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If we pass current through a gas, like in the discharge tube, the electrons will accelerate in the electric field. The accelerated electrons will collide with gas molecules, and transfer some of their energy to them. Thus, producing gas molecules in the excited state. The excited molecule can come to ground state by emitting its excess of energy in the form of light. Thus, we can see gasses emitting light when current is passed through them.

Will the current carrying solid conductors (example,copper wire) emit light similar to that of gases?

$\large{\color{red}{\checkmark}}$This particular question is about the current carrying conductors which are not insulated (seen in everyday life) and so as to why they don't emit light. If we consider other cases, there will be radiation emitted from the conductor as discussed in this particular question.

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    $\begingroup$ Blackbody radiation is the answer. $\endgroup$
    – jinawee
    Commented Oct 20, 2013 at 19:51
  • $\begingroup$ i somehow managed to edit your question and wrote superconductor instead of conductor to make some sense. As normal conductors always heat up and heat means IR. $\endgroup$
    – user28737
    Commented Oct 21, 2013 at 8:27
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    $\begingroup$ @WaqarAhmad That's an invalid edit as it radically changes the original intent of the question. Please see the FAQ and the two policy blog posts, here and particularly here. $\endgroup$
    – Emilio Pisanty
    Commented Oct 21, 2013 at 9:34
  • $\begingroup$ Corona discharge may be relevant then. $\endgroup$
    – user28737
    Commented Dec 14, 2013 at 8:50

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No, it won't.

In a gas, you can treat the electrons and atoms as simple particles. They collide, and the collisions happen with sufficient energy to knock some of the atoms electrons to higher energy bands. When these electrons fall back, the energy is released as light.

In a metal, the atoms are so close together that their outer electrons can freely move around. And since electrons are indistinguishable, all these outer electrons together form a single "cloud" between the nuclei. This cloud, since it's not tied to a single atom, can easily move around. This is why metals are so conductive.

However, this means that the electric current in a metal is just the cloud as a whole moving. There are no collisions which knock electrons into higher bands (those simply don't exist/would overlap the neighboring atoms), so there's no associated radiation.

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Metals can emit radiation in a similar way, but they must also be in gas form. Typically they are encased in glass, and heated up with current until vaporized--then these excited gaseous atoms relax, emitting photons. Your common CFLs and fluorescent lights use Mercury specifically. Street lamps use sodium, and high power stadium lights may use iodides or bromides of these same metals.

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    $\begingroup$ Not exactly what was asked. $\endgroup$
    – Ignacio Vergara Kausel
    Commented Oct 20, 2013 at 20:39
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Blackbody radiation (as commented above) is because of the temperature and present in any solid body. But this is not analogous to the electronic transitions described in the question.

The closest thing I can think of in a solid object are LEDs, but they're semiconductors and not conductors.

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