Explanation

(b) Energy possessed by a photon is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = hv = \frac{hc}{\lambda}$
If power of each photon is $P$ then energy given out in $t$ second is equal to $Pt.$ Let the number of photons be $n$, then
$n = \frac{Pt}{E} = \frac{Pt}{(hc/\lambda)} = \frac{Pt\lambda}{hc}$
For red light, $n_{R} = \frac{Pt\lambda_{R}}{hc}$
For violet light, $n_{V} = \frac{Pt\lambda_{V}}{hc}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{n_{R}}{n_{V}} = \frac{\lambda_{R}}{\lambda_{V}}$
As $\lambda_{R} > \lambda_{V}$
So, $n_{R} > n_{V}$