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In an incandescent light bulb, electrons flow through wires to the bulb, where is passes through a filament, and the filament lights up.

Does this work because of several high energy electrons losing some of their energy to disturb the electromagnetic field, creating photons? The electrons just leave the filament afterward and are replaced by more high-energy electrons from the negative terminal. Or does it work differently?

As a secondary question, is there a type of light bulb that directly converts electron flow to photons?

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The filament will be made of a material with very high resistivity - such as tungsten. Tungsten has a resistivity of $52.8\ \Omega m$ and compared to a substance such as copper which is typically used for wires and has a resistivity of $1.72\times 10^{-8}\ \Omega m$ that is very high.

This means that the motion of the electrons through the wire is resisted, and when they hit atoms they transfer their kinetic energy to them in the form of thermal energy. This causes the wire to heat up.

As things heat up they begin radiating light - and when they become hot enough they being to radiate visible light which is what is seen in incandescant light bulbs.

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  • $\begingroup$ Is there any light bulb that directly converts the electron flow to light? $\endgroup$ – Aditya Radhakrishnan Sep 3 '17 at 10:32
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    $\begingroup$ The closest is an LED $\endgroup$ – Jim Garrison Sep 4 '17 at 3:42
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    $\begingroup$ I doubt it. Electrons are charged. Light is not. Charge is maintained. $\endgroup$ – Hennes Sep 4 '17 at 5:03

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