(b)

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

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} $

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