Answer
(a) $E=2.97\times 10^{-19}J$, $P=9.89\times 10^{-28} Kg.m/s$
(b) $N=3.4\times 10^{15}\frac{Photons}{s}$
(c) $F_{avg}=3.36\times 10^{-12}N$
Work Step by Step
(a) We can find the required energy and momentum of photon as follows:
$E=\frac{hc}{\lambda}$
We plug in the known values to obtain:
$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{670\times 10^{-9}}$
$\implies E=2.97\times 10^{-19}J$
and now the momentum is given as
$P=\frac{h}{\lambda}$
$\implies P=\frac{6.26\times 10^{-34}\times 3\times 10^8}{670\times 10^{-9}}$
$\implies P=9.89\times 10^{-28}Kg.m/s$
(b) The required number of photons emitted can be determined as follows:
$N=\frac{P}{E}$
We plug in the known values to obtain:
$N=\frac{1\times 10^{-3}}{2.97\times 10^{-19}}$
This simplifies to:
$N=3.4\times 10^{15} \frac{photons}{s}$
(c) The required average force can be calculated as follows:
$F_{avg}=N\frac{\Delta P}{Delta t}$
We plug in the known values to obtain:
$F_{avg}=3.4\times 10^{15}\times \frac{9.89\times 10^{-28}}{1}$
This simplifies to:
$F_{avg}=3.36\times 10^{-12}N$
