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(B) Given: Wavelength $\lambda = 300 \ m$,Power $P = 10 \ kW = 10^4 \ W$.Energy of one photon $E = \frac{hc}{\lambda}$.The number of photons emitted p

Vedclass · Accepted Answer

$1.5 	imes 10^{31}$

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(B) Given: Wavelength $\lambda = 300 \ m$,Power $P = 10 \ kW = 10^4 \ W$.Energy of one photon $E = \frac{hc}{\lambda}$.The number of photons emitted per second $n = \frac{P}{E} = \frac{P \lambda}{hc}$.Substituting the values: $n = \frac{10^4 	imes 300}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$.$n = \frac{3 	imes 10^6}{19.89 	imes 10^{-26}} \approx 0.1508 	imes 10^{32} = 1.508 	imes 10^{31}$.Thus,the number of photons emitted per second is approximately $1.5 	imes 10^{31}$.

$A$ radio station broadcasts at a wavelength of $300 \ m$. If the radiating power of the transmitter is $10 \ kW$,then the number of photons emitted per second is:

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