Consider an incandescent bulb having a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. The bulb is powered at constant voltage.

Can we solve this situation quanititatively for the temperature gradient or the resistance varitaion with temperature? Before the wire breaks, will it emit light of shorter wavelength? Is there any equation or concept regarding rate of evaporation of solids? Is it possible to predict these variations qualitatively atleast? Also, can tungsten evaporate? It has the highest melting point among all elements after all. And usually, isn't that vapour thing coming off of heated tungsten due to oxidation of tungsten?

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    $\begingroup$ At the operating temperature of the tungsten filament, tungsten has significant vapour pressure, so it evaporates very slowly. $\endgroup$
    – Gert
    May 12 '19 at 9:35

The non-uniform evaporation probably originates from a small variation in thickness of the filament, or possibly local variations in the degree of perfection of the crystal structure. Now thinner parts of the filament will have a greater than average resistance per unit length, and so will dissipate more power ($P=I^2R$). So they will get hotter, as they will tend to the steady state when power in = Power out, that is $$I^2R=\sigma A_{\text{surf}} T^4.$$ [Here we're assuming that the 'hot spot' in the filament radiates as a black body.] Because it's hotter the peak wavelength of emission is lower (Wien's law: $\lambda_{\text{peak}} \propto T^{-1}$), which I hope answers one of your questions.

Now the key point to note is this: we have an unstable situation... Because the thin parts of the filament are hotter, tungsten evaporates (sublimes) faster from them than from the rest of the filament. So they become thinner still, get hotter still – and so on.

Using a simple approximate formula linking sublimation rate to temperature, we could no doubt develop these ideas quantitatively.

  • $\begingroup$ There is also the emissivity to consider. It may vary locally do to surface roughness and corrosion. This may raise the temperature locally. $\endgroup$
    – my2cts
    May 12 '19 at 11:28
  • $\begingroup$ Yes indeed. Though, for the little it's worth, if I were trying for a quantitative treatment, I'd probably start with thickness as the only difference, and work with the black body approximation. $\endgroup$ May 12 '19 at 11:34
  • $\begingroup$ It all depends on how quantitative the treatment should be. Such an approach would not be adequate for a real light bulb. Because of the 4th power in T, 4% thickness variation only gives 1% temperature variation. In the end of course this will create instability and failure as you describe but effects of corrosion, convection, tungsten vapour transport all will play a role in the wire burning through. $\endgroup$
    – my2cts
    May 12 '19 at 12:13
  • $\begingroup$ I'm not claiming adequacy; just a simple starting approach. $\endgroup$ May 12 '19 at 12:16
  • $\begingroup$ Cracks may also appear. It would take a literature study, there should be plenty of it, to find out what limits the life time of the wire. Failure usually involves multiple causes combined. Failure is a very interesting and hot topic. $\endgroup$
    – my2cts
    May 12 '19 at 12:21

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