# A certain ruby laser emits 1.00 Joule pulse of light whose wavelength is 694 nm. What is the minimum number of Cr3+ ions in ruby?

Khimraj
3007 Points
4 years ago
h = 6.63*10-34  c = 3*108   $\lambda$ = 694*10-9
The energy of a 694 nm photon is
E = hc/$\lambda$ = 2.866*10-19 J
In each pulse of light, there are
n = 1/E photons, because 1 joule per pulse divided by joules per photon gives photons per pulse.
n = 3.489*1018 photons
There must be at least that many chromium 3+ ions in the crystal.
Hope it clears........................