# Does Planck's relation apply to radio waves?

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## Main Question or Discussion Point

I have some doubts about whether Planck's relation (E=hf) applies to radio waves. This has been bugging me because trying to apply Planck's relation to radio frequency results in some inconsistencies that I've been unable to resolve. BTW, I have no physics training, so please go easy on me Radio waves can be generated by alternating current. The frequency of the current is usually determined by a crystal or a LC circuit. For example, a 28 MHz crystal will generate an alternating current at 28 MHz (hopefully). When used in a radio transmitter, the resulting RF is 28 MHz.

Is there a relationship between the frequency of this radio wave and its energy? In other words, does Planck's relation work here? And if so, why? Assuming it does, would the energy of this radio wave be 1.86e-26 J.s?

As I understand it, Planck's relation only works with electron transition levels. Here's where I see a problem with applying Planck's relation to RF: We can generate RF with very specific frequencies, which translates into specific energies. But we know the energies from electrons jumping down orbitals are quantized. They can emit energy at only a finite number of frequencies, so I don't see how it's possible for radio waves to have an infinite number of frequencies.

But for the sake of argument, let's say there it is possible for an atom to emit energy at any arbitrary frequency. How can the alternating current frequency (28 MHz) cause electrons in an atom to jump up and down to the right orbital to emit energy at that precise frequency?

So far the only conclusion that makes sense is that Planck's relation has nothing to do with radio waves, and that comparing electron energy levels with RF energy is comparing apples and oranges.

Thanks!

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PeroK
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Any accelerating charge emits EM radiation. I don't know anything about radio engineering, but an accelerating charge emits radiation according to Maxwell's equations. It isn't solely due to electron transitions in an atom.

However, when EM radiation interacts with matter, it does so in discrete amounts of energy related to its frequency. To understand this goes beyond the classical theory of EM and you need the quantum theory of electrodynamics.

• Ben DA and sophiecentaur
I don't know anything about radio engineering, but an accelerating charge emits radiation according to Maxwell's equations. It isn't solely due to electron transitions in an atom.
Thanks. Wouldn't that then imply that Planck's relation can't be applied to radio waves?

PeroK
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Thanks. Wouldn't that then imply that Planck's relation can't be applied to radio waves?
No. Planck's relation applies to all EM radiation, however it is generated.

I imagine that ultimately in a radio transmitter the radiation at the QM level is created by transitions of a bound system. Which needn't be an atom.

In any case, $E = hf$ applies to all EM radiation, whatever the source.

• Ben DA
sophiecentaur
Gold Member
Thanks. Wouldn't that then imply that Planck's relation can't be applied to radio waves?
Why shouldn't it? The charges in the aerial are accelerating back and forth along a dipole. The energy of the photons is much much lower because, unlike the few eV of energy for optical energy level changes for electrons bound in atoms, the energies required for a single electron to move about in a metal wire is much less than μV.

• Ben DA
The energy of the photons is much much lower because, unlike the few eV of energy for optical energy level changes for electrons bound in atoms, the energies required for a single electron to move about in a metal wire is much less than μV.
If I understood the math it would probably make more sense. I'm assuming that the AC oscillation frequency is the same as the frequency of the emitted photons. But perhaps that's not the case.

Staff Emeritus
As I understand it, Planck's relation only works with electron transition levels.
There's your problem. That's not true.

• Ben DA
sophiecentaur
Gold Member
Again - why not? The electrons in the wire are not part of any particular atom. They are 'dissociated' conduction electrons and they interact with the whole structure. The energy transitions are tiny and the frequency of the wave is the frequency of the transmitter. (Obvs??? )

• Ben DA
PAllen
note, to the extent that a radio signal is modeled well by a classical approximation, the number of photons is enormous. Thus, the signal energy has essentially nothing to do with the photon energy, which is what the Planck equation gives. You can and should forget about photons in an ordinary radio transmission and reception problem. But this does not in any way mean the Planck equation isn’t true at the essentially unobservable quantum level.

Again - why not? The electrons in the wire are not part of any particular atom.
I suppose because some of the energy is lost to heat so the RF would have to be less than the AC oscillation frequency.

Perhaps at the risk of muddying the waters, here's a slightly different example: AC power to homes in the US oscillates at about 60 Hz, but it can be used to power a transmitter that produces RF in much higher frequency ranges (kHz, MHz, GHz, etc.) This leads me to think that "frequency" when speaking of an AC oscillation can't be the same as frequency when applied to radio waves. Hence my suspicion that I'm comparing apples and oranges.

Staff Emeritus
here's a slightly different example: AC power to homes in the US oscillates at about 60 Hz, but it can be used to power a transmitter that produces RF in much higher frequency ranges (kHz, MHz, GHz, etc.)
Sure. You can even power RF with DC. Your cell phone does that. And of course the power supply is a different frequency as the transmitter. Otherwise you'd need a different power supply for every channel.

his leads me to think that "frequency" when speaking of an AC oscillation can't be the same as frequency when applied to radio waves.
Where do you get that? All that means is that the power supply doesn't need to be the same frequency as the radio itself.

As Mark Twain said, “What gets us into trouble is not what we don't know. It's what we know for sure that just ain't so.”

• Ben DA
Where do you get that? All that means is that the power supply doesn't need to be the same frequency as the radio itself.
I'm drawing that conclusion from the (probably wrong) math of using a 60 Hz current to power a radio transmitter transmitting at 100 MHz:

For 60 Hz, E1= 60h = 3.9756e-32 J
For 100 MHz, E2= 100,000,000h = 6.626e-26 J

E1 < E2 so where is the extra energy coming from?

As Mark Twain said, “What gets us into trouble is not what we don't know. It's what we know for sure that just ain't so.”
I know I'm missing something here. I just don't know what it is yet.

Last edited:
PAllen
I'm drawing that conclusion from the (probably wrong) math of using a 60 Hz current to power a radio transmitter transmitting at 100 MHz:

For 60 Hz, E1=60h = J.s = 3.9756e-32 J
For 100 MHz, E2=100,000,000h = 6.626e-26 J

E1 < E2 so where is the extra energy coming from?

I know I'm missing something here. I just don't know what it is yet.
You are missing that you are completely misunderstanding and misusing the Planck equation. You can see from both photon energies that they are effectively zero on the energy scale of the transmission process. The photons simply have nothing to do with this problem.

As to your (silly) extra energy question, a silly analogy is: I take a million grains of sand and put them into a box, and I do this for many boxes. The mass of the boxes is in no way limited by the mass of a sand grain.

You are missing that you are completely misunderstanding and misusing the Planck equation.
Thanks for the correction. This is why I started by asking how to properly use Planck's relation.

You can see from both photon energies that they are effectively zero on the energy scale of the transmission process. The photons simply have nothing to do with this problem.
The energies I used were not photon energies. They were AC oscillation frequencies. If I understand you, I should not be using Planck's relation with AC oscillation frequencies.

PAllen
The energies I used were not photon energies. They were AC oscillation frequencies. If I understand you, I should not be using Planck's relation with AC oscillation frequencies.
Planck's equation gives photon energies. When these are infinitesimal compared to the energies at issue, then they are irrelevant, and you use continuous methods.

• PeroK and Ben DA
Planck's equation gives photon energies. When these are infinitesimal compared to the energies at issue, then they are irrelevant, and you use continuous methods.
Thank you! That answers my question.

davenn
Gold Member
I'm drawing that conclusion from the (probably wrong) math of using a 60 Hz current to power a radio transmitter transmitting at 100 MHz:

yes, that is wrong you have totally ignored all the technology that goes into making a transmitter
of any frequency.
As V50 said, "your mobile phone uses a DC supply " ... how do you think that happens ?
.... and it generates frequencies around 1000 MHz, 1GHz, plus or minus a few 100 MHz depending on the band it's operating on.

I know I'm missing something here. I just don't know what it is yet.
this .......
There are oscillator circuits for the transmitter and receiver stages within the mobile phone or for that matter, any other radio transceiver circuit that generate all the different RF frequencies

The frequency of the mains power 50/60 Hz is totally irrelevent

Dave

• sophiecentaur
sophiecentaur
Gold Member
I suppose because some of the energy is lost to heat so the RF would have to be less than the AC oscillation frequency.
I can see that you are trying to make 'connections' between all the concepts you are aware of. Not a bad thing in principle but it can lead you up a gumtree if you don't know the whole story. The thermal energy from the passage of the electrons is nothing to do with the radiated RF energy. It's a different interaction; there is interaction with free space and there is interaction with the metal lattice. There is no way that any different RF frequency from the input from the transmitter can be involved. I think you are confusing the photon energy with the energy from the power supply.
The mains frequency is also totally irrelevant- the same transmitter can be battery powered or use US mains 60Hz or European mains 50Hz. No supply current needs to flow through the antenna - again, why would it be needed?

Note:
You must avoid bringing the simple Hydrogen Atom ideas into this model. But, even if you do, consider a gas discharge tube, producing monochromatic light. It will involve an electron beam which will cause ionised atoms to emit photons and also heat up the low pressure gas inside the tube. This will not significantly change the frequency of the light photons. (This effect would be 'measurable' but it's another, secondary consideration - before someone else jumps on my head)

• Ben DA