A beam of light has two wavelengths of \(4972\,\mathop A\limits^o \) and \(6216\,\mathop A\limits^o \) with a total intensity of \(3 .6 \times  10^{- 3}\,\,Wm^{-2}\) equally distributed among the two wavelengths. The beam falls normally on an area of \(1\,cm^2\) of a clean metallic surface of work function \( 2.3\,eV.\) Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in \(2\,s\) is approximately

and \({\lambda _2} = 6216\,\mathop {\text{A}}\limits^o \)

and \(I=3.6 \times 10^{-3}\, \mathrm{Wm}^{-2}\)

Intensity associated with each wavelength

\(=\frac{3.6 \times 10^{-3}}{2}=1.8 \times 10^{-3} \,\mathrm{Wm}^{-2}\)

Work function \(\phi=h v=\frac{h c}{\lambda}\)

\( = \frac{{\left( {6.62 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}}{\lambda }\)

\(=\frac{12.4 \times 10^{3}}{\lambda} \mathrm{ev}\)

for different wavelengths

\({{\phi _1} = \frac{{12.4 \times {{10}^3}}}{{{\lambda _1}}}\, = \,}\) \({\frac{{12.4 \times {{10}^3}}}{{4972}} = 2.493\,\,{\text{eV}}}\)

\(=3.984 \times 10^{-19}\, \mathrm{J}\)

\({\phi _2} = \frac{{12.4 \times {{10}^3}}}{{{\lambda _2}}}\) \({ = \,\,\frac{{12.4 \times {{10}^3}}}{{6216}} = 1.994\,\,{\text{eV}}}\)

\(=3.184 \times 10^{-19}\, \mathrm{J}\)

Work function for metallic surface \(\phi=2.3\,eV\) (given)

\(\phi_{2}<\phi\)

Therefore, \(\phi_{2}\) will not contribute in this process.

Now, no. of electrons per \(\mathrm{m}^{2}-\mathrm{s}=\) no of photons per \(m^{2}-s\)

no. of electrons per \(m^{2}-s\)

\(=\frac{1.8 \times 10^{-3}}{3.984 \times 10^{-19} \times 10^{-4}}\)

\(\left(\because 1 \mathrm{cm}^{2}=10^{-4} \mathrm{m}^{2}\right)=0.45 \times 10^{12}\)

So, the number of photo electrons liberated in \(2\) sec.

\(=0.45 \times 10^{12} \times 2\)

\(=9 \times 10^{11}\)