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(B) Given: Power $P = 10 \ kW = 10^4 \ W$,Wavelength $\lambda = 500 \ m$.Energy of one photon is $E = \frac{hc}{\lambda}$.The number of photons emitte

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$10^{30}$

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(B) Given: Power $P = 10 \ kW = 10^4 \ W$,Wavelength $\lambda = 500 \ m$.Energy of one photon is $E = \frac{hc}{\lambda}$.The number of photons emitted per second $n$ is given by $n = \frac{P}{E} = \frac{P \lambda}{hc}$.Substituting the values: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$.$n = \frac{10^4 	imes 500}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8} = \frac{5 	imes 10^6}{19.89 	imes 10^{-26}} \approx 0.251 	imes 10^{32} = 2.51 	imes 10^{31}$.The order of magnitude is $10^{31}$,which is closest to $10^{30}$ among the given options.

$A$ $10 \ kW$ transmitter emits radio waves of wavelength $500 \ m$. The number of photons emitted per second by the transmitter is of the order of:

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