The beam of light has three wavelengths 4144 Ã, 4972 Ã and 6216 Ã...

As we know, threshold wavelength \begin{array}{c} \left(\lambda_{0}\right)=\frac{\mathrm{hc}}{\phi} \\ \Rightarrow \lambda_{0}=\frac{\left(6.63 \times 10^{-34}\right) \times 3 \times 10^{8}}{2.3 \times\left(1.6 \times 10^{-19}\right)}=5.404 \times 10^{-7} \mathrm{~m} \end{array} $\Rightarrow \lambda_{0}=5404 Ã…$ Hence, wavelength $4144 Ã…$ and $4972 Ã…$ will emit electron from the metal surface. For each wavelength energy incident on the surface per unit time = intensity of each $\times$ area of the surface wavelength $$=\frac{3.6 \times 10^{-3}}{3} \times(\mathrm{lcm})^{2}=1.2 \times 10^{-7} \text{joule }$$ Therefore, energy incident on the surface for each wavelength in $2 \mathrm{~s}$ $$\mathrm{E}=\left(1.2 \times 10^{-7}\right) \times 2=2.4 \times 10^{-7} \mathrm{~J}$$ Number of photons $\mathrm{n}_{1}$ due to wavelength $4144 Ã…$ $$n_{1}=\frac{\left(2.4 \times 10^{-7}\right)\left(4144 \times 10^{-10}\right)}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}=0.5 \times 10^{12}$$ Number of photons $\mathrm{n}_{2}$ due to the wavelength $4972 Ã…$ $$\mathrm{n}_{2}=\frac{\left(2.4 \times 10^{-7}\right)\left(4972 \times 10^{-10}\right)}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}=0.572 \times 10^{12}$$ Therefore total number of photoelectrons liberated in $2 \mathrm{~s}$ \begin{aligned} \mathrm{N} &=\mathrm{n}_{1}+\mathrm{n}_{2} \\ &=0.5 \times 10^{12}+0.575 \times 10^{12} \\ &=1.075 \times 10^{12} \end{aligned}