Author Topic: I you heat an orange up to 1 billion degrees and let it cool off in the air...  (Read 2948 times)

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Offline BeaminTopic starter

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You would have an explosion equal to a nuclear bomb right? Or maybe not a ball of water but rather a ball of lead the size of an orange. If its at this temperature it will gives off its thermal energy in xrays that I do know. Maybe even gamma rays (is their a mechanism for photons to be produced by electrons descending to a lower energy level where the potential is great enough to give off gamma? I know the balmer series or one of the orbital decay series gives off Xrays which is how we make them in xray machines) . It seems this is where the destructive power comes from.  :scared: :bullshit:
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Offline T3sl4co1l

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Hot stuff gives off all kinds of radiation, yes. ;D

Suppose you've got a ball of really hot stuff, that's somehow confined in that space so it can be considered thermalized.  Each particle banging into the next has an average collision energy in the ~MeV, say.  Not only will the nuclei be ~fully stripped of electrons (making for some very intense points of electric field), but everything zipping around will give rise to pair production, whether from electron-electron, electron-photon, photon-photon, or electron-nucleus* interactions.

*AFAIK, nuclei aren't important to the process, but their intense electric field is, and the nucleus serves as reaction mass, which is where the momentum balances for the "new" particle(s) produced.

Because of these interactions, temperature is not just the average energy, but the average mass-energy, and the sum must include the fraction of antimatter present, as well as the kinetic energy.  The heat capacity of such a plasma should be quite high indeed, because more energy leads to more probability of pair production, and more pair production leads to higher intensity gammas, as well as all the bremsstrahlung making good old x-rays (except they're gammas too, because of the energy scale).

I think.  I don't know of pair production being relevant to very hot electron-degenerate matter (white dwarfs).  They're only in the millions of K, though.  They're also gravitationally bound.  And, their density is hyperbolic with energy, collapsing into a denser form of matter (i.e., neutron stars) when the electrons become relativistic, so at the same point where pair production might become relevant, gravity is too strong to matter and it blips into something else.

In any case, the spectrum would be continuum, because of the Doppler shift (you wouldn't expect to see 511keV gammas, but a very broad peak smeared out by the average kinetic energy), excess collision energy, and bremsstrahlung.  There might be some \$\textrm{K}\alpha\$ lines (or gaps, for that matter) due to heavy, nearly-bare nuclei having atomic transitions (relevant to the case with lead, perhaps).

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Offline free_electron

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bremsstrahlung.. the germans have such nice words ....  radiation caused by slowing down of particles. literally : brakingradiation
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Offline T3sl4co1l

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And it's one of the longer German words that I manage to remember the spelling of. ;D

Also, name relevant ^^ 8)

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Offline TimFox

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At the University of Chicago, we had a guest lecturer from Darmstadt who was a little worried about his English, and arranged for an American student with good German to sit in the front row to help with any language problem.  At one point, he asked "Wie sagt man 'Bremsstrahlung' auf Englisch", to which the student replied "Bremsstrahlung".
 

Offline BeaminTopic starter

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So to make gammas from electrons jumping down orbitals how long is the jump? I thought the limit was xrays as there just weren't any higher orbital to jump from.

Same thing with breaking radiation: electrons (from an external source crash into other electrons(in a different atom) kicking one out, then jumping into a lower orbital but once again where does this higher valance band live considering its crashing into air (O2 N2) or sand (Si O and Mg)?

I remember in inorganic chem we went over this and no one could figure out why the electrons left the atom in B- decay and weren't absorbed into the outer shell. I don't think the professor knew to well because the simple answer was it was too many keV to be captured.
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Offline Yansi

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And it's one of the longer German words that I manage to remember the spelling of. ;D

Also, name relevant ^^ 8)

Tim

There seems to be some people liking making fun out of German language.
So what about this one?  :-DD
http://www.catb.org/jargon/html/graphics/gefingerpoken.jpg


I sometimes use that sign out of pure stupid fun printed and taped to some delicate (and quite dangerous to some rubbernekken) stuff like 7kWh 400V accumulator pack.  :D  ;)
 
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Offline T3sl4co1l

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So to make gammas from electrons jumping down orbitals how long is the jump? I thought the limit was xrays as there just weren't any higher orbital to jump from.

Same thing with breaking radiation: electrons (from an external source crash into other electrons(in a different atom) kicking one out, then jumping into a lower orbital but once again where does this higher valance band live considering its crashing into air (O2 N2) or sand (Si O and Mg)?

I remember in inorganic chem we went over this and no one could figure out why the electrons left the atom in B- decay and weren't absorbed into the outer shell. I don't think the professor knew to well because the simple answer was it was too many keV to be captured.

Take note of the condition I added -- that heavy atoms are involved. :)

The first electron to bind to a nucleus is hydrogenic, i.e., its wave function is equivalent to the single proton, single electron, hydrogen atom case, but with the nucleus being much stronger (both in mass and in charge).  This scales the atomic levels proportionally, so that a +20 charge ion binds at 20 * 13.6eV, and so on.

If you have lead nuclei floating in such a soup, you'll have most electrons too energetic to bind (hence the atom is fully ionized, or nearly), and that first binding level (or two or three) is very high energy (>100keV).  Hence you might expect an absorption or emission line of characteristic spectra.

On a related note, this is part of the reason why atoms beyond Z ~ 137 are hypothesized not to exist: the s1 orbital energy is greater than the electron rest mass, therefore the electron is effectively sucked into the nucleus, where it undergoes reverse neutron decay (e + p --> n), thus limiting Z.

As for beta decay, indeed, most are too energetic to remain.  Obviously, positron betas won't remain at all*.  Some atoms do sometimes decay through electron capture events (see above, but Z < 137), in which case the emitted gamma(s) has to do with the cascade of electrons jumping down to the just-now-vacant s1 (or whatever) orbital that got captured.

*Heh, I suppose if you had a negative ion, a positron could bind, if it were cooled to a nice low energy along the way.  The captured positron would quickly locate an electron in the atom, then later, annihilate; but it could be stable for picoseconds at a time, maybe even nanoseconds.  Positronium (the hydrogenic atom formed between an electron and positron) has a lifetime in fractional nanoseconds (same spins) to hundreds of ns (opposite).

Tim
« Last Edit: August 25, 2017, 07:13:35 am by T3sl4co1l »
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Offline BradC

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So what about this one?  :-DD
http://www.catb.org/jargon/html/graphics/gefingerpoken.jpg

I showed that to a German girl I was dating. Didn't matter how many times I flipped the paper over, she could never see the funny side of it. I should have taken that as a sign.
 

Offline Halcyon

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So what about this one?  :-DD
http://www.catb.org/jargon/html/graphics/gefingerpoken.jpg

Well done!

Reminds me of a movie, Top Secret! with Val Kilmer, well worth seeing if you like that subtle yet obvious humour. You can buy it on DVD fairly cheaply.

Here are some relevant scenes:



 

Offline John Coloccia

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What's the question?
 

Offline SeanB

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If you heat it up and pop it into any arbitarary room, there will, within what in almost all measures is nearly instantly, a very large and rapidly expanding cloud of plasma, mostly composed of the contents of the room, the air in there along with some really energetic radiation ( xrays will certainly be at the lower end of this radiation, kind of like the nearly supercold radiation) propagating out with the blast front. It will, in almost every respect aside from the orange part, be equivalent to a 50 kiloton nuclear bomb going off there.
 

Offline T3sl4co1l

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Similar question, rather more impressive result:
https://what-if.xkcd.com/1/

Tim
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Offline TimFox

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Note:  gammas and x rays are both examples of photons.  The names refer to their origin:  x rays are produced by electron transitions between atomic states, including Bremsstrahlung, but gammas are produced in nuclear transitions.  For high-energy radiography, I used to use linear-accelerator x-ray sources that generated nominal 6 or 9 MeV x rays (broad energy spectrum), as well as Co-60 radioactive-decay sources that produced gammas at approximately 1.2 and 1.3 MeV (two-energy spectrum).
 

Offline Rick Law

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If you heat it up and pop it into any arbitarary room, there will, within what in almost all measures is nearly instantly, a very large and rapidly expanding cloud of plasma, mostly composed of the contents of the room, the air in there along with some really energetic radiation ( xrays will certainly be at the lower end of this radiation, kind of like the nearly supercold radiation) propagating out with the blast front. It will, in almost every respect aside from the orange part, be equivalent to a 50 kiloton nuclear bomb going off there.

Trouble is --- Can you still call a swamp of plasma an orange?

Way before it hit a billion degree, what was an orange is now vapor.  Molecules that was part of that orange will dissociate...
 


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