Answer
a) $n=1.59\times 10^{16}$ photons/sec
b) $\Delta p=-1.05\times 10^{-27}Kg\frac{m}{s}$
c) $F=1.67\times 10^{-11}N$
Work Step by Step
(a) We know that
$Energy=\frac{hc}{\lambda}$
$Energy=\frac{6.63\times 10^{-34}\times 3\times 10^8}{632.8\times 10^{-9}}=3.14\times 10^{-19}J$
As $Power=n\times Energy$
$\implies 5\times 10^{-3}=n\times 3.14\times 10^{-19}$
$\implies n=1.59\times 10^{16}$ photons/sec
(b) As $p=\frac{h}{\lambda}$
$\implies p=\frac{6.63\times 10^{-34}}{632.8\times 10^{-9}}=1.05\times 10^{-27}Kg\frac{m}{s}$
We can find the change in momentum as
$change \space in \space momentum=final\space momentum-initial momentum=0-1.05\times 10^{-27}=-1.05\times 10^{-27}Kg\frac{m}{s}$
(c) We can find the required force as
$F=\Delta p\times n$
We plug in the known values to obtain:
$F=1.05\times 10^{-27}\times 1.59\times 10^{16}$
$F=1.67\times 10^{-11}N$
