A transmitter of power 10 text{ kW} emits rad | vedclass

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(B) The power $P$ of the transmitter is given by $P = \frac{n E_{photon}}{t}$,where $n$ is the number of photons emitted in time $t$ and $E_{photon} =

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$10^{31}$

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(B) The power $P$ of the transmitter is given by $P = \frac{n E_{photon}}{t}$,where $n$ is the number of photons emitted in time $t$ and $E_{photon} = \frac{hc}{\lambda}$.Thus,the number of photons emitted per second is $\frac{n}{t} = \frac{P \lambda}{hc}$.Given: $P = 10 	ext{ kW} = 10^4 	ext{ W}$,$\lambda = 500 	ext{ m}$,$h = 6.63 	imes 10^{-34} 	ext{ J s}$,and $c = 3 	imes 10^8 	ext{ m/s}$.Substituting the values:$\frac{n}{t} = \frac{10^4 	imes 500}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$$\frac{n}{t} = \frac{5 	imes 10^6}{19.89 	imes 10^{-26}}$$\frac{n}{t} \approx 0.251 	imes 10^{32} = 2.51 	imes 10^{31}$.Therefore,the order of magnitude is $10^{31}$.

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

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