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Grade: 12th pass
        
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?
one year ago

Answers : (1)

Khimraj
3008 Points
							
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........................
one year ago
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