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Question

(A) The power $P$ of the transmitter is the energy emitted per second. The energy of a single photon is given by $E = h
u$,where $h = 6.6 	imes 10^{

Vedclass · Accepted Answer

$1.72 	imes 10^{31}$

Vedclass · Answer

(A) The power $P$ of the transmitter is the energy emitted per second. The energy of a single photon is given by $E = h
u$,where $h = 6.6 	imes 10^{-34} \, J \cdot s$ is Planck's constant and $
u$ is the frequency.If $n$ is the number of photons emitted per second,then the total power is $P = n 	imes E = n 	imes h
u$.Therefore,$n = \frac{P}{h
u}$.Given: $P = 10 \, kW = 10^4 \, W$ and $
u = 880 \, kHz = 880 	imes 10^3 \, Hz$.Substituting the values:$n = \frac{10^4}{(6.6 	imes 10^{-34}) 	imes (880 	imes 10^3)}$$n = \frac{10^4}{5808 	imes 10^{-31}}$$n = \frac{10^4}{5.808 	imes 10^{-28}}$$n \approx 1.72 	imes 10^{31}$ photons per second.

$A$ radio transmitter operates at a frequency of $880 \, kHz$ and a power of $10 \, kW$. The number of photons emitted per second is:

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