Question

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 1

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

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

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(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:
\(P=nE\)
\(∴n=\frac{P}{E}\)
\(=\frac{9.42×10^{-3}}{3.141×10^{-19}}≈3×10^{16}\, photon/s\)

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