Question

A certain Ruby Laser emits 100J pulses of light whose wavelength is 6940 A0 . What is
the minimum no of Cr3+ ions in the laser?

Answer 1

Given:

A certain Ruby Laser emits 100J pulses of light whose wavelength is 6940 A0 .

To find:

What is  the minimum no of Cr3+ ions in the laser?

Solution:

From given, we have,

A certain Ruby Laser emits 100J pulses of light whose wavelength is 6940 A0.

E = 100 J

λ  = 6940 °A = 6.94 × 10^{-7} m

h = 6.63 × 10^{-34}

c = 3 × 10^{8}

we use the formula,

E = n hc/λ

⇒ n = Eλ/hc

n = (100 × 6.94 × 10^{-7})/(6.63 × 10^{-34} × 3 × 10^{8})

n = (6.94 × 10^{-5})/(1.989 × 10^{-25})

n = 3.489 × 10^{20}

Therefore, the minimum number of Cr3+ ions in the laser are 3.489 × 10^{20}