A cylindrical rod of some laser material 5 x 10^-2 m long and 10^-2 m in diameter contains 2 x 10²⁵ ions per m³. On excitation all the ions are in the upper energy level and de-excite simultaneously emitting photons in the same direction contained in a pulse of radiation of wavelength 6.6 x 10^-7 m. If the pulse lasts for 10^-7s, calculate the average power of the laser (in MW) during the pulse.

Total number of ions in the rod

N = number of ions per unit volume x volume of rod 

$\Rightarrow \mathrm{N}=\left(2\times {10}^{25}\right)\times \left\{3.14\times (0.5\times {10}^{-2}{)}^2\times \left(5\times {10}^{-2}\right)\right\}=7.85\times {10}^{19}$

The number of photons emitted in one direction is equal to the total number of ions because all ions are excited. Now excited energy 

E = number of excited photons $\times$ energy of photon

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

Average power

$\mathrm{P}=\frac{\text{ Energy }}{\text{ Time }}=\frac{23.55\text{ joule }}{{10}^{-7}\text{ second }}=23.55\times {10}^7\mathrm{W}=235.5\mathrm{MW}$