The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600

Question

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

Accepted Answer

The power of a source is given as $P = \frac{E}{t} = \frac{n}{t} \left(\frac{hc}{\lambda}ight)$. Therefore, $\frac{n}{t} = \frac{P}{(hc/\lambda)}$. Here $\frac{n}{t}$ is number of photons emitted per second. $\frac{n}{t} = \frac{3.3 	imes 10^{-3} 	imes 6 	imes 10^{-7}}{6.6 	imes 10^{-34} 	imes 3 	imes 10^8} = 10^{16}	ext{ photons per second}$.

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