The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of $3.3 \times 10^{-3}$ watt will be: ($h=6.6\times 10^{-34}$Js)

(1) $10^{18}$

(2) $10^{17}$

(3) $10^{16}$

(4) $10^{15}$

*Solution*

Let there be n number of photons per second.

Total energy per second of n number of photons = $n. \frac {hc}{\lambda} = 3.3 \times 10^{-3}$

$\therefore n=\frac {3.3 \times 10^{-3} \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} =10^{16}$

Answer: (3)