High-level (?) physics question

[Vegitto]

Hey all. I know I should be asking this in highly technical, but I don't know if I'll get a response there..

This is quite a story to me, and I'll probably say a couple of things wrong, but please bear with me:

A gas discharge tube is a tube with a cathode, an anode and a low-pressure gas inside. My textbook says this:

When a current is applied to the cathode, it starts releasing electrodes.
(Makes sense.)

These electrodes want to travel to the positively charges anode.
(Makes sense.)

These electrodes can collide with the gas-particles while traveling to the anode.
(Makes sense.)

When these two particles collide, the electrons transfer energy to the atom, which in turn causes the atom to ionize.
(This doesn't completely make sense. What kind of energy is transferred? Is the ionized atom positive or negative?)

Both sets of electrons, the original electron and the newly gained electrons then travel to the anode. Underway, more of these reactions can take place, causing an avalanche-effect. (Okay, so this must mean that the ionized atoms are positive, right? Since they lost some electrodes and they were neutral first, they should be negative now.)

In time, the atoms will release the energy they gained from the collision in the form of radiation.
(Where is the extra energy? Where did it come from? Didn't the impact from the electron knock loose some other electrons and that's that? And what would be the correct way to describe this radiation? Since we use this system in lightbulbs, it must be low-intensity gamma radiation, in other words, light, right?)

Please explain, I feel like an idiot.

So the questions are:
-What kind of energy is transferred when an electron meets a particle?
-What will happen with this particle, what will be it's charge and will there be any energy left?
-What would the best way be to describe the reaction that takes place when the ionized particle releases its energy in the form of radiation?

 

[dighn]

- kinetic, which causes the electron(s) in the atom to either become excited or break free
- the atom becomes ionized, so there's a positive part (the bulk of it) and the free electrons (negative)
- spontaneous emission would be how an excited atom (atom that has excited electrons) releases its excess energy in the form of photons. recombination of electrons and ions is probably different.

see this: http://en.wikipedia.org/wiki/Gas-discharge_lamp

 

[Vegitto]

-Kinetic, that makes a lot of sense..
-Okay, that seems obvious.
-Spontaneous? Could you describe it in the form of a formula? I know that for example, a neutron of a radioactive particle can transfer into a proton and an electron (and gamma radiation, but this is emitted in the form of a foton which doesn't have any influence on the particle itself, right? Well, there's the foto-electric effect, but that's not an issue in this case, right?)
By formula, I mean this:
1,0(n) --> 1,1(p) + 0,-1(e)
A neutron, mass 1u, charge 0 becomes a proton, mass 1u, charge +1 and an electron, mass 0u, charge -1.

Thanks a lot, this helped.

 

[dighn]

in spontaneous emission, the atom itself does not change in its make up, only that the energy state of its electrons decreases and emit a photon in the process.
during recombination, you might have something like A+ + e- = A, ion + electron = original atom

 

[alkemyst]

waiting for the wiki physics experts to chime in

 

[dighn]

Originally posted by: alkemyst
waiting for the wiki physics experts to chime in

umm too late?

 

[Vegitto]

Originally posted by: dighn
in spontaneous emission, the atom itself does not change in its make up, only that the energy state of its electrons decreases and emit a photon in the process.
during recombination, you might have something like A+ + e- = A, ion + electron = original atom

Hmm, okay. Thanks, sounds sound to me..

 

[WildHorse]

send a PM to DrPizza he can help you out he's a Science teacher

 

[mfs378]

Originally posted by: Vegitto
-Kinetic, that makes a lot of sense..
-Okay, that seems obvious.
-Spontaneous? Could you describe it in the form of a formula? I know that for example, a neutron of a radioactive particle can transfer into a proton and an electron (and gamma radiation, but this is emitted in the form of a foton which doesn't have any influence on the particle itself, right? Well, there's the foto-electric effect, but that's not an issue in this case, right?)
By formula, I mean this:
1,0(n) --> 1,1(p) + 0,-1(e)
A neutron, mass 1u, charge 0 becomes a proton, mass 1u, charge +1 and an electron, mass 0u, charge -1.

Thanks a lot, this helped.

In Dirac notation the equation would look something like:

(a_n' a_p)|psi_1>|n_k> --> |psi_0>|n_(k+1)>

where |psi_1> is the excited electronic state and |psi_0> is the ground state.

|n_k> is the system with k photons in the optical mode indexed by n into which your photon goes, and |n_(k+1)> is the system with k+1 photons, with the extra one coming from the spontaneous emission event.

a_n' is the creation operator which adds a photon to the optical mode n, and a_k is the ladder annihilation operator which takes an electron from the excited state to the next lower state.

 

[CallMeJoe]

Vegitto: A couple of technical points:
You do not apply current to a cathode; you apply voltage, and the voltage difference between anode and cathode induces current flow through the tube.
The cathode and anode are the electrodes. It is electrons that flow in the circuit and excite the gas (you use the term "electron" correctly later in your post, but use "electrode" in its place in your questions).