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(B) The power $P$ of the light source is given by $P = n \cdot E_{photon}$,where $n$ is the number of photons emitted per second and $E_{photon} = \fr

Vedclass · Accepted Answer

$10^{20}$

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(B) The power $P$ of the light source is given by $P = n \cdot E_{photon}$,where $n$ is the number of photons emitted per second and $E_{photon} = \frac{hc}{\lambda}$.Thus,$n = \frac{P \cdot \lambda}{hc}$.Given: $P = 100\,W$,$\lambda = 5000 	imes 10^{-10}\,m$,$h = 6.63 	imes 10^{-34}\,J\cdot s$,and $c = 3 	imes 10^8\,m/s$.Substituting the values:$n = \frac{100 	imes 5000 	imes 10^{-10}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$$n = \frac{5 	imes 10^{-5}}{19.89 	imes 10^{-26}} \approx 0.25 	imes 10^{21} = 2.5 	imes 10^{20}$.Therefore,the rate of emission of photons is of the order of $10^{20}$.

$A$ $100\,W$ light source is emitting radiation of wavelength $5000\,\mathring{A}$. The rate of emission of photons is of the order of:

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