Explanation

(a) Energy of photon
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \frac{hc}{\lambda}$
Given, $\ \ \ \ \ \ \ \ \ \ \ \ \lambda = 5000\ \mathring{\mathrm{A}}\ = \ 5 \times 10^{- 7}m$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{5 \times 10^{- 7}}$
$= \ 3.96 \times 10^{- 19}\ J$
Energy received per second = $10^{- 8}\ Js^{- 1}$
$\therefore$ Number of photon’s received per second
= $\frac{Energy\ received\ per\ second\ }{Energy\ of\ one\ photon\ }$
= $\frac{10^{- 8}}{3.96 \times 10^{- 19}} = 2.5 \times 10^{10}$