Energy flux of sunlight = Total energy per square metre per second = 1.388 x 10³ Wm^–2 

Energy of each photon is given by,

                       $\boxed{\mathrm{E} = \frac{\mathrm{hc}}{\mathrm{\lambda }}}$ 
Therefore, 
 $$\begin{gathered} \mathrm{Power} \mathrm{of} \mathrm{each} \mathrm{photon}, \mathrm{P} = \mathrm{nE} \\ = \frac{6.63 \times {10}^{-34}\times 3 \times {10}^8}{550 \times {10}^{-9}}\mathrm{J} \\ = 3.62 \times {10}^{-19}\mathrm{J} \end{gathered}$$ 
Number of photons incident on earth's surface per square metre per second is, 

$= \frac{\mathrm{Total} \mathrm{energy} \mathrm{per} \mathrm{square} \mathrm{metre} \mathrm{per} \mathrm{second}}{\mathrm{Energy} \mathrm{of} \mathrm{one} \mathrm{photon}}$ 

$$\begin{gathered} = \frac{1.388 \times {10}^3}{3.62 \times {10}^{-19}} \\ = 3.8 \times {10}^{21}. \end{gathered}$$

Given,
Power, P (power) = 100 W
Wavelength of sodium light, λ = 589 x 10^–9 m 

(a) Energy of each photon assosciated with the sodium light,
                $$\begin{gathered} \mathrm{E} = \mathrm{hv} = \frac{\mathrm{hc}}{\mathrm{\lambda }} \\ =\frac{6.63 \times {10}^{-34} \times 3 \times {10}^8}{589 \times {10}^{-9}}\mathrm{J} \end{gathered}$$ 

$\Rightarrow$           $\mathrm{E} = 3.38 \times {10}^{-19}\mathrm{J}$ 

(b) Number of photons delivered to sphere per second, 
As,              P= nE

                 $\mathrm{n} = \frac{\mathrm{Energy} \mathrm{radiated} \mathrm{per} \mathrm{second}}{\mathrm{Energy} \mathrm{of} \mathrm{each} \mathrm{proton}}$ 

                 $$\begin{gathered} \mathrm{n} = \frac{100}{3.38 \times {10}^{-19}} \\ = 3 \times {10}^{20} \mathrm{photon}/\mathrm{s}. \end{gathered}$$