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The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm600\text{ nm}, when it delivers the power of 3.3×10−3 watt3.3 \times 10^{-3}\text{ watt} will be (h=6.6×10−34 J sh = 6.6 \times 10^{-34}\text{ J s})

A

101810^{18}

B

101710^{17}

C

101610^{16}

D

101510^{15}

Step-by-Step Solution

The power of a source is given as P=Et=nt(hcλ)P = \frac{E}{t} = \frac{n}{t} \left(\frac{hc}{\lambda}\right). Therefore, nt=P(hc/λ)\frac{n}{t} = \frac{P}{(hc/\lambda)}. Here nt\frac{n}{t} is number of photons emitted per second. nt=3.3×10−3×6×10−76.6×10−34×3×108=1016 photons per second\frac{n}{t} = \frac{3.3 \times 10^{-3} \times 6 \times 10^{-7}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 10^{16}\text{ photons per second}.

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