Naked Science Forum
On the Lighter Side => New Theories => Topic started by: sorincosofret on 26/10/2008 13:11:16
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Electric charge movement and gas tube discharge
Background
A gas tube, in principle, is a very simple device. It consists of an evacuated glass tube fitted at each end with a metal terminal called electrode. The simplest tube contains inside, a small amount of highly purified inert gas or a mixture of compounds for more complex one. Connected to the two electrodes is a source of high-voltage electrical power able to deliver a limited current. When the current is turned on, the tube glows.
In the absence of a potential difference, the gas inside the tube is formed by neutral particles and the current does not pass through. In the presence of a high voltage potential difference, a process of ionization takes place and the current can flow through tube.
It is admitted the existence of a small quantity of charge in a specific medium (gas) due to cosmic radiation background. These charges are not enough to insure a flow of an electric current. Besides this charge, when a high enough voltage is applied on gas, the electrons of gas medium are stripped away from their respective nuclei, creating conditions of ionization. In these conditions, with a higher concentration of free electrons able to migrate from a region to another, gas becomes a relatively good conductor of electricity.
In a simplified description, the electrons coming from electric current, having a minimum specific energy for every gas, are powerfully enough to hit the gas molecules, to generate new charges between cathode and anode and the low rarefied gas become conductor of electricity.
Because like electrical charges repel one another and unlike charges attract each other, a free electron is strongly attracted to any nearby positive ion. This attraction leads to the rapid combination of the positive ions and electrons into neutral particle, and this process is responsible of light production and its color depends upon on gas type.
Thus, to produce a gas discharge and tube glows, electrons must be removed from neutral molecules and recombined with positive ions to form other neutral molecules. The practical way of producing this ionization is by passing a current through the gas.
When voltage is applied to the electrodes, cathode emits electrons and anode attracts these electrons. If the voltage is high enough, the electrons will be attracted with tremendous force and will accelerate toward the positive electrode reaching speeds of tenth of km per second.
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Figure 1 Action of electrons and positive ions in gas-filled tube.
During this trip toward anode, electrons collide with a neutral gas molecule that lies in its path. It hits this molecule with such force that one or more (usually more) electrons are liberated. These secondary electrons, once free, starts moving toward anode. They soon collide with other neutral molecules, generating in a cascade process more charge into gas. Shortly after high voltage is applied to the tube, there is a general process of ionization in entire volume of gas. Electrons are liberated from molecules, free electrons combine with positive ions, giving off light, and then are blasted apart again.
For our discussion it is necessary to remind the fact that gas current that flow through the tube depends largely on the pressure. In the low-pressure gas, the atoms are fewer in number and farther apart from one another; therefore, the free electron has a longer distance in which to get up speed before it hits a neutral atom. As a consequence, when the free electron hits, it hits much harder, and more free electrons are liberated than in case of higher pressure. Hence, as the pressure is lowered, the current will increase.
If, however, the pressure is reduced so much that a nearly perfect vacuum forms inside the tube, then there are so few atoms available that the current will decrease due to lack of electrons and ions, even though the electron speeds are very high.
Depending on the gas composition, it is known from practice that gas tubes present specific colors:
Gas Color
Helium Whitish orange
Neon Red-orange
Argon Violetish pale - lavender blue
Experimental part
Long time ago, I’ve found that a gas tube approached by a charged Van der Graff device produce light. There was only a curiosity at that time, but after that I was thinking of the mechanism of tube discharge. At that time, the tube was moved toward VDG sphere with metallic electrode directed to sphere. At a distance of about 1 cm between VDG sphere and tube electrode, a spark appear in the space between, and simultaneously an electric discharge appear in the tube (fig.1).
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Figure 1. Glows of a gas tube near a VDG device
For the beginning, in a trial to elucidate the mechanism of tube lightening, a gas tube in from of U is used. The tube is moved close to the VDG sphere, keeping the electrodes outside, as in fig. 2.
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Figure 2. Lightening of a U form gas tube near a VDG device
Contrary to normal expectations, the tube starts lightning even there is no metallic electrode able to be charged and to emit light. The tube is lightening for minutes (with my small VDG device) in the absence of an apparent source of electrons. In order to block a eventually electron charge displacement from air, both tube electrodes are put in short circuit and covered with a insulating material as in fig. 3 and the experiment repeated.
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Figure 3. U form gas tube variant
When this modified U tube is moved closed to VDG sphere, the tube is lightening as in precedent case.
Let’s analyze a little bit these results of the experiments from actual orthodox theory.
In every manual of physics VDG sphere is considered positively charged. So when a gas tube is closed to the VDG sphere, there is no electron flux able to excite the gas in the tube. In fig. 1 the second electrode of the tube is not connected so it is impossible to have an electric current (an electron flux) flowing into circuit. The tube is observed to glow even the actual theory is not able to explain this.
In the fig. 2 and 3, a U gas tube is glowing, even the electric field acts directly on the gas and not on the electrodes. How is possible this? From where is directed this current in gas tube and in what direction is going on?
I leave to actual elite physicists to formulate a coherent explanation.
In proposed theory, the ionization phenomenon is not the principal factor responsible for gas tube glowing. It can be showed the contradiction between ionization phenomena, gas tube glowing and predominance of one color lightening in the spectrum, all these in frame of quantum hypothesis.
Considering a simple gas tube with a single gas inside, let’s analyze the process of ionization taken in discussion the quantum hypothesis and the experiment Franck –Hertz (an experiment fundamental for quantum theory).
In Franck Hertz experiment, it was demonstrating that, in a tube filled with mercury vapors (the results are valid for every other gas, only the values of currents are different), atoms could absorb (and are excited) only by specific amounts of energy (quanta).
At low potentials, the accelerated electrons suffer purely elastic collisions with mercury atoms in the tube. This is due to the prediction of quantum mechanics that an atom can absorb no energy until the electron kinetic energy is equal with energy necessary to lift an electron into a higher energy state.
The fundamental question is: An electron lift in a higher energy state means all the time an ionization process?
In case of Franck Hertz experiment there isn’t an ionization process effectively, because an ionization process means for colliding electron to have enough energy able to lift the electron from excited state to other higher excited states until the electrons becomes free as indicated in fig.4 -b.
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This thing does not happened, and emission spectra of excited mercury give exact indication about type of excitation, eventually ionization. The spectra of mercury vapors emission gives only a single line in ultraviolet (= 254 nm) in case of Franck Hertz experiment, and this means only excitation and not ionization ( fig. 4-a details).
By analogy, let’s analyze the case of simple glowing gas tubes. In case of a recombination process between electrons with kinetic energy of tenth of km per second and positive ions, there should be observed a multi lines spectrum according to fig. 5. In the fig. 5 only a part of entire possible spectrum is figured out. At their speeds, electrons are able to produce a large range of photons from infrared up to x-rays (of course are absent the electromagnetic waves or in actual representation radio and microwave photons). With this complex spectrum, the yields of produced visible light should be lower then a normally thermal source.
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Figure 5. Theoretical emission spectra of a gas tube
But the reality is completely different, and a gas tube is very specific. In normal condition, the emission spectrum is composed by a single line or combinations of very few lines.
In the same time it must be highlighted that a direct collision between a free electron with high energy and a positive ion, should determine the generation of primary X-ray photons and in an indirect manner a visible photons.
Despite this prediction, most of common used tubes (mercury is an exception) works directly with visible photons and this means the impossibility of existence of an ionized gas, but only an excited gas. Even in case of mercury the line transition in UV does not mean ionization, it is only excitation, with a greater gap between ground level and excited level. In case it is admitted that mercury line produced in UV is characteristic for an ionized medium, the Franck Hertz experiment must receive a new more fantastic explanation…..
In proposed theory (and if the Van der Graff device produce a high positive potential on the sphere tower), the light emitted by a gas tube has another fundament.
A neutral atoms found in a strong positive potential, becomes excited due to the electron jump from ground state to a excited state. In this case, great parts of the atoms arrive on the single excited state, and their emission spectrum is formed by a single line. In case of excessive excitation energy, there is the possibility to appear more line in the emission spectrum. The low pressure existent in tube prevent the direct collision between excited atoms so, in comparison with a thermal source, a gas tube can be classified as a cold light source.
Of course a small number of atoms can become ionized, but ionization is not the basic fundament of light emission in case of gas tubes.
In the proposed theory an excited gas atom can leave the electric current to pass through that medium. Admitting a time for a electron jump from ground state to excited state and the same time for jump between excited state back to ground level, equal with 10-8 s, it can be considered that electron magnetic moment is available for electric current propagation during this interval of 10-8 s.
Admitting the actual speed of electromagnetic propagation c= 3*108 m/s, it can be observed that an electric current can pass through a 3 m depth of a excited gas without any problem. In reality this depth of gas is greater because any time a fraction of gas atoms are in excited state and another in ground state. There is a continuous permutation between these two states and this ensure a possibility of electric current to pass through on longer distance, of course with a consume of electric energy.
In order to prove the proposed explanation, new and simple experiments are made.
The first experiment is a repetition of a previous U form gas tube experiment with a direct source current attached to it as in fig. 6. When tube is moved closed to the VDG sphere, with the base U part directed to sphere, a flash of light appear in the tube and a electric current pass through tube circuit, and this current is observed on the ammeter.
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Figure 6. U gas tube with DC current experiment
So a single positive potential is enough to excite the low pressure gas and the current flow in the circuit.
The third experiment uses a modified CRT electron circulation connected to a gas tube as is indicated in fig 8.
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Figure 8 Modified CRT electron circuit for the proposed gas tube discharge experiment
In order to obtain this modified electron flow, the high potential cable mounted on the neck of tube is removed and this potential is connected to a foil of aluminum; the aluminum foil is then fixed on the external part of the screen.
At the connection point situated on the neck of tube, an ammeter and a gas tube of different length (different about 10 W power are connected in series. The other connection of this circuit is to a null point (from heating home installation). The CRT alimentation is started and the indication of ammeter and the ,,current” passing through gas tube is counted.
Strange enough there is a small ,,electric current”, approx. 0.1 micro A flowing between electrodes. At this stage of experiments it is difficult to attribute this charge conductibility to internal gas or to material which form the tube.
Probably there are both, because if a tube is connected with a metallic electrode to circuit and the connection coming from the ammeter is touching the glass tube, a ,,electric current” is counted. So the glass is a charge conductor, but as is well known has an insulator comportment for a real electric current.
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"Contrary to normal expectations"
No Sorin, only contrary to your expectations. The rest of us know that insulators don't work at very high voltages.
"In every manual of physics VDG sphere is considered positively charged. "
Wrong but not important. Plenty of V de Gs are run in pairs, one negative the other positive, to get twice the potential difference.
"I leave to actual elite physicists to formulate a coherent explanation. "
I don't know about the elite physicists- I guess they are busy so I will have a go.
The system works because the air round a V De G is ionised by the high potenetial gradient.
What else limits the output voltage?
You can test this idea if you like. Hold the tube so your hand (which is earthed) is near the top of the V De G. It doesn't light or at lesat it's much less bright. This is because the charge carried friom the generator is shunted away to earth through your hand.
Similalry, if you take a long thin insulating thread and use that to hold the tube near the generator it doesn't light. That's because there's no conducting path to complete the circuit.
OK, now for those who are not familiar with it there's an illustration of the experiment here.
http://hyperphysics.phy-astr.gsu.edu/hbase/frhz.html
" An electron lift in a higher energy state means all the time an ionization process?"
No, there are lots of excited states, in fact an infinite number, between the ground state and ionisation.
Now the web page above shows the bands of light corresponding to the ionisation of neon. How else can you explain them and what's wrong with the conventional explanation Sorin?
Re the other experiments with the CRT; I see you still don't understand that "insulator" is a relative term not an absolute one.
Glass does conduct- badly.
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New toy. New theory. New Bull***t.
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How else can you explain them and what's wrong with the conventional explanation Sorin?
For those who consider physics a science of circumstances there is nothing to be explained;
As another English physicist (100 years ago) said: Monday, Wednesday and Friday light is wave; Tuesday, Thursday and Saturday is corpuscle; Sunday is silent and mystery, for you the physics means inconsequence and arbitrary.
Based on the Franck Hertz experiment it can be seen that a excited, but not ionised mercury can produce a ultraviolet line.
In a gas tube the same mercury produce the same line, but in this case it is necessary to exist a ionised mercury in order to ensure the gas conductibility.
One of these affirmation must be wrong because in the frame of the same theory ( in fact in any theory) they enter in contradiction.
Its up to you what you consider more plausible: an excited atom can be a conductor or an ionised atom produce a single line instead of a entire spectrum.
I supposed only that central VDG sphere is positively charged, and for the moment it was not necessary to determine its sign. With advancement in the experiments this think will be made.
There are not a infinity of excited state, because in this case there will be a infinity of emitted photon energy ( with different probability) and the spectrum of an element it will be a continuous spectrum.
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A strong E field will strip an electron from an atom, involving one energy jump (hundreds or thousands of eV). When the electron re combines, it will go through a number of energy changes. The optical / UV energy changes are only in the region of a very few eV. For a given element the result will be a spectral line. There will be other EM waves also emitted - but you don't see them. There is no contradiction.
Sorin - yet again the actual NUMBERS count in the real explanation of a Scientific phenomenon. You really ought to read more before you start to write.
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"Based on the Franck Hertz experiment it can be seen that a excited, but not ionised mercury can produce a ultraviolet line.
In a gas tube the same mercury produce the same line, but in this case it is necessary to exist a ionised mercury in order to ensure the gas conductibility"
Most of that is at least close to a correct description of what happens.
It takes 10.4 electron volts of energy to ionise an atom of mercury from the ground state.
Mercury also has an excited state about 5EV abof the ground state.
An electron with an energy of 5eV which colides with a mercury atom is likely to raise the atom to an excited state. That atom will probaly release that energy as a photon of UV light with an wavelength of about 254nm.
You can do all this without ionising the mercury.
However, if you want to get a measurable current through the mercury you need to ionise it.
There are two different processes happening here. They require different energies (one happens to be about twice the other).
Depending on the circumstances one, the other, or both may happen.
So what?
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A strong E field will strip an electron from an atom, involving one energy jump (hundreds or thousands of eV). When the electron re combines, it will go through a number of energy changes. The optical / UV energy changes are only in the region of a very few eV. For a given element the result will be a spectral line. There will be other EM waves also emitted - but you don't see them. There is no contradiction.
After your description, a gas tube should emit more radio, microwave and after that less infrared and visible. The energy of a radio or microwave photon is smaller then a VIS or IR photon. In this case maybe is better to put a microwave protection, because the working temperature of a gas tube is about 30 degree, and the maximum of emission is in microwave.
Far away from your dreams, a gas tube emits only few (or one ) line in visible. The emission in IR is very small and in radio and microwave is nonexistent.
For the other comment, I strongly recommend to read at least Logic of Aristotle.
It is a simple problem of logic.
In actual theory an excited atom from a material does not permit the electric current passing and in the same time an excited atom has a simple spectrum ( few line).
If the atom become ionised, in actual theory, the medium become a conductor, but the spectrum is a complex one (there is explained already).
In experiment there is a simple line spectrum, but the material is conductor.
Which sentence is wrong?
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have you measured any other radiations from a flourescent tube than visible? Have you shown there is no RF?
A domestic tube produces UV more than visible - hence the phosphor coating inside the tube.
Bearing in mind the poor match as microwave radiator, you might get little RF from a tube.
Where is the implication in what I said about relative levels of different frequencies? They are quite efficient so there can't be a lot of energy apart from uv and visitble.
The energy of the moving electrons willdepend on themean free path and the voltage gradient along the tube. The MFP will be chosen (by the chosen pressure) to give the electrons the chance to achieve a few eV before they collide with an atom and cause, evevtually, the emission of a UV photon.
The line specta will be expected because, whatever energy the conduction electron has, the gas atom will only have it's natural transitions.
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"Far away from your dreams, a gas tube emits only few (or one ) line in visible. The emission in IR is very small and in radio and microwave is nonexistent"
Bollocks
One of the most practical ways to test the gadgets you can buy to see if your microwave oven is leaking, is to hold it up to a fluorescent lamp- they generate readily measurable amounts of microwaves.
Fluorescent lights often interfere with AM radio too because they emit radio waves (and they did before the advent of HF starters)
Once again Sorin's "theory" is at odds with reallity. Why don't we have a rubbish bin for threads like this?
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One operator should move this post to ,,new theory" section. I have tried to do myself, but It seems I don't have this right.
All experiments regarding electrostatic need reinterpretation. The concept of ,,charge" is taken into discussion.
I will measure the spectra of a gas tube in IR, in radio and microwave.
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No new theory has been put forward. No evidence of a problem with the current theory has been produced.