# Is the energy of a $1$ Volt $300$ MHz EM wave greater than a low frequency $200$ V EM wave? [closed]

A DC voltage source supplies electrons that move in a single direction, producing a current $I$ distributed through the conductor's cross-section.

As per Ohm's law, a change in voltage changes the force on the electrons, causing an increase in the speed (and hence the current) of the electrons.

Electrons under the influence of a DC voltage don't move in a straight path, as they are attracted by the positive charge of atomic nuclei in their path.

When we apply a sinusoidal AC voltage to a conductor, electrons alternately accelerate, slow, stop, and reverse direction.

When we apply high frequencies, like $300$ MHz, I assume that the alternation of voltage direction is so fast that electrons don't move, they only hesitate in one place. But if this is true, that electrons are standing still, then how do antennas oscillate current to create $EM$ waves?

If we look at the plot (below), with two different frequency waves, it appears that the electrons get to the same speed (or amplitude) with different frequencies, but they just reach it faster at higher frequencies.

What is the reason for the energy difference between the higher and lower frequency waves?

• Possibly the difference is due to the energy of a quantum of each $EM$ wave since its energy depends on its frequency $E=h\nu$?
• Possibly the difference is due to Special Relativity effects. Maybe the $E$ field of the oscillating electron increases and compresses with the increase in velocity of the electron?
• In my example (below), the frequency doesn't change the amplitude of the wave. Is it the change in electron speed that increases the EM wave energy? Or, is it the peak voltage upon which the EM wave energy depends?

Example - illustrating my question:

If I generate a $300$ MHz $EM$ wave through an antenna with a $1$ V voltage source, will that EM wave have higher energy than a much lower frequency wave produced by the same antenna but with a $200$ V amplitude? If so, why? • Your assumption is incorrect. 300MHz is not 'too fast' for an electron. – Jon Custer Jul 25 '18 at 17:57
• @JonCuster Really? I thought, electrons moves slowly, than snail, in current. You think it's wrong? – user202366 Jul 25 '18 at 18:02
• Remember, it is scattering a lot - there is a difference between how quickly it can respond to field changes, and how far it can move in some period of time. – Jon Custer Jul 25 '18 at 18:22
• @JonCuster, so according to my complete question, what can You say? – user202366 Jul 25 '18 at 18:25

Electrons are incredibly tiny (their diameter, if they even have one, is less than $10^{-18}$ m). And have very little mass (on the order of $10^{-30}$ kg).

So the motion of the electrons is not what limits our ability to transmit high frequency signals over wires.

There are two more important limits on transmitting high frequency signals over wires:

• The skin effect which causes the current to crowd near the outer surface of the wire as the frequency increases. This increases the effective resistance of the wire.

• Radiation, meaning that at very high frequencies even a short wire starts to act as an antenna and radiate energy from your signal away into space.

• I don't say about wires, I wrote about oscillating into space instead. Ok, I edited my question, can You answer on it? – user202366 Jul 25 '18 at 18:05
• @Artur, EM waves propagate perfectly well in empty space with no electrons present at all. So there's no need to worry about the motion of electrons when thinking about an EM wave in space. – The Photon Jul 25 '18 at 18:18
• Come on, I did not meant space, like out of earth. I mean just out of wire. But did You read my edition? – user202366 Jul 25 '18 at 18:23
• If you are asking about a wire, then my answer applies. The momentum of the electron is not the limitation on the maximum frequency. The skin effect is a major limitation. If you are asking about an antenna, then the limitation is that at very high frequencies you can't deliver the signal to or from the antenna in a wire. – The Photon Jul 25 '18 at 18:42

The $E$ in $E = h\nu$ is photon energy. Each photon carries $E$ joules of energy.

The power of a radio transmission depends on how many photons per second are radiated. We could work it out: for example say we have 1 kW transmitter at 300 MHz, and we assume the antenna, feedline, and other practicalities are 100% efficient. 1 kW is 1000 watts, and 1 watt is 1 joule per second. So the sum of all the energy of the radiated photons in one second must add up to 1000 joules.

At this frequency the photon energy is:

$$(6.63 \times 10^{-34}) (300 \times 10^{6}) = 1.99 \times 10^{-15} \:\mathrm{joules \over photon}$$

The transmitter power multiplied by the reciprocal of the photon energy gives photons per second:

$$\require{cancel} 1000\:\mathrm {\cancel{joules} \over second} \times {1\:\mathrm{photon} \over 1.99 \times 10^{-15} \:\cancel{\mathrm{joules}}} = 5.03 \times 10^{27} \:\mathrm{photons \over second}$$

A higher transmitter frequency would mean more energy per photon, so assuming transmitter power is held constant, there must be fewer photons per second radiated.