\(\left(\lambda_0\right)=\frac{ hc }{\phi}=5404\,A\)

For each wavelength energy incident on the surface per unit time

= intensity of each \(\times\) area of the surface wavelength \(=1.2 \times 10^{-7}\) joule

\(E=\left(1.2 \times 10^{-7}\right) \times 2=2.4 \times 10^{-7}\,J\)

Number of photons \(n _1\) due to wavelength \(4144\,A\)

\(n_1=\frac{\left(2.4 \times 10^{-7}\right)\left(4144 \times 10^{-10}\right)}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^8\right)}=0.5 \times 10^{12}\)

Number of photons \(n_2\) due to the wavelength \(4972 \mathring  A\)

\(n _2=\frac{\left(2.4 \times 10^{-7}\right)\left(4972 \times 10^{-10}\right)}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^3\right)}=0.572 \times 10^{12} \)

\(N = n _1+ n _2=1.075 \times 10^{12}\)