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I know that the secondary coil used in the Tesla coil radiates electromagnetic waves in the surroundings. This radiation makes the electrons in the fluorescent material of the light oscillate, and they emit photons.

My question is,

  1. Does this oscillating electromagnetic field (i)excite the electrons in the fluorescent material directly, or (ii)does it excite the gas to emit UV, which in turn, excite the fluorescent material?

  2. Tesla coils generate an electromagnetic field with frequency in the radio wave region. If (i) (in Q1) is true, then how does low-frequency radio waves excite the fluorescent material to emit visible light (of higher frequency)?

  3. If (ii) is true, then how can radio waves excite gas molecules to generate UV?

  4. If (i) is true, then can we use Tesla coil to excite proteins with fluorescence tag, or proteins like Green Fluorescent Protein, so that they emit without any light of higher frequency exciting them? (Actually, in case (i) is true, I want to do this experiment, that's why I am asking)
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The radio frequency photons emitted by a Tesla coil are much too low energy to directly excite atoms to emit visible (~2 eV) or UV photons (~6 eV) . A 1 MHz radio photon ($h\nu=4\times 10^{-9}eV$) is also way too low in energy to ionize an atom which requires ~10 eV.

What actually happens is that the near electric field from the Tesla coil accelerates any free electrons in the tube's low pressure gas. These electrons pick up enough energy before they collide with another gas atom that they ionize that gas atom. The ionized electrons are then accelerated and ionize further atoms. When electrons fall back onto the ionized atoms, visible light (as in a neon filled tube) or UV light (from mercury vapor in a fluorescent tube) is emitted. The phosphor coating on the inside of the fluorescent tube is excited by the UV photons or by the accelerated electrons directly.

For some numbers, suppose the peak voltage between the ends of a 1 meter long tesla coil is $10^5$ volts. The near electric field is then $10^5 volts/meter$. The mean free path of an electron in the low pressure gas (pressure=100 um Hg) is ~1 mm. Thus the electron is accelerated to ~100 eV which is more than enough energy to ionize an atom.

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  • $\begingroup$ This makes sense. I used to wonder why such low energy photons can ionize the gas. But there are already some small number of free electrons due to thermal energy. $\endgroup$ – Archisman Panigrahi Apr 29 at 2:24
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The oscillating electric field generated by a Tesla coil, excites the gas, commonly, mercury vapor, inside the fluorescent light, which produces a UV light, which, in turn, excites the fluorescent material, commonly, phosphor coating, which produces a visible light.

The molecules of the fluorescent material can be excited by high energy (hν) UV, but cannot not be excited by low energy RF.

So the answer to your first question is (ii).

As for your third question, the mechanism of exciting gas molecules inside a fluorescent bulb by a Tesla coil is similar to the mechanism behind the normal operation of the fluorescent light: it is a strong electric field, which ionizes gases.

In the normal operation, a relatively low AC voltage is directly applied to the electrodes of the bulb.

With a Tesla coil a very high AC voltage is applied to a circuit that includes the bulb and very small parasitic capacitance between the coil, the bulb and other objects in the neighborhood. As a result, only a small fraction of the coil voltage is applied to the bulb, but, considering that even a small Tesla coil can generate voltages on the order of hundreds of kilovolts, its small fraction could be sufficient to ionize the gas.

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  • $\begingroup$ Even if the electric field is strong, how can high intensity radio wave ionize an electron of the gas? The ionization energy corresponds to the UV range $\endgroup$ – Archisman Panigrahi May 5 '18 at 3:51
  • $\begingroup$ AC field produced by a Tesla coil does not directly excite electrons in mercury to produce UV. Instead, it ionizes the gas and fast moving ions of the ionized gas collide with mercury atoms and excite mercury electrons. In the fluorescent coating, the charges cannot be sped up to cause ionization and collisions (because it is not a gas), so we rely on UV produced by mercury to excite electrons in the fluorescent coating. $\endgroup$ – V.F. May 5 '18 at 13:12
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  1. If (ii) is true, then how can radio waves excite gas molecules to generate UV

ii) is what happens, i) is unlikely to happen, since radio frequencies usually do not excite strong higher frequency emissions, and electric breakdown in solids requires much higher electric field strength.

Tesla coil does produce radio waves, but these are not what excites gas molecules. It is the electric field near the tesla coil, where it is strong enough to induce electric breakdown$^*$ of the rarified gas.

Nearby the Tesla coil the electric field is more like radial electric field coming from the coil than transversal field of a radio EM wave. If you place a mercury lamp farther from the coil, the electric field similar to plane EM wave will be present, but its strength will be much lower, the breakdown will not occur and the lamp won't emit light.

$^*$ During breakdown, molecules get excited and some of them tear into charged components - quasi-free electrons and positive ions. When these eventually get back together (recombine), EM radiation with line spectrum is generated. Some of these lines are in UV region (visible and UV radiation is typical for electrical discharges in gases).

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  • $\begingroup$ Your answer does not explain how the dielectric breakdown of the gas create the UV. Also, is it true that (ii) happens, not (i) ? $\endgroup$ – Archisman Panigrahi Apr 29 '18 at 15:02
  • $\begingroup$ @ArchismanPanigrahi, I have edited my answer. $\endgroup$ – Ján Lalinský Apr 29 '18 at 20:08

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