**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?

**Solution:**

Wavelength of the monochromatic light, *λ* = 632.8 nm = 632.8 × 10−9 m

Power emitted by the laser, *P* = 9.42 mW = 9.42 × 10−3 W

Planck’s constant, *h* = 6.626 × 10−34 Js

Speed of light, *c* = 3 × 108 m/s

Mass of a hydrogen atom, *m* = 1.66 × 10−27 kg

(a)The energy of each photon is given as:

$E=\frac{h c}{\lambda}$

$=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{632.8 \times 10^{-9}}=3.141 \times 10^{-19} \mathrm{~J}$

The momentum of each photon is given as:

$P=\frac{h}{\lambda}$

$=\frac{6.626 \times 10^{-34}}{632.8}=1.047 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

(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:

$P=n E$

$\therefore n=\frac{P}{E}$

$=\frac{9.42 \times 10^{-3}}{3.141 \times 10^{-19}} \approx 3 \times 10^{16}$ photon/s

(c) Momentum of the hydrogen atom is the same as the momentum of the photon, $p=1.047 \times 10^{-27} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$ Momentum is given as:

$p=m v$

Where,

*v* = Speed of the hydrogen atom

$\therefore v=\frac{p}{m}$

$=\frac{1.047 \times 10^{-27}}{1.66 \times 10^{-27}}=0.621 \mathrm{~m} / \mathrm{s}$

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