A radio transmitter radiates 1 text{ kW} powe | vedclass

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(C) The power $P$ radiated by the transmitter is given by $P = \frac{E}{t}$, where $E$ is the total energy emitted in time $t$.Since $E = n 	imes E_{

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$10^{30}$

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(C) The power $P$ radiated by the transmitter is given by $P = \frac{E}{t}$, where $E$ is the total energy emitted in time $t$.Since $E = n 	imes E_{	ext{photon}} = n 	imes \frac{hc}{\lambda}$, where $n$ is the number of photons emitted per second, we have $P = \frac{nhc}{\lambda}$.Given: $P = 1 	ext{ kW} = 10^3 	ext{ W}$, $\lambda = 198.6 	ext{ m}$, $h = 6.63 	imes 10^{-34} 	ext{ J s}$, and $c = 3 	imes 10^8 	ext{ m/s}$.Substituting the values: $10^3 = \frac{n 	imes 6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{198.6}$.$n = \frac{10^3 	imes 198.6}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8} \approx \frac{198600}{19.89} 	imes 10^{26} \approx 10^{30}$.Thus, the number of photons emitted per second is $10^{30}$.

$A$ radio transmitter radiates $1 \text{ kW}$ power at a wavelength of $198.6 \text{ m}$. How many photons does it emit per second?

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