Radio signals are being transmitted in a frequency of $ 8.4 \times 10^9 \text{s}^{-1} $ and being received by an antenna that is capable of receiving power of $ 4 \times 10^{-21} \text{Watt} $ ($ 1 \, \text{Watt} = 1 \, \text{J s}^{-1} $ ) .

Estimate that number of photons per second of this electromagnetic radiation that this antenna is capable of receiving.

I can easily calculate the number of photons per second that this radio produces, dividing the emitted power by the energy of one photon,

$\hbar\omega=h\nu = h\times8.4 \times 10^9 s^{-1}$

but have no idea how to calculate the number of photons that this antenne can receive.


How can you calculate the number of produced photons if the only known thing is the frequency?

To calculate the number of photons received (or emitted) per second, you have to divide the received (emitted) power by the energy of one photon $\hbar\omega=h\nu$.

  • $\begingroup$ But how does this depend on the frequency of the radio signals? I can't understand it... Can you please detail a bit? Thanks ! $\endgroup$ – joshua Sep 17 '12 at 18:44
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    $\begingroup$ Frequency of the electromagnetic wave $\nu$ is the frequency of photons constituting this wave. $\endgroup$ – Vladimir Kalitvianski Sep 18 '12 at 11:38

It doesn't matter how strong the transmitting station is. The receiving antenna is only absorbing a tiny amount of power. The idea of the calculation is to show that even in a "classical" situation like a receiving antenna, at some level the "granular" nature of electromagnetic energy makes itself apparent.

At least that seems to be the way the numbers are set up. It seems to me to be a very dishonest calculation. I don't believe there is any way to observe the quantum nature of radiation by looking at the current in a radio receiver.

  • $\begingroup$ So what you're actually saying is that we don't really care about the radio signals being transmitted? Only about the power of the antenna? Thanks! $\endgroup$ – joshua Sep 18 '12 at 11:22
  • $\begingroup$ Technically yes, but again, it's not a physically realistic problem. A real antenna receives more power depending on what the transmitter puts out, but in this problem it only says the antenna is "capable" or receiving so and so many watts, without regard to the transmitted power. That's not physically realistic. $\endgroup$ – Marty Green Sep 18 '12 at 12:35
  • $\begingroup$ @MartyGreen: The indicated power is really small so it is obviously the minimum power the antenna is capable of receiving. $\endgroup$ – Vladimir Kalitvianski Sep 19 '12 at 14:05
  • $\begingroup$ Vladimir, I'm quite sure that's not the intention of the question. There's no such thing as the minimum or maximum power of an antenna. The power received is always proportional to the incident power. I'm quite sure the person who designed the question chose a very small power in a misguided attempt to demonstrate that quantum effects can be demonstrated in a classical situation. In fact, even if we accept the numbers given, there is no way to detect the supposed granularity of the received power by measuring the antenna current. $\endgroup$ – Marty Green Sep 19 '12 at 18:06
  • $\begingroup$ @MartyGreen: You are right, of course. This is maybe the minimum distinguishable power with respect to the receiver own noise. $\endgroup$ – Vladimir Kalitvianski Sep 19 '12 at 19:16

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