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Q62P

Expert-verifiedFound in: Page 1250

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 $1.1\times {10}^{6}\mathrm{W}$.

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

The output pulse last, $\u2206\mathrm{t}=2\mathrm{\mu s}$

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,$h=6.63\times {10}^{-34}J\cdot s$

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

**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,

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

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

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

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.

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

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

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