| Topic: |
Science > Physics |
| User: |
"PlanckQ" |
| Date: |
13 Oct 2005 04:38:47 PM |
| Object: |
Blackbody radiation VS Spectral line emissions |
Suppose you have a blackbody at temperature T and a pure low-density
gas at the same temperature. For T, the blackbody radiation spectrum
can be plotted using Planck's theory. I also understand that the
spectrum of the gas is zero for all wavelengths except for certain
"emission bands".
If you were to superimpose the gas spectrum onto the blackbody
spectrum, do the gas emission bands ever exceed the blackbody curve, or
is the latter curve the limit?
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| User: "Androcles Androcles@ MyPlace.org" |
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| Title: Re: Blackbody radiation VS Spectral line emissions |
13 Oct 2005 06:23:33 PM |
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|
"PlanckQ" <engpete@hotmail.com> wrote in message
news:1129239527.745032.261340@g14g2000cwa.googlegroups.com...
| Suppose you have a blackbody at temperature T and a pure low-density
| gas at the same temperature. For T, the blackbody radiation spectrum
| can be plotted using Planck's theory. I also understand that the
| spectrum of the gas is zero for all wavelengths except for certain
| "emission bands".
|
| If you were to superimpose the gas spectrum onto the blackbody
| spectrum, do the gas emission bands ever exceed the blackbody curve,
or
| is the latter curve the limit?
I don't understand your question. Perhaps this will help.
Emission:
http://users.forthnet.gr/ath/jgal/spectroscope/amici.html
Absorption:
http://outreach.atnf.csiro.au/education/senior/astrophysics/spectral_class.html
Androcles
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| User: "Timo Nieminen" |
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| Title: Re: Blackbody radiation VS Spectral line emissions |
13 Oct 2005 06:42:41 PM |
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On Fri, 13 Oct 2005, PlanckQ wrote:
Suppose you have a blackbody at temperature T and a pure low-density
gas at the same temperature. For T, the blackbody radiation spectrum
can be plotted using Planck's theory. I also understand that the
spectrum of the gas is zero for all wavelengths except for certain
"emission bands".
If you were to superimpose the gas spectrum onto the blackbody
spectrum, do the gas emission bands ever exceed the blackbody curve, or
is the latter curve the limit?
For a blackbody, emissivity = absorptivity = 1 for all wavelengths. For
anything, absorptivity = emissivity at each wavelength. At a given
temperature, radiated power at each wavelength is proportional to the
emissivity.
Therefore, your question is equivalent to asking whether the emissivity =
absorptivity at an wavelength can be greater than 1 for the gas. Since the
absorptivity is what fraction of incident power is absorbed, it cannot be
greater than one. Therefore the blackbody curve is the limit.
That's assuming that the gas is in local thermodynamic equilibrium (LTE,
which means that level populations are given by the Boltzmann
distribution), which is implied by saying the gas is at the temperature T.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "PlanckQ" |
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| Title: Re: Blackbody radiation VS Spectral line emissions |
13 Oct 2005 09:13:53 PM |
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Hi Timo,
You base your answer on the assumption that a blackbody will absorb all
the EM radiation at a certain wavelength and re-emit it at the same
wavelength for thermal eqbm.
I think this is wrong because the blackbody will attain thermal eqbm by
re-emitting the absorbed energy into a full range of wavelengths (in
the Planck form) NOT just the absorbed wavelength.
Pete
Timo Nieminen wrote:
On Fri, 13 Oct 2005, PlanckQ wrote:
Suppose you have a blackbody at temperature T and a pure low-density
gas at the same temperature. For T, the blackbody radiation spectrum
can be plotted using Planck's theory. I also understand that the
spectrum of the gas is zero for all wavelengths except for certain
"emission bands".
If you were to superimpose the gas spectrum onto the blackbody
spectrum, do the gas emission bands ever exceed the blackbody curve, or
is the latter curve the limit?
For a blackbody, emissivity = absorptivity = 1 for all wavelengths. For
anything, absorptivity = emissivity at each wavelength. At a given
temperature, radiated power at each wavelength is proportional to the
emissivity.
Therefore, your question is equivalent to asking whether the emissivity =
absorptivity at an wavelength can be greater than 1 for the gas. Since the
absorptivity is what fraction of incident power is absorbed, it cannot be
greater than one. Therefore the blackbody curve is the limit.
That's assuming that the gas is in local thermodynamic equilibrium (LTE,
which means that level populations are given by the Boltzmann
distribution), which is implied by saying the gas is at the temperature T.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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| User: "Timo Nieminen" |
|
| Title: Re: Blackbody radiation VS Spectral line emissions |
13 Oct 2005 10:56:50 PM |
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|
On Fri, 13 Oct 2005, PlanckQ wrote:
You base your answer on the assumption that a blackbody will absorb all
the EM radiation at a certain wavelength and re-emit it at the same
wavelength for thermal eqbm.
I think this is wrong because the blackbody will attain thermal eqbm by
re-emitting the absorbed energy into a full range of wavelengths (in
the Planck form) NOT just the absorbed wavelength.
Yes, of course it will. The "temperature" of the incident radiation
doesn't need to be the same as the temperature of the blackbody. All that
is required is that the _absorptivity_ equals the _emissivity_ at each
wavelength; this is _not_ the same as saying that the absorbed power is
the same as the radiated power at each wavelength.
But the question of whether a gas in LTE can emit more at some wavelength
than a blackbody doesn't depend on that. Just that at any wavelength, the
maximum absorptivity is 1, and therefore the maximum emissivity is 1.
If you want more than a rather hand-waving explanation, it is possible to
show in more detail that the blackbody spectrum is the limit for emission
by a gas of uniform temperature. If you're keen, and know about the
Einstein coefficients A and B for absorption and emission:
http://www.physics.uq.edu.au/people/nieminen/papers/thesis/chapter3_spectrum.pdf
Eqn (3-11) shows that if the source function S_lambda is constant, then
that's the maximum intensity at that wavelength. Section 3.3.4 shows that,
for a gas in LTE, S_lambda = the Planck function. Thus, the blackbody
spectrum is the limit.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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| User: "Uncle Al" |
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| Title: Re: Blackbody radiation VS Spectral line emissions |
13 Oct 2005 08:39:00 PM |
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PlanckQ wrote:
Suppose you have a blackbody at temperature T and a pure low-density
gas at the same temperature. For T, the blackbody radiation spectrum
can be plotted using Planck's theory. I also understand that the
spectrum of the gas is zero for all wavelengths except for certain
"emission bands".
At any given wavelength, blackbody emissivity must exactly equal
absorptivity - First Law of Thermodynamics. A colorless gas cannot
emit were it lacks optical absorption. Electronic emission must be
dipole ungerade. That doesn't give an argon atom many options until
you pump in enough energy to pump electronic transitions in the atom.
A dissociated plasma, OTOH, can be opaque and brilliantly emissive:
the sun's "surface."
If you were to superimpose the gas spectrum onto the blackbody
spectrum, do the gas emission bands ever exceed the blackbody curve, or
is the latter curve the limit?
Do you think the mercury 254 nm line or sodium D-line (doublet) is
limited to thermal amplitudes? Ever see a neon light?
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
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| User: "PlanckQ" |
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| Title: Re: Blackbody radiation VS Spectral line emissions |
13 Oct 2005 09:56:01 PM |
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Al,
I believe the 1st law of thermo for a blackbody does not have to be
satisfied by absorption AND emission AT THE SAME wavelength. As I
mentioned to Timo, I think re-emission is distributed over all
wavelengths for a blackbody, whilst still satisfying the 1st law.
RE your neon light example, it's not the same situation as my thought
experiment in which I mean to assume no external sources of excitation
of the gas molecules. Nevertheless, can you provide proof that the neon
temp is not high enough to be the limiting factor of its emission
bands? ie. the gas temp may be high enough to raise the blackbody limit
to levels emitted by the excited gas.
Pete
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| User: "Androcles Androcles@ MyPlace.org" |
|
| Title: Re: Blackbody radiation VS Spectral line emissions |
14 Oct 2005 03:56:16 AM |
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|
"PlanckQ" <engpete@hotmail.com> wrote in message
news:1129258561.776432.174730@g14g2000cwa.googlegroups.com...
| Al,
|
| I believe the 1st law of thermo for a blackbody does not have to be
| satisfied by absorption AND emission AT THE SAME wavelength.
Correct.
Google for Balmer, Lyman, Paschen, Pfund, Rydberg, Rayleigh-Jeans.
Androcles.
| As I
| mentioned to Timo, I think re-emission is distributed over all
| wavelengths for a blackbody, whilst still satisfying the 1st law.
|
| RE your neon light example, it's not the same situation as my thought
| experiment in which I mean to assume no external sources of excitation
| of the gas molecules. Nevertheless, can you provide proof that the
neon
| temp is not high enough to be the limiting factor of its emission
| bands? ie. the gas temp may be high enough to raise the blackbody
limit
| to levels emitted by the excited gas.
|
| Pete
|
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