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Does a (classical) radio wave with a given amplitude carry more power if that wave is at a higher frequency than at a lower frequency?

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Energy density in EM field is given by $$ u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2 \mu_0} B^2 $$ and energy flux (energy crossing unit area per unit time is) $$ {\bf S} = {\bf E} \times {\bf H} = \frac{1}{\mu_0} {\bf E} \times {\bf B} $$ where the second version applies in vacuum.

It follows that the answer to your question is no: there is no dependence on frequency if the amplitude is given.

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Perhaps you need to ask a similar question on Amateur Radio StackExchange. There they could explain to you the observable details of how a given antenna rod and a given power from a wave generator change the output of the antenna. I will give you an idea of how the achievable output power of a radio wave changes at constant input power.

The power of a radio wave depends from the number of involved surface electrons on the antenna rod, the rate of emission from these electrons per time and the wavelength of the emitted photons.

Depending from the length of the rod there is an optimal frequency of the wave generator. For this frequency the radiated energy has its maximum. For lower frequency the electrons get accelerated forth and back less often and the number of emitted photons decreases. For higher frequency among other things the heat losses increases and the radiated power goes down.

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The frequency is proportional to the energy of single photons according to $E=h \cdot f$. However, if the amplitude is given and fixed, less photons would be emitted from the radio having higher frequency. So the power would be the same.

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  • $\begingroup$ @Andrew Steane: Cool, didn't know that. How about a lamp? If you have two lasers emitting the same amplitude but different wavelength. Do they differ in emitted optical power? Should behave similarly I think...? $\endgroup$ Commented Jun 2, 2021 at 8:53
  • $\begingroup$ @Andrew Steane: I edited the answer, is it correct now? Many thanks for your help! $\endgroup$ Commented Jun 2, 2021 at 9:03
  • $\begingroup$ Yes it is correct now so I have removed my previous comment (I did not vote either way). $\endgroup$ Commented Jun 2, 2021 at 9:33
  • $\begingroup$ Thanks, Andrew Steane! I really like learning things (and actually therefore I even like making mistakes...) $\endgroup$ Commented Jun 2, 2021 at 9:35

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