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(B) The power $P$ of the transmitter is given by $P = n \frac{hc}{\lambda}$,where $n$ is the number of photons emitted per second.Given:Power $P = 10\

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$10^{31}$

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(B) The power $P$ of the transmitter is given by $P = n \frac{hc}{\lambda}$,where $n$ is the number of photons emitted per second.Given:Power $P = 10\, kW = 10^4\, W$Wavelength $\lambda = 500\, m$Planck's constant $h \approx 6.63 	imes 10^{-34}\, J\cdot s$Speed of light $c = 3 	imes 10^8\, m/s$Rearranging the formula for $n$:$n = \frac{P \lambda}{hc}$Substituting the values:$n = \frac{10^4 	imes 500}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$$n = \frac{5 	imes 10^6}{19.89 	imes 10^{-26}}$$n \approx 0.251 	imes 10^{32} \approx 2.51 	imes 10^{31}$Therefore,the order of magnitude is $10^{31}$.

$A$ $10\, kW$ 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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