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(B) The power $P$ of the transmitter is given by the total energy radiated per unit time,$P = \frac{E}{t}$.Since the energy of a single photon is $E_{

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$1.5 	imes 10^{31}$

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(B) The power $P$ of the transmitter is given by the total energy radiated per unit time,$P = \frac{E}{t}$.Since the energy of a single photon is $E_{photon} = \frac{hc}{\lambda}$,the total energy radiated by $n$ photons in time $t$ is $E = n 	imes \frac{hc}{\lambda}$.Therefore,the number of photons radiated per second is $\frac{n}{t} = \frac{P \lambda}{hc}$.Given: $P = 10 \ kW = 10^4 \ W$,$\lambda = 300 \ m$,$h = 6.6 	imes 10^{-34} \ J \cdot s$,and $c = 3 	imes 10^8 \ m/s$.Substituting the values:$\frac{n}{t} = \frac{10^4 	imes 300}{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}$$\frac{n}{t} = \frac{3 	imes 10^6}{19.8 	imes 10^{-26}}$$\frac{n}{t} = \frac{3}{19.8} 	imes 10^{32} \approx 0.1515 	imes 10^{32} = 1.515 	imes 10^{31} \approx 1.5 	imes 10^{31}$ photons per second.

An $AIR$ station is broadcasting waves of wavelength $300 \ m$. If the radiating power of the transmitter is $10 \ kW$,then the number of photons radiated per second is:

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