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Found in: Page 1250

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

Ruby lasers are at a wavelength of 694 nm. A certain ruby crystal has Cr ions (which are the atoms that lase). The lasing transition is between the first excited state and the ground state, and the output is a light pulse lasting ${\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{\mu s}}$. As the pulse begins, 60.0% of the Cr ions are in the first excited state and the rest are in the ground state. What is the average power emitted during the pulse? (Hint: Don’t just ignore the ground-state ions.)

The average power emitted during the pulse is $1.1×{10}^{6}\mathrm{W}$.

See the step by step solution

Step 1: The given data

The wavelength of the ruby lasers,$\lambda =694nm$

The output pulse last, $∆\mathrm{t}=2\mathrm{\mu s}$

As the pulse begins, 60% are in a first excited state and 40% in the ground state.

Consider the known data as below.

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

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

Step 2: Understanding the concept of average power and Plank’s relation:

The rate of energy flow in every pulse is called average power.

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 equation of Plank's relation, to get the energy of the photon that is being emitted by the laser. Now, using this value in the power equation considering the criteria of emitted photons, to find the average power of the emitted pulse.

Formulas:

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

$E=\frac{hc}{\lambda }$ ….. (1)

The average power of the emitted pulse is,

${P}_{avg}=\frac{\left({N}_{1}-{N}_{2}\right){E}_{photon}}{\Delta t}$ ….. (2)

Step 3: Calculation of the average power:

At first, the energy of the photon is calculated using the given data in equation (1) as follows:

$E=\frac{\left(6.63×{10}^{-34}J.s\right)\left(3×{10}^{8}m/s\right)}{694×{10}^{-9}m}\phantom{\rule{0ex}{0ex}}=2.87×{10}^{-19}\mathrm{J}$

Now, according to the problem consider the known data due to the absorption of the emitted photons of the Cr ions in the excited state by the ions in the ground state.as below.

The number of atoms in the excited state,${N}_{1}=0.6$ and

The number of atoms in the ground state, ${N}_{2}=0.4$

Thus, the average power emitted during the pulse is given using equation (2) and, the given data is as follows.

${P}_{avg}=\frac{\left(0.6-0.4\right)\left(2.87×{10}^{19}J\right)}{0.2×{10}^{-6}s}\phantom{\rule{0ex}{0ex}}=1.1×{10}^{6}J/s\phantom{\rule{0ex}{0ex}}=1.1×{10}^{6}W$

Hence, the value of the average power is $1.1×{10}^{6}W$.