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.

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

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