Q 12-11-011NEETNEET 2021Top questionEasy
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 : ($h = 6.6 \times 10^{-34}$ J s)
Answer: (D) $10^{16}$
Energy of one photon:
$$E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}} = 3.3 \times 10^{-19}\ \text{J}$$
$$N = \frac{P}{E} = \frac{3.3 \times 10^{-3}}{3.3 \times 10^{-19}} = 10^{16}\ \text{per second}$$
Solution by Sreeraj P, M.Sc Physics