0
$\begingroup$

According to the European Space Agency:

Every object with a temperature above absolute zero (that corresponds to 0 K, or -273 degrees C) emits electromagnetic (EM) radiation over virtually all wavelengths.

Say I have a tank of hydrogen gas, that is in itself in a vacuum chamber (so the tank itself only loses negligible amounts of energy to its surroundings via conduction/convection).

According to Purdue University, if I were to heat it by passing an electric current through it it would emit blue light:

When an electric current is passed through a glass tube that contains hydrogen gas at low pressure the tube gives off blue light. When this light is passed through a prism (as shown in the figure below), four narrow bands of bright light are observed against a black background. These narrow bands have the characteristic wavelengths and colors shown in the table below.

Wavelength (nm): Color
656.2: red
486.1: blue-green
434.0: blue-violet
410.1: violet

NIST provides even more detail on the spectral bands.

What if instead of passing an electric current through the gas, I were to heat it by a heating element on the bottom, that causes the gas to heat due to conduction and thereafter convection?

Say the tank reaches an equilibrium temperature of 10$^{\circ}$C. Will it:

  1. Emit just the same frequencies as it would with the electric current; or
  2. Emit just the same frequencies of light that any object at 10$^{\circ}$C would (i.e. in the infrared range);
  3. Emit some other combination of frequencies?

If #1, what is causing it to behave differently than another similarly-temperatured object? Why wouldn't it emit EM over "virtually all wavelengths"? (i.e. in this case, what makes it not an instance of "Every object" as in the first quote?)

If #2, how is it able to emit light in the infrared range? Doesn't the molecular structure of hydrogen only allow it to emit light in certain frequency ranges (i.e. from electron transitions(?)).

If #3, what's causing it to emit those other frequencies?

I believe #3 is the case but I am a bit unclear on the details of how IR emission happens vs. emission along the spectral lines.

$\endgroup$
1
  • $\begingroup$ Please see my answer and spectra here: physics.stackexchange.com/a/768678/313612. In the electrically excited hydrogen discharge tube, the tube has a plasma consisting of hydrogen molecules, hydrogen atoms, electrons, ions and various excited and not excited such species. As shown in my spectra, you get the Balmer lines from excited hydrogen atoms that de-excite. You also get, e.g., the Fulcher alpha band emission from excited hydrogen molecules that de-excite. Dissociating hydrogen molecules by heating them requires fairly high temperatures. $\endgroup$
    – Ed V
    Commented Jul 3, 2023 at 0:09

1 Answer 1

0
$\begingroup$

(An attempt to answer my own question, but not sure if this is correct.)

My understanding currently is that the answer is #3.

If we heat the gas via conduction/convection, when the gas as a whole is 10C, it'll emit infrared radiation because of the vibration of the molecules. Whatever the gas would be (helium, argon, oxygen, nitrogen), it will emit basically the same wavelengths of infrared at this temperature. These vibrational effects don't ionize the gas so it won't emit the blue light. Thermal equilibrium is when it's emitting as much energy via infrared radiation as energy it's receiving via the heating element.

If we heat the gas via the electric current only, then this ionizes the gas and therefore it does emit the blue light. As per Andrew Stean's answer here:

When you heat a gas by electromagnetic radiation with the molecules absorbing the incident radiation, what happens is that energy which at first is internal to each molecule then gets shared among molecules by collisions, so what what was an internal excitation energy gets transferred into other forms of energy such as motion of the molecule as a whole. All this happens on a timescale of nanoseconds at ordinary pressure and temperature. The molecules will also emit much of the radiation so overall the heating of the gas is a result of an average over many such processes. Since each process is so quick the gas overall heats up in a way quite like the way it does when heated by some other mechanism such as a flame warming the bottom of the tank.

That is to say, some of the ionized gas will just heat the rest of the gas, while some will de-ionize and emit the blue light.

Thermal equilibrium will be reached when the energy loss via infrared radiation of the molecule vibration plus the energy loss via the deionization equals the energy gain of the electric current. This means it will emit less infrared than the gas heated vibrationally, since it's losing the same amount of energy but via two mechanisms instead of one.

$\endgroup$
0

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.