Ok I'm totally out of my depth here as I'm only a student but I'm confused about the relationship between the emissivity and black-body characteristics of tungsten. I'm not sure if my fundamental understanding of the topics is incorrect but here is my research:

Spectral emissivity is the ratio of energy radiated from a real material's surface compared with that radiated from a perfect black-body at the same temperature, wavelength and observing angle. The closer this ratio is to one, the closer the material is to a good black-body

Tungsten has a spectral emissivity that averages about 0.4 depending on temperature, age etc.

Tungsten filaments in incandescent lights are very good representations of black bodies.

But how can the spectral emissivity of tungsten be so low but an incandescent bulb be almost a perfect black-body?


3 Answers 3


If the emissivity does not strongly depend on frequency in the relevant range the emitted spectrum will be like that of a black body, just with lower intensity.


Don’t think of emissivity as a fundamental material property in the same way as other physical properties.

It is actually a little complicated since it depends on the surface of a material as well as what the material is made of.

A smooth polished metal surface will have a lower emissivity than a rough or textured metal surface. Usually a rough oxidized surface will have a higher emissivity. The emissivity can also depend on the angle and wavelength especially for textured surfaces.

Thus you measure the emission from the smooth metal surface with a radiation thermometer it will appear to have a lower temperature than the rough surface. For the same temperature if you point the radiation thermometer at a black body of the same temperature of a metal with a low emissivity, it will read a higher temperature from the black body.

if you have a heat transfer calculation the emissivity is a useful parameter in your equations. In general radiation thermometers is calibrated by the black body radiation.

Note that like mentioned in another answer. The black body spectrum is the radiation emitted as a function of wavelength and when people talk about it they are generally referring to the general shape of the spectrum as a broad band light source who’s peak will shift with temperature.

So when you heat up the tungsten filament and look at the white light spectrum from a light bulb it will have the same shape as other black body radiators at the same temperature.

But if you have a polished tungsten plate at some temperature and an oxidized copper plate at the same temperature, you would find the the copper plate would have a higher emissivity if you measured the temperatures with a radiation thermometer calibrated to a true black body source.


According to Stefan-Boltzmann equation(ε =σT^4), emissive power is directly proportional to fourth power of temparature. So,(i) Tungsten has to be heated to very high temparature which is assisted by its high boiling point. According to Wien displacement law(max(λ)=b/T),the maximum frequency of spectral distribution for which the radiation has the greatest intensity is directly proportional to the temparature of the (nearly) black body. But Tungsten has a low average emissive power,so it can emit only low frequency radiation like red and orange on incandescence.


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