What happens if you heat an object until it glows in the xray spectrum?

Suppose that you had a lump of stuff in the middle of an evacuated chamber and were able to heat it continuously. Would there be a point where it was giving off nothing but invisible radiation?


It’d probably come apart.

X-rays go up to about 10 nm. Using the Wien displacement Law, a black body with a peak at 10 nm would have to be at about 300,000K. the highest substance melting point is about 7500 . Things vaporize well before 300,000 (although it’s short of plasma temperatures).

If you could heat something upo to that high a temperature and have it behave like a blackbody, there will still be an exponential tail extending into the visible. One common problem in optics is to derive the locus of blackbody points on a CIE chromaticity diagram. You can carry it out to infinitely high temperatures, and it tends toward a bluish point (“blue heat”, greater than “white heat”)


As you raise the temperature, the peak of the spectrum shifts to higher energy (shorter wavelength), but that doesn’t mean it stops radiating at lower energies. In fact, at a specific wavelength, the hotter the object, the more it radiates. (A white-hot surface emits more red light than a red-hot surface. It emits way more yellow & blue light, so on balance, it looks white.) So if you heat up an object to the point where it radiates X-rays, it becomes much brighter in visible light.

Of course, in the real world (even in vacuum), the object will turn into plasma before it gets that hot. And at that point, it doesn’t behave very much like a blackbody. But I don’t think it will stop emitting visible light.

Having it emit x-rays is easy. You’re emitting some x-rays right now at your current temperature, just not very much.

Having the spectrum peak in x-rays is harder. As CalMeacham notes, very high temperatures are involved.

Having it emit nothing but x-rays is damn-near impossible. Assuming a black-body spectrum, the radiation emitted at every wavelength increases with temperature. The position of the peak changes only because it increases faster for some frequencies that it does for others. Your hypothetical x-ray-peaking object is also going emitting tonnes of visible light – even more than it was when it was peaking in the visible range.

I’d say the turkey is done. Take it out of the oven already!

But the little wax thing hasn’t popped up yet!

The visible light a blackbody emits is proportional to its temperature to the first power, providing that it is hot enough that its peak emission wavelength is much shorter than visible light.

I can see its bones!! :eek:

BTW, I have never understood what exactly a blackbody is and how it differs from real stuff.

There’s a coupling factor between vibration in a substance and electromagnetic radiation in space, and it is always between zero and one. A blackbody is a mythical object with a value of one for this coupling factor, which is called emissivity

Some materials come pretty close. Humans are 97% of the way there, if it’s our skin that shows.

Whatever the emissivity of a flat exposed surface is, if you make a container whose interior is lined with that substance, and give it a big interior and a small opening to the outside, then the difference between one and its emissivity is the same difference for the flat exposed surface divided by the square of the ratio of the diameters of the opening and the interior. In other words, if you create a great big cave with a tiny porthole, the porthole itself has an emissivity very near 1 and is very close to a blackbody.

An entirely sealed cavity, as seen from its inside, is actually a blackbody (if the observer has zero size).

Depending, of course, on what wavelength band you’re looking at.

A blackbody recieves ratdiation at exactly the same rate, at exactly the same spectrum, as it emits it.

What effect, if any, does this have on the emission spectrum? If I put a notch absorbtion element in the chamber of a black-body emission device, what effect would it have?

Well, first, the rates can be different, depending on the temperature of the blackbody and the environment. I think you’re making the point that the emissivity as a function of wavelength is the same in transmitting mode as in receiving mode, right?

If you add any element (including a notch absorber) to the inner surface of the blackbody chamber, as long as that entire chamber wall including the element is at the same temperature (a restriction I forgot to mention above), it makes no difference. If you add the absorption element as a real physical object with nonzero size, it breaks the magic spell of an entirely closed isothermal surface.

A good way to think about the black body cavity is that any ray of radiation you want to follow is going to hit the surface, and it can either be absorbed, or it can bounce; and if it bounces, then it is going to hit the surface and can either be absorbed or bounce, etc etc. The only way it can end for the ray is to be absorbed, whether on the first bounce or the millionth. So, the absorption is complete and total. If your proposal changes that, you’ve changed the situation so it’s not a blackbody cavity.

If there is an environment, there isn’t blackbody radiation. My understanding of the reason for the hole-in-the-wall apparatus is to get an approximation of theoretical black-body radiation.

I’m asking now about the situation where the entire wall of the chamber is NOT at the same temperature. It is at the same temperature EXCEPT there is a notch in the radiation pattern (not the absorption pattern, because I’ve postulated a seperate notch absorber).

Obviously, the spectrum comming out of the hole in the wall will be different if the notch absorbtion devices occludes the hole, but I’m not interested in that.

My question is, what happens to the transmission spectrum of a hot body if it is absorbing not black body radiation, but a notched emission spectrum.

Does it re-radiate with a matching notch?

No. The whole point of a blackbody is that it absorbs 100% of the radiation incident on it (hence “black”) and emits radiation only due to its own properties. The emission spectrum of a black body of a given temperature looks exactly the same no matter what radiation is being used to raise it to that temperature.

Contrast this with ordinary objects that reflect some of the radiation incident on it (i.e. most normal things you look at). The spectrum you get off them definitely depends on what kind of light is being used to illuminate them.

I’m not sure of the rationale behind this statement. A black hole, for instance, is certainly a blackbody, and radiates as such, no matter what environment you put it in. And there are countless other examples for which it’s still an extremely good approximation.

I had to ask the question : How do you heat the body in the OPs question ? most of the heating methods will act as cooling methods far below the “X-ray temperature” that the OP desires!!

You bombard it with a bunch of energy, of course. What kind of energy matters almost not at all. Contrary to what you say, it’s really easy to heat things up, and quite difficult to cool them.