91Ó°ÊÓ

Volume of the laser constructed of semiconductor,$V=200\mathrm{μ}{m}^3$

Power of the laser,$P=5mW$

Wavelength of laser,$\mathrm{λ}=0.8\mathrm{μ}m$

Consider the known data as below.

The Plank’s constant, $h=6.63×{10}^{-34}Jâ‹\dots s$

Speed of light,$c=3×{10}^8\frac{m}{s}$

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.

Use the concept of Planck's relation to finding the value of the minimum wavelength that permits the removal of an electron from the lowest energy level. Again, the K line is produced only when an electron jumps from K to L energy level. Thus, the energy difference between these levels will provide the required energy. Similarly, the energy difference for the K line for the K to M energy jump can be calculated.

Formulas:

The rate of the photon emission,

$Rate=\frac{P}{E}$ ….. (1)

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

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

Here, is the energy of the photon, h is the Plank’s constant, and c is the speed of light.

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

$$\begin{gathered} Rate=\frac{P\mathrm{λ}}{hc} \\ =\frac{\left(5×{10}^{-3}W\right)\left(0.8×{10}^{-6}m\right)}{\left(6.63×{10}^{-34}J.s\right)\left(3×{10}^{8}m/s\right)} \\ =2×{10}^{16}{s}^{-1} \end{gathered}$$

Hence, the value of the rate of photon emission is $2×{10}^{16}{s}^{-1}$.