Question

A helium–neon laser emits red light at wavelength $\mathit{\lambda }\mathbf{=}\mathbf{633}\mathbf{}\mathit{n}\mathit{m}$in a beam of diameter 3.5 mm and at an energy-emission rate of 5.0 mW. A detector in the beam’s path totally absorbs the beam. At what rate per unit area does the detector absorb photons?

Short answer

The rate of absorption of photon per unit area is$1.7\times {10}^{21}\mathrm{photons}/{\mathrm{m}}^2.\mathrm{s}$

Step-by-step solution

Describe the expression of energy of the photon

The energy Eof a photon of wavelength$\lambda$ is given by,

$E=\frac{hc}{\lambda }$

Here, h is the Planck’s constant, and c is the speed of light.

Determine the rate of absorption of photon per unit area 

Assume that the protons are emitted by a rate R from the sodium lamp. Then, the power P of the sodium lamp is equal to the product of rate R and the energy of each photon E.

$\begin{array}{c}P=RE\\ P=R\left(\frac{hc}{\lambda }\right)\\ R=\frac{p\lambda }{hc}\end{array}$…… (1)

Divide the rate R by area of the beam.

$\begin{array}{c}\frac{R}{A}=\frac{P\lambda }{hc\times \frac{\pi }{4}{d}^2}\\ =\frac{4P\lambda }{\pi d^{2}hc}\end{array}$…… (2)

Therefore, the rate of absorption of photon per unit area is${\mathrm{1.7\times 10}}^{21}\mathrm{photons}/{\mathrm{m}}^2.\mathrm{s}$