Given, $\lambda_1=4972 Å$
and $\lambda_2=6216 Å$
and $\mathrm{I}=3.6 \times 10^{-3} \mathrm{Wm}^{-2}$
Intensity associated with each wavelength
$$
=\frac{3.6 \times 10^{-3}}{2}=1.8 \times 10^{-3} \mathrm{Wm}^{-2}
$$
Work function $\phi=\mathrm{hv}=\frac{\mathrm{hc}}{\lambda}$
$$
\begin{aligned}
&=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^8\right)}{\lambda} \\
&=\frac{12.4 \times 10^3}{\lambda} \mathrm{ev}
\end{aligned}
$$


for different wavelengths
$$
\begin{array}{r}
\phi_1=\frac{12.4 \times 10^3}{\lambda_1}=\frac{12.4 \times 10^3}{4972}=2.493 \mathrm{eV} \\
=3.984 \times 10^{-19} \mathrm{~J} \\
\phi_2=\frac{12.4 \times 10^3}{\lambda_2}=\frac{12.4 \times 10^3}{6216}=1.994 \mathrm{eV} \\
=3.184 \times 10^{-19} \mathrm{~J}
\end{array}
$$
Work function for metallic surface $\phi=2.3$ $\mathrm{eV}$ (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 $\mathrm{m}^2-\mathrm{s}$
no. of electrons per $\mathrm{m}^2-\mathrm{s}$
$$
\begin{aligned}
&=\frac{1.8 \times 10^{-3}}{3.984 \times 10^{-19}} \times 10^{-4} \\
&\left(\because 1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2\right)=0.45 \times 10^{12}
\end{aligned}
$$
So, the number of photo electrons liberated
in 2 sec.
$=0.45 \times 10^{12} \times 2$
$=9 \times 10^{11}$