## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$3.47\times 10^{18}~~$ atoms undergo stimulated emission.
We can find the energy of each photon: $E = \frac{hc}{\lambda}$ $E = \frac{(6.63\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{690\times 10^{-9}~m}$ $E = 2.8826\times 10^{-19}~J$ We can find $E_p$, the energy emitted during the pulse: $E_p = (100\times 10^6~W)(10\times 10^{-9}~s) = 1.0~J$ We can find the number of photons emitted during this pulse: $\frac{1.0~J}{2.8826\times 10^{-19}~J} = 3.47\times 10^{18}$ $3.47\times 10^{18}~~$ atoms undergo stimulated emission.