The average number of photons emitted per second by a laser of power $6.6 \times 10^{-3} \mathrm{~W}$…
The average number of photons emitted per second by a laser of power $6.6 \times 10^{-3} \mathrm{~W}$ producing a light of wavelength $600 \mathrm{~nm}$ is (Planck's constant, $h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )
$2 \times 10^{16}$
$3 \times 10^{16}$
$4 \times 10^{16}$
$6 \times 10^{16}$
Solution
Photons emitted per second
$n=\frac{\text { Power }}{\text { Energy of } 1 \text { photon }}$
$=\frac{P}{\left(\frac{h c}{\lambda}\right)}=\frac{P \lambda}{h c}$
Note $\therefore h c=1240 \mathrm{eV}-\mathrm{nm}$
$1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$
As given that, $\lambda=600 \mathrm{~nm}$
so, $\frac{h c}{\lambda}=\frac{1240 \mathrm{eV}-\mathrm{nm}}{600 \mathrm{~nm}}$
$\begin{aligned} & =2.067 \mathrm{eV} \\ & =2.067 \times 1.6 \times 10^{-19} \mathrm{~J}=3.3 \times 10^{-19} \mathrm{~J}\end{aligned}$
So, number of photons / sec
$=n=\frac{P}{\left(\frac{h c}{\lambda}\right)}=\frac{6.6 \times 10^{-3}}{3.3 \times 10^{-19}}$
$=2 \times 10^{16}$ photons per second.