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Physics Question on Dual nature of radiation and matter

Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser.The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b)How many photons per second, on the average, arrive at a target irradiated by this beam? ( Assume the beam to have uniform cross-section which is less than the target area ), and
(c)How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

Answer

Wavelength of the monochromatic light, λ=632.8nm=632.8×10−9mλ=632.8nm=632.8×10^{−9}m
Power emitted by the laser, P=9.42mW=9.42×10−3WP=9.42mW=9.42×10^{−3}W
Planck’s constant, h=6.626×10−34Jsh=6.626×10^{−34} Js
Speed of light,c=3×108m/s ,c=3 ×10^8m/s
Mass of a hydrogen atom,m=1.66×10−27kgm=1.66×10^{−27}kg

(a)The energy of each photon is given as:
E=hcλE=\frac{hc}{λ}
=6.626×10−34×3×108632.8×10−9=3.141×10−19J=\frac{6.626×10^{-34}×3×10^8}{632.8×10^{-9}}=3.141×10^{-19}J
The momentum of each photon is given as:
P=hλP=\frac{h}{λ}
=6.626×10−34632.8=1.047×10−27kg ms−1=\frac{6.626×10^{-34}}{632.8}=1.047×10^{-27}kg \,ms^{-1}

(b)Number of photons arriving per second,at a target irradiated by the beam=n Assume that the beam has a uniform cross-section that is less than the target area.Hence,the equation for power can be written as:
P=nEP=nE
∴n=PE∴n=\frac{P}{E}
=9.42×10−33.141×10−19≈3×1016 photon/s=\frac{9.42×10^{-3}}{3.141×10^{-19}}≈3×10^{16}\, photon/s

(c)Momentum of the hydrogen atom is the same as the momentum of the photon,
p=1.047×10−27kg ms−1p=1.047×10^{-27}kg\, ms^{-1}
Momentum is given as:
p=mvp=mv
Where,
V=Speed of hydrogen atom
∴v=Pm∴v=\frac{P}{m}
=1.047×10−271.66×10−27=0.621m/s=\frac{1.047×10^{-27}}{1.66×10^{-27}}=0.621m/s