
#1
Feb1113, 07:36 PM

P: 1,362

Should it be straightforward to estimate the number of radio wavelength photons emitted per electron per cycle (or is it number of electron cycles per photon) of a half wavelength radio antenna running at 100MHz and 1000 watts power output? Is it many photons per cycle per electron or is it many cycles per photon per electron? Or how far does the typical electron "travel" up and down the antenna before it emits a photon? Here there are 100,000,000 cycles per second.
Hope I'm clear. Thanks for any help! 



#2
Feb1113, 09:04 PM

Sci Advisor
PF Gold
P: 2,247

Hi there
Trillions of photons would be emitted. The number of photons is so huge that it makes no sense to talk about individual photons or even the collective numbers of them. Rather we measure the RF EM field strength in eg. uV / metre etc Dave 



#3
Feb1113, 09:21 PM

PF Gold
P: 11,058

At 100 MHz each photon has energy equal to 6.62148x10^{26} joules.
One watt is equal to one joule per second, so 1000 watts is 1000 joules per second. This equates to 1000/6.62148x10^{26}, or about 2x10^{24} photons per second. As Davenn said, this number is so large that we don't even use it. One trillion is only 10^{12}, so there is literally 2 trillion trillion photons! (2 x 1 trillion x 1 trillion) 



#4
Feb1213, 06:10 AM

P: 1,362

Estimate of photons/electron/cycle for radio antenna?
We know roughly how many conducting electrons there are in the antenna, fewer if we account for the skin effect, for a given power output we know the number of photons emitted per cycle, so could we not come up with an estimate that I wanted in the first post? Just looking for an estimate.
Thanks! 



#5
Feb1213, 06:28 AM

Mentor
P: 28,841

Zz. 



#6
Feb1213, 11:02 AM

Sci Advisor
PF Gold
P: 11,399

@Spinnor
I can sympathise with the fact that you want to relate something 'new' to something 'old' that you already feel you understand but, as has been said already, it really has no relevance to practical reality. On Photon is not related to one Electron because the energy that is transferred between an EM wave and a piece of metal is due to an interaction with the metal as a whole and not to individual atoms. But you now have your figure of 10^{26} Photons per second and you can find a number for the density of free electrons in a metal in this link. Choose a size and material for your dipole, assume that all the electrons along its length are involved in the emission of the Photons and then you will have the number you want. (I can't be bothered, personally, but have a go, by all means.) 



#7
Feb1313, 10:31 AM

P: 1,362

Let λ = 1m, then nu = 3E8 cycles per second. Let the antenna have a length of λ/2, be made of steel, and have a diameter, D, of 5E3m. The density of free electrons, ρ, is about 1E29 electrons/m^3. The skin effect results in only the electrons in a surface layer, δ, of about 3E6m being accelerated. The volume of the antenna that gives rise to photons is about, V = ∏Dδλ/2 ≈ 3.14x5E3x3E6x1/2 ≈ 74E9 ≈ 8E8m^3 An estimate for the number of electron "involved" in "producing" photons is ρV ≈ 1E29x8E8m^3 = 8E21 electrons For this antenna let Vmax = 73 volts, then Imax = V/R = 1amp. The max power is about 73 watts so the average power is about half that, call it 40 watts. Each photon has an energy, E = hnu ≈ 6E34x3E8 ≈ 2E25 J. The number of photons emitted per second is about 40/2E25 = 2E26 photons per second or 2E26/3E8 ≈ 6.66E17 photons per cycle. Putting this all together it looks like we have, 8E21 electrons giving rise to 6.66E17 photons per cycle or about 10,000 cycles per electron per photon? Does that sound reasonable? Thanks for any help! 



#8
Feb1313, 10:40 AM

Sci Advisor
PF Gold
P: 11,399

Fair enough, you have emerged with a number. Whether or not it's relevant, is another thing. Afaics, it could just be a bit of numerology unless you can show a mechanism that actually connects RF photons with individual electrons. I thought that the interaction between RF frequency EM was on the bulk material and not individual electrons. (The Hydrogen Atom ideas definitely do not apply in the solid state.)



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