0c{m}^2=10^{-4}{m}^2)$ is
$\rho =1.2\times 10^{-3}\times 10^{-4}$
$=1.2\times 10^{-7}J/s$
Let $n_1$ be the number of photons incident per unit time in the given area corresponding to first wavelength. Then,
$n_1=\frac{\rho }{E_1}=\frac{1.2\times 10^{-7}}{2.99\times 1.6\times 10^{-19}}$
$=2.5\times 10^{11}$
Similarly,
$n_1=\frac{\rho }{E_2}=\frac{1.2\times 10^{-7}}{2.49\times 1.6\times 10^{-19}}$
$=3.0\times 10^{11}$
Since, each energetically capable photon ejects electron, total number of photoelectrons liberated in 2 s.
$=2(n_1+n_2)=2(2.5+3.0)\times 10^{11}$
$=1.1\times 10^{12}$