Question: 16. 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 : ( \mathrm{h}=6.6 \times 10^{-34} \mathrm{Js} )
(1) 10^{18}
(2) 10^{17}
(3) 10^{16}
(4) 10^{15}
Answer: Option (3)
Explanation:
The energy of a single photon is given by E=\frac{hc}{\lambda}.
The given wavelength is \lambda = 600\,\text{nm} = 600 \times 10^{-9}\,\text{m}.
Substituting the values h = 6.6 \times 10^{-34}\,\text{Js} and c = 3 \times 10^{8}\,\text{m s}^{-1},
E = \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{600 \times 10^{-9}}.
Simplifying, E = 3.3 \times 10^{-19}\,\text{J}.
The power of the source is the energy emitted per second,
so the number of photons emitted per second is given by N = \frac{P}{E}.
Given power P = 3.3 \times 10^{-3}\,\text{W},
N = \frac{3.3 \times 10^{-3}}{3.3 \times 10^{-19}} = 10^{16}.
Therefore, the number of photons emitted per second is 10^{16},
and the correct answer is Option (3).