5
$\begingroup$

Related: Which green spectral line(s) are emitted in a Thomson tube?

After reading Lisa Lee’s OP on an electron deflection tube, although she had some misunderstandings on its operation, I still believe that her question is still relevant. If one looks at the Thomson e/m tube filled with helium gas (made by Pasco), the tracer light is created by de-excited electrons, not by an electron beam interacting with a phosphor as in Lisa Lee's OP. That is, the accelerated electrons scatters off of a low pressure helium gas and emit a tracer line to follow the electrons path, as shown below.

enter image description here

Now if you look at the bright line spectrum of helium,

enter image description here

One can clearly see that there are two “cyan” color lines around 500 nm. Following in Lisa Lee’s footsteps, there is a series of question one can ask: (1) which line is emitted by the Helium gas atoms? Or are both lines emitted? (2) How can a variable accelerating voltage for the electrons always produce the same tracer line color? In theory, I assume that if I decreased the energy of the electron beam, I could produce red (around 670 nm) or yellow (around 590 nm) tracer lines, but that doesn’t happen. Somehow, it appears to me that the e/m tube is “tuned” just right so that the emitted light is always this “cyan” color. Why?

$\endgroup$
5
  • $\begingroup$ @JohnRennie: I think this is the question that Lisa Lee thought she was asking. $\endgroup$
    – Carlos
    Oct 3, 2013 at 2:49
  • $\begingroup$ Btw, that ping won't notify John Rennie. See meta.stackexchange.com/questions/43019/… $\endgroup$
    – user10851
    Oct 3, 2013 at 4:24
  • $\begingroup$ Saw it anyway :-) I have to dash but I'll look at the question later. $\endgroup$ Oct 3, 2013 at 8:08
  • $\begingroup$ Thanks for the link to Pasco - some interesting toys there! $\endgroup$ Sep 12, 2014 at 8:18
  • 1
    $\begingroup$ Nice question. Have you tried holding a diffraction grating in front of your eye and observing the diffracted circle? $\endgroup$
    – user4552
    Oct 13, 2014 at 15:57

3 Answers 3

2
$\begingroup$

The electrons in this experiment are accelerated through a voltage of 150 - 300V, according to the manual. That's about 100 times higher than the energy in a visible photon. It's also much more than the (comparable) ionization energy of Helium. Considering the energy of the electrons is only varying by a factor 2 and it is far greater than the obvious energy scales involved, one wouldn't expect the spectrum to change significantly with the voltage.

Presumably, all of the lines are emitted in a proportion such that they look cyan overall.

$\endgroup$
7
  • $\begingroup$ If I understand correctly, the minimum 150V effectively "tunes" the apparatus to produce more cyan light than red or yellow, but red and yellow will still be observed. I believe this makes sense because I would expect the whole spectrum of helium to be emitted, just more of the cyan-colored lines appearing. Our school currently doesn't have an e/m tube so I can't check it with a diffraction grating. $\endgroup$
    – Carlos
    Oct 3, 2013 at 11:19
  • 1
    $\begingroup$ @Carlos: excitation by (relatively) high energy electrons is a messy process and you'll end up with He atoms in lots of different excited states. As Mark says, I doubt small tweaks to the electron energy would make much difference, though changing the energy by a factor of ten might. Adjusting the He pressure will make some difference to the colour because collisions between excited He atoms can transfer energy between them. I think He tends to give a violet colour at high(ish) pressure and blue/green at low(ish) pressure. $\endgroup$ Oct 3, 2013 at 13:46
  • 2
    $\begingroup$ @JohnRennie: Great, I do understand better now after yours and Mark comments. Question: how does one get a feel for the energy required to make a difference in the emitted light? Since the jump in visible light (400-700 nm) energies do not change much (range of 1.8 – 3.1 eV) compared to the incoming electrons (150-300 eV) energies, it seems that this range is tiny and I must be missing something? Continued below. $\endgroup$
    – Carlos
    Oct 3, 2013 at 17:46
  • 1
    $\begingroup$ In addition to the explanation made in this answer, here's another thing to consider. Many of these lines probably are not transitions to the ground state. Therefore the wavelengths of the lines do not have to match up in any way with the excitation energy of the initial state. For example, the red line could be emitted in a transition from a very high energy state to another high-energy state close below it. $\endgroup$
    – user4552
    Oct 13, 2014 at 15:58
  • 1
    $\begingroup$ Is there really no satisfactory answer for this? $\endgroup$ Mar 22, 2018 at 21:49
0
$\begingroup$

It is the 492 nm line. It is monochromatic. Only a single spectral line is observed through a visual inspection using a grating, and the spectroscope gives this value to better than 0.2 nm at this point. I do not know why only one line should be produced.

$\endgroup$
-1
$\begingroup$

In an esoteric sense, the helium has enough components to be helium but there isn't quite enough room, due to the surrounding pressures. The variable density of electrons is 'simply pressuring the helium' so after the helium is sufficiently annoyed ( and starts producing the cyan output) increasing the density doesn't increase the helium's output, it's only annoyed or calm. Now why the output is specifically cyan? That theory can wait >:)

$\endgroup$
1
  • 1
    $\begingroup$ "Now why the output is specifically cyan?" is not a good answer to the OP's question, which exactly what s/he is asking. Furthermore what is an "annoyed" helium atom? $\endgroup$
    – Gonenc
    Jun 6, 2015 at 22:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.