Question

A certain ruby laser emits 1.00-J pulses of light whose wavelength is 694 nm. What is the minimum number of Cr³+ ions in the ruby?

Answer 1

Energy of the ions:

$$E=\frac{hc}\lambda=2.86·10^{-19}\text{ J}.$$

One pulse is 1 J. So, there are n ions in the 1-J pulse:

$$n=\frac{E_p}{E}=\frac{1}{2.86·10^{-19}}=3.49·10^{18}.$$