Question

A ruby laser emits a $100MW$, $10-ns$-long pulse of light with a wavelength of $690nm$. How many chromium atoms undergo stimulated emission to generate this pulse?

Short answer

The number of chromium atoms undergo stimulated emission to generate this pulse are $3.49\times {10}^8$

Step-by-step solution

Given information

We are given that power of ruby laser is $100MW$,wavelength is $690nm$, time $10ns$.

We need to find that how many chromium atoms undergo stimulated emission to generate this pulse

Simplify

To calculate number of photons , we will use equation ,$∆E=N.E_p$where $N$is number of photons and $E_p$is energy of photons . For that matter we need to calculate ,$∆E$;

We will use formula of power , $P=\frac{∆E}{∆t}\to ∆E=P∆t$

Substituting the values we get,$∆E=\left(1\times {10}^{8}W\right)\left(1\times {10}^{-8}s\right)=1J$

We will also use $E_p=\frac{hc}{\lambda }$

$N=\frac{∆E}{{\displaystyle \frac{hc}{\lambda }}}=\frac{∆E\lambda }{hc}$

Substituting the values we get,$N=\frac{1\times 690}{\left(6.63\times {10}^{-34}Js\right)\left(3\times {10}^{8}m/s\right)}=3.49\times {10}^{18}$