Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: theThinker on 16/03/2024 04:06:10
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In the gas discharge tube experiment for H2, we have four emission lines in the visible spectrum. The explanation is that they are due to excited electrons that radiating according to the
formula:1/λ = R(1/nf? - 1/ni?) for nf= 2, ni = 3,4,5,6.
But QM requires that the electrons could only be excited by the exact frequency light (red, blue, violet, violet) in order to reradiate the four colors. But the source of the energy to the discharge tube is electrical, so how in the first place could the electrons get excited and reradiate. There are no visible light entering the discharge tube in the first place.
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The initial excitation is by collision with cations (in this case protons) streaming from the anode.
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Hi.
The initial excitation is by collision with cations (in this case protons) streaming from the anode.
Some if it is but there are also always going to be some free electrons to be found.
For one reason or another, there will be some atoms in the gas tube that are ionised so that we have a free electron and a positive ion (a "cation"). This might be a consequence of cosmic rays striking some atoms or something else. One way or another you have a tiny number of cations and free electrons in the gas, even though it might be a very small number.
Most gas discharge tubes work by causing an "avalanche effect". See https://en.wikipedia.org/wiki/Townsend_discharge for a description of the "Townsend Avalanche".
(https://upload.wikimedia.org/wikipedia/commons/thumb/a/ac/Electron_avalanche.gif/1280px-Electron_avalanche.gif)
Basically if you have a small number of free electrons to start with and then accelerate them in an electric field, they have collisions with other atoms and quickly start to liberate more electrons and leave more positive charged ions. Under the right conditions the numbers of electrons increase geometrically (typically doubling on each collision). So fairly quickly you have a large number of free electrons and positive ions in the gas tube. There is a more detailed discussion in the Wikipedia page that was linked to.
Now let's go back to your ( @theThinker ) comments:
But QM requires that the electrons could only be excited by the exact frequency light
That's half right. If electrons were going to be moved from one orbit to another by absorbing a photon then those photons should have exactly the right amount of energy corresponding to the energy gap between the orbits. However, QM never said that absorbing a photon was the ONLY way to move electrons from one orbit to some other (or in this case to liberate them from the atom completely).
It is quite possible to liberate electrons through impact between the atom and some other particle. In our case, we have free electrons (and also some cations which can also be accelerated and get involved) these collide with the (un-ionised) atoms and liberate electrons from them.
Actually, if you just want to completely free the electron then you CAN also do this with photons and it's a lot less restrictive then trying to move them from one orbit precisely to another. See "the photo electric effect", https://en.wikipedia.org/wiki/Photoelectric_effect , provided the photons have energy greater than the ionisation energy then the electron can be freed and the surplus energy will usually just become kinetic energy that the electron leaves the atom with. However, that isn't what is done in gas discharge tubes - we just have electrons liberated as a consequence of collisions and impact energy.
In the gas discharge tube experiment for H2, we have four emission lines in the visible spectrum.
Once you have quite a few free electrons these will start to fall back into the ionised atoms. They don't usually go from being free electrons straight to being bound in the lowest energy ground state of the atom. Indeed they will usually start at some higher orbit energy and steadily fall further down to lower energy orbits. Exactly as you have stated, some of these transitions will be to the final level n=2 from higher energy orbits (this is the Balmer series of emission lines). These are the ones which produce visible light. More discussion and diagrams available here: https://www.chemguide.co.uk/atoms/properties/hspectrum.html
(https://www.chemguide.co.uk/atoms/properties/tieup4.gif)
I hope that helps.
Best Wishes.
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Hi.
The initial excitation is by collision with cations (in this case protons) streaming from the anode.
Some if it is but there are also always going to be some free electrons to be found.
For one reason or another, there will be some atoms in the gas tube that are ionised so that we have a free electron and a positive ion (a "cation"). This might be a consequence of cosmic rays striking some atoms or something else. One way or another you have a tiny number of cations and free electrons in the gas, even though it might be a very small number.
Most gas discharge tubes work by causing an "avalanche effect". See https://en.wikipedia.org/wiki/Townsend_discharge for a description of the "Townsend Avalanche".
(https://upload.wikimedia.org/wikipedia/commons/thumb/a/ac/Electron_avalanche.gif/1280px-Electron_avalanche.gif)
Basically if you have a small number of free electrons to start with and then accelerate them in an electric field, they have collisions with other atoms and quickly start to liberate more electrons and leave more positive charged ions. Under the right conditions the numbers of electrons increase geometrically (typically doubling on each collision). So fairly quickly you have a large number of free electrons and positive ions in the gas tube. There is a more detailed discussion in the Wikipedia page that was linked to.
Now let's go back to your ( @theThinker ) comments:
But QM requires that the electrons could only be excited by the exact frequency light
That's half right. If electrons were going to be moved from one orbit to another by absorbing a photon then those photons should have exactly the right amount of energy corresponding to the energy gap between the orbits. However, QM never said that absorbing a photon was the ONLY way to move electrons from one orbit to some other (or in this case to liberate them from the atom completely).
It is quite possible to liberate electrons through impact between the atom and some other particle. In our case, we have free electrons (and also some cations which can also be accelerated and get involved) these collide with the (un-ionised) atoms and liberate electrons from them.
Actually, if you just want to completely free the electron then you CAN also do this with photons and it's a lot less restrictive then trying to move them from one orbit precisely to another. See "the photo electric effect", https://en.wikipedia.org/wiki/Photoelectric_effect , provided the photons have energy greater than the ionisation energy then the electron can be freed and the surplus energy will usually just become kinetic energy that the electron leaves the atom with. However, that isn't what is done in gas discharge tubes - we just have electrons liberated as a consequence of collisions and impact energy.
In the gas discharge tube experiment for H2, we have four emission lines in the visible spectrum.
Once you have quite a few free electrons these will start to fall back into the ionised atoms. They don't usually go from being free electrons straight to being bound in the lowest energy ground state of the atom. Indeed they will usually start at some higher orbit energy and steadily fall further down to lower energy orbits. Exactly as you have stated, some of these transitions will be to the final level n=2 from higher energy orbits (this is the Balmer series of emission lines). These are the ones which produce visible light. More discussion and diagrams available here: https://www.chemguide.co.uk/atoms/properties/hspectrum.html
(https://www.chemguide.co.uk/atoms/properties/tieup4.gif)
I hope that helps.
Best Wishes.
Your reply is very helpful. I like to note that GDT works with AC/DC.
I have a related question on the ionization part.The first process in gas discharge is ionization. It is easy to explain classically as the electric field applied pulls the nuclues and the electrons apart in opposite direction thus separating them.
But if we want a QM viewpoint, it may be difficult. An electron could be free of the nucleus only if it absorbs a photon raising its energy beyond the transition limit. But where do such an initial photon come from?
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This " An electron could be free of the nucleus only if it absorbs a photon raising its energy beyond the transition limit. " is missing the point of a gas discharge lamp.
The voltage gradient is big enough to pull the electrons out of the atom.
In some cases they have a hot filament which emits electrons and those can be accelerated by the field and when they hit an atom (or molecule) they ionise it and start a Townsend Avalanche.
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Hi.
Thank you for your reply, @Bored chemist . I was wondering if I had got this slightly wrong.
This is what I think:
(i) There are devices where causing an electrical breakdown in the gas, i.e. stripping some electrons out of an atom by applying a large voltage gradient across the gas is important. These devices tend to be called Fluorescent tubes (or other names).
(ii) There are devices where electrical breakdown is not so important. This includes devices that tend to be called gas discharge tubes and are the thing you ( @theThinker ) asked about.
An electrical breakdown of some small region of the gas, maybe even just a few atoms, can be something that accounts for the first few or initial electrons and cations BUT the ejection of more electrons as a consequence of an avalanche effect is much more important.
Please read this section again:
QM never said that absorbing a photon was the ONLY way to move electrons from one orbit to some other (or in this case to liberate them from the atom completely).
It is quite possible to liberate electrons through impact between the atom and some other particle.
In further reply to this point you made:
The first process in gas discharge is ionization. It is easy to explain classically as the electric field applied pulls the nuclues and the electrons apart in opposite direction thus separating them.
Actually it's not so easy to explain or understand. An electron orbits at a radius approximately equal to the Bohr radius from the nucleus. If you put the numbers in for the electronic charges of the nucleus and electron with this typical distance, then you can estimate the Coloumb force between electron and nucleus.
F = e2/ (4πε0r2), set this equal to Einternal . e so that Einternal represents the Electric field in which the electron moves inside the atom, it's the internal electric field that the electron sees. This internal Electric field is about 105 MV/m, i.e. it is massive.
You can then determine the external Electric field, E, that you are applying. Air breaks down in field strength of about 3 MV/m, I can't find the breakdown point for Hydrogen gas but it'll be similar, maybe 10 MV/m.
Anyway what you have is a difference of about 4 orders of magnitude.
To paraphrase this: If you were going to look at the situation classically, then there is no contest in the tug of war between the nucleus and the external electric field over who will control the electron. The nucleus is a Titan, while the external electric field is just an insect. Classically, you will not get the atom to fall apart and liberate an electron just by applying such a pathetic external electric field. You can say "the electric field applied pulls the nuclues and the electrons apart in opposite direction thus separating them" exactly as you have done and indeed this simple explanation can be found in some school level texts - BUT the numbers just are not there to back this up.
Fortunately, under Quantum Mechanics, the electron isn't always orbiting this close to the nucleus and the possibility of liberating an electron has to be dealt with differently.
Anyway, with the help of a cascade or avalanche effect, where even just a few electrons are liberated that is enough. A different process takes over where more electrons can now be liberated as a consequence of collisions between these few free electrons and the other atoms.
Best Wishes.
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Hi ES, I would have said 1Mv/m for normal air but I could be wrong. Many discharge tube use additional methods to start the initial ionisation, prior to avalanche occurring such as heated filaments in florescent lamps, a high voltage external conductor in xenon flash tubes and an auxillary electrode very close to one of the main electrodes in mercury vapour lamps(to start a local ionisation). High pressure sodium lamps and tin halide lamps use a high voltage source in series with one of the electrode supplies. The gas pressure plays a major role in the voltage needed to initiate ionisation, referred to as the strike voltage: the higher the pressure, the higher the necessary voltage and a lot of discharge lamps have to cool down after power interruption before reuse as the pressure is higher when hot. Many other factors may account for the discrepancies in the calculations: random cosmic rays as you suggested or low level radionuclides. Field emission as raised by Alancalverd may play a role(wild assed guess).
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Hi.
Thanks for the extra information @paul cotter .
I would have said 1Mv/m for normal air
I just took a value from the first two sites reported by Google. 3 Mv/m was what they both said and all I needed was the order of magnitude.
Best Wishes.
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The simpler Geissler-type tube doesn't have a filament. The initial event is field ionisation at the anode.
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Looking at my last post(which I posted while quite tired last night) I must apologise: my remark about the breakdown of air is petty and not relevant to the argument. Another factor to consider is the purity of the gas: no substance exists in 100% purity and stray atoms with lower ionisation potential are sure to be present--in principle a single ion being accelerated could trigger the necessary avalanche.
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High voltage from Tesla coils can cause the gas in the tube to glow. Microwave can also do the job.