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LocationX
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Why do we see a continuous spectrum of radiation for blackbodies when we know that energy in atoms are quantized thus only emitting discrete energy levels?
LocationX said:Why do we see a continuous spectrum of radiation for blackbodies when we know that energy in atoms are quantized thus only emitting discrete energy levels?
LocationX said:You have a macroscopic piece of solid carbon. Each carbon atom has the same energy levels thus the emission lines should correspond to the same discrete energy levels. Would we still see a continuous spectra?
When we say 100 atoms, does this include any atoms? What about mercury gas? Or Hydrogen gas? When we heat up mercury gas or hydrogen gas and look at it through a diffraction grating we see the emission spectra for that specific element. Thus, is it necessary to mandate that the blackbody must be multi-elemental in order to get a continuum?
ansgar said:no solids have energy BANDS not energy lines.
you know how a black body works so that suggestion is not correct ;)
LocationX said:Where do energy bands come from? And also what about the second part of that question?
I think the harmonious_oscillator model can explain this phenomenon well. And frequecy of oscillation is not the same!LocationX said:Why do we see a continuous spectrum of radiation for blackbodies when we know that energy in atoms are quantized thus only emitting discrete energy levels?
ansgar said:1) detectors smear the energy
2) energy lines in atoms are smeared as well "time-energy uncertainty relation"
genneth said:Also:
3) The energy spectrum of a macroscopic body is not simply the atomic spectra
4) Putting two and three together, let's assume we have 100 atoms, each with 4 reasonably accessible energy levels. That gives 4^100 states. Generously, these will live within 100 eV of each other. That mean the mean energy spacing is about 6*10^-59 eV, or 10^-77 joules. Using the energy-time uncertainty relation, to resolve this energy difference, you would need to make an observation for about 3*10^35 years. The current probably age of the universe is 1.4*10^10 years. There are many more than 100 atoms in a typical macroscopic sample, and a lot more than 4 levels per atom.
LocationX said:You have a macroscopic piece of solid carbon. Each carbon atom has the same energy levels thus the emission lines should correspond to the same discrete energy levels. Would we still see a continuous spectra?
When we say 100 atoms, does this include any atoms? What about mercury gas? Or Hydrogen gas? When we heat up mercury gas or hydrogen gas and look at it through a diffraction grating we see the emission spectra for that specific element. Thus, is it necessary to mandate that the blackbody must be multi-elemental in order to get a continuum?
ansgar said:no solids have energy BANDS not energy lines.
you know how a black body works so that suggestion is not correct ;)
Sillyboy said:I think the harmonious_oscillator model can explain this phenomenon well. And frequecy of oscillation is not the same!
No, The black body radiation should be explained by the harmonious oscillation of the atoms.LocationX said:Why does the detector not smear the energy for hydrogen gas when it is heated? We see the emission lines very clearly thought a diffraction grating.
3. Is a bunch of hydrogen not a macroscopic body? By mandating that it has to be macroscopic, the we shouldn't see spectra lines for hydrogen gas when heated in a tube.
4. So let me get this straight. We have 100 hydrogen atoms. However, if we blast the heck out of these atoms, they will jump into higher and higher energy states. In your example, you're saying that the energy spacing is within 100eV of each other. However, for H gas, the energy spacing is much less for hydrogen, just several eV. If you blast it too much with thermal energy, it will ionize thus making plasma (?). So here's the thing, even though hydrogen gas has much lower energy spacing than 100eV, the energy spectra does not smear to the detector (the human eye). We see discrete energy levels corresponding to the emission spectra of hydrogen. Why does this not spread and a blackbody would spread the energy spectra? I really feel like I'm missing the big picture.
The wiki article says:
"It is due to the diffraction of the quantum mechanical electron waves in the periodic crystal lattice."
Now, do they mean the electrons bound to the atoms? The free electrons? The electrons in the valance band? Conduction band? doped electrons?
Assuming these energy bands represent the electrons bound to the lattice, then why are they bands? As genneth says, " let's assume we have 100 atoms, each with 4 reasonably accessible energy levels. That gives 4^100 states." Then i can understand there will be a smear of different energy states that make it look like a continuous energy distribution. However, in realize, these are actually very packed discrete energy levels. Is this the right picture?
Then why are there bands? What causes the regions where no electrons can have that energy level? Wikipedia says "the diffraction of the quantum mechanical electron waves". Why would bound electrons diffract off of anything? Unless this was talking about an electron traveling inside the lattice, which makes it even confusing as to why that would effect the various energy levels of the electrons bound to the lattice.
And why doesn't a blackbody have energy bands if solids have energy bands?
A harmonic oscillator describes the bound electron to the nucleolus right? However, how would a simple model like that exhibit band structure?
jrosen13 said:Thats right, it is not transitions between atomic levels that leads to blackbody radiation, but instead if we have a collection of harmonic oscillators at some temperature and we ask how the different states will be populated. Each state corresponds to a different frequency of vibration, hence a different wavelength of light. The reason we do not observe the discreet changes in frequency is that we are talking about atoms, not electrons, and they are many orders of magnitude heavier, so the level spacing is so small that we can consider it to be continuous.
LocationX said:Why do we see a continuous spectrum of radiation for blackbodies when we know that energy in atoms are quantized thus only emitting discrete energy levels?
LocationX said:"Each state corresponds to a different frequency of vibration, hence a different wavelength of light." What would cause photons to be emitted from the blackbody? Unlike the absorption and radiation from electrons going up and down energy states in an atom, what do these "frequency of vibrations" do to emit photons? Are they constantly wanting to go back to the ground state thus the phonons emit photons?
LocationX said:If a blackbody was heated to exactly the same temperature, completely uniformly, then wouldn't we see ONE frequency for photon emission from the blackbody?
f95toli said:Real atoms do NOT have "discrete" energy levels. Real energy levels -be it in atoms or in solids etc.- always have a finite width. Even if you take the simplest case of a single atom you will find that the energy levels are "smeared" due to the coupling to the vacuum (the eigen- energies you get from solving the SE corresponds to the centres of Lorentzian distributions).
Note also that the usual mathematical relation between bandwidth and lifetime applies, excited states with a finite lifetimes (=all real states) will always correspond to levels with finite width.
Who said I had a bunch of bosons? We keep a piece of steal at a temperature that makes it glow red, thus we have just confined it to emit a certain wavelength. What's wrong with that?Again, we heat to a very specific temperature T=X>>0K. Boltzmann distribution tells us the number of states possible at T=X. So am I correct to say that blackbodies obey the Boltzmann distribution at high temperature and thus we have lots and lots of energy states that could be occupied? But that is only for ONE temperature. If it could occupy so many different states with just one temperature, then why does a piece of metal glow red at a specific temperature and not have a huge spectrum of various photon energies at T=X?jrosen13 said:This is definitely not correct, since at some finite temperature we have a thermal population of the levels according the the Bose-Einstein distribution which is similar to the Boltzmann distribution at high temperatures.
ansgar said:"Why does the detector not smear the energy for hydrogen gas when it is heated? We see the emission lines very clearly thought a diffraction grating."
"very clearly" is not equal to "infinite thin"... they are smeared, but not much. This is the issue of building good detectors, compare how good a germanium detector is to resolve gamma rays compared with a scintillator...
LocationX said:So a Sodium atom does not have discrete energy levels? Even if NIST has measured these very energy levels to precise eV INCLUDING spin orbit coupling and hyperfine splitting? What you're saying doesn't make sense. How would you even include the coupling to vacuum in the Hamiltonian? Are you talking about coupling to virtual particles?
LocationX said:Who said I had a bunch of bosons?
LocationX said:We keep a piece of steal at a temperature that makes it glow red, thus we have just confined it to emit a certain wavelength. What's wrong with that?
LocationX said:Again, we heat to a very specific temperature T=X>>0K. Boltzmann distribution tells us the number of states possible at T=X. So am I correct to say that blackbodies obey the Boltzmann distribution at high temperature and thus we have lots and lots of energy states that could be occupied? But that is only for ONE temperature. If it could occupy so many different states with just one temperature, then why does a piece of metal glow red at a specific temperature and not have a huge spectrum of various photon energies at T=X?
LocationX said:Why do we see a continuous spectrum of radiation for blackbodies when we know that energy in atoms are quantized thus only emitting discrete energy levels?
Dickfore said:Because blackbodies are not individual atoms.
jrosen13 said:I think I did.
Nothing, depending on what you mean by certain.
It is clear that your understanding of a distribution is not very good. The reason it is called a blackbody distribution is that it is a probability distribution for all the wavelengths that will be emitted. This is basic stuff, can you see IR? Would you be able to see the part of the spectrum emitted in IR or UV? Not with your eyes. So your "gut" feeling for red meaning one wavelength is just simply not correct. Its a lot of different wavelengths. As it gets gotter, does it go like the rainbow ROYGBIV? Not exactly. The sun is a blackbody, does it emit at just one frequency? No. Its a distribution! When it gets really hot, it will just look white, i.e. "white hot", because it has a lot of different wavelengths.
LocationX said:and neither is hydrogen gas... but we see discrete spectra
LocationX said:Once again, my fundamental question has not been answered. It's is obvious that I do not understand blackbody radiation. I know what it describes, but those are only words without substance. Saying a blackbody is a probability distribution for all wave lengths is the same as saying BE distribution is a distribution for bosons. It means nothing. Let me explain again why I clearly do not understand blackbodies.
We take hydrogen gas. We thermally excite hydrogen gas. There are lots of atoms, that all excite and then drop back down thus producing the discrete spectra that we see. The explanation for ONLY discrete spectra lines are from the electrons having discrete energy levels in their various orbitals.
Now let's think about a blackbody. A blackbody is also made of atoms. Let's for example make all the atoms identical, thus its all the same element. We excite these atoms and they should in principle also emit discrete energy lies. But why do they not? What is physically happening in the atoms that causes them to produce a continuous spectra? That's the other thing. Why doesn't the electrons in the atoms ionize in a blackbody?
Blackbody radiation refers to the electromagnetic radiation emitted by an object at a certain temperature. This radiation is produced due to the thermal motion of particles within the object and follows a specific distribution known as the blackbody radiation curve.
Blackbody radiation was initially a mystery to scientists because classical physics could not fully explain its behavior. However, with the development of quantum mechanics, it was discovered that the distribution of blackbody radiation could be accurately described by using quantum principles.
The blackbody radiation curve shows the distribution of energy emitted by an object at different wavelengths. As the temperature of the object increases, the peak of the curve shifts towards shorter wavelengths, meaning that more high-energy radiation is emitted.
The Planck constant, denoted by the symbol h, is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. In the context of blackbody radiation, it is used to calculate the amount of energy emitted at a particular wavelength, which is crucial in understanding the behavior of blackbody radiation.
Blackbody radiation has numerous practical applications, especially in the field of thermal imaging. By analyzing the radiation emitted by objects at different temperatures, scientists and engineers can create images that reveal temperature differences, which can be useful in fields such as medicine, astronomy, and materials science.