A cylindrical rod of some laser material  $5\times {10}^{-2}m$ long contains $2\times {10}^{25}$ ions per $m^3$. On excitation all the ions are in the upper energy level and de-excite simultaneously emitting photons in the same direction, creating a pulse of radiation of wavelength $6.6\times {10}^{-7}m$. If the pulse lasts for ${10}^{-7}$ second, calculate the average power of the laser during the pulse? (Diameter of the rod = 10 mm)

Total number of ions in the rod is

$N=\left(Number of ions per unit volume\right)\times \left(Volume of the rod\right)$

$\Rightarrow N=\left(2\times {10}^{25}\right)\times \left\{\mathrm{\pi }\times (5\times {10}^{-3}{)}^2\times \left(5\times {10}^{-2}\right)\right\}$

$\Rightarrow N=7.85\times {10}^{19}$

As all the ions de-excite simultaneously, the number of photons emitted in the same direction is also $7.85\times {10}^{19}$.

So, the energy contained in a pulse of radiation of wavelength $6.6\times {10}^{-7}m$ is

$E=N\left(\frac{hc}{\lambda }\right)$

$\Rightarrow E=\left(7.85\times {10}^{19}\right)\times \frac{\left(6.6\times {10}^{-34}\right)\times \left(3\times {10}^8\right)}{\left(6.6\times {10}^{-7}\right)}=23.55 \mathrm{J}$

Average power $P=\frac{Energy}{Time}$

$P=\frac{23.55}{{10}^{-7}}=23.55\times {10}^{7}W=235.5 \mathrm{MW}$