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Expert-verified Found in: Page 1250 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A helium–neon laser emits laser light at a wavelength of 632.8 nm and a power of 2.3 mW. At what rate are photons emitted by this device?

The rate at which the device emits photons is $7.3\mathrm{x}{10}^{15}{\mathrm{s}}^{-1}$.

See the step by step solution

## Step 1: The given data:

The wavelength of the laser light, $\mathrm{\lambda }=632.8\mathrm{nm}$

Power of the laser beam, P = 2.3 mW

Consider the known data as below.

The Plank’s constant, $\mathrm{h}=6.63×{10}^{-34}\mathrm{J}.\mathrm{s}$

Speed of light, $\mathrm{c}=3\mathrm{x}{10}^{8}\mathrm{m}/\mathrm{s}$

## Step 2: Understanding the concept of rate of emission and Plank’s relation:

The output power is equal to the number of photons emitted per second which is multiplied by the power of each photon. Divide the energy released by the energy of each photon, to calculate the rate of photon emission.

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

Using the given data and Planck's relation in the equation of the rate of the emitted photons, calculate the required value of the rate at which the photons are emitted by the device.

Formulae:

The rate of the photon emission,

$\mathrm{Rate}=\frac{\mathrm{P}}{\overline{\mathrm{E}}}$ ….. (1)

The energy of the photon due to Planck’s relation,

$\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$ ….. (2)

## Step 3: Calculation of the rate at which the device emits photons:

Substituting the value of given data and equation (2) in equation (1), the rate of the photon emission by the device is as follow.

$\mathrm{Rate}=\frac{\mathrm{P\lambda }}{\mathrm{hc}}\phantom{\rule{0ex}{0ex}}=\frac{\left(2.3×{10}^{-3}\mathrm{W}\right)\left(632.8×{10}^{-9}\mathrm{m}\right)}{\left(6.63×{10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(3×{10}^{8}\mathrm{m}/\mathrm{s}\right)}\phantom{\rule{0ex}{0ex}}=7.3×{10}^{15}{\mathrm{s}}^{-1}$

Hence, the value of the rate of photon emission is $7.3×{10}^{15}{\mathrm{s}}^{-1}$. ### Want to see more solutions like these? 