Re: Temperature and radiation absorption
From: Old Man (nomail_at_nomail.net)
Date: 09/25/04
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Date: Fri, 24 Sep 2004 22:25:55 -0500
"Doodedski" <billybilly54321@hotmail-dot-com.no-spam.invalid> wrote in
message news:4153c433$1_2@127.0.0.1...
> 2 unrelated questions that no one at school can answer clearly.
> First, why is the definition of temperature not just simply a constant
> times the kinetic energy (rotation, vibration, translation...) of
> particles instead of dQ/dS which I don't quite understand the
> physical sense?
Temperature is defined WRT an equilibrium distribution of
photon energies, Planck's black body distribution, or WRT an
equilibrium distribution of gas particles, Maxwell's distribution.
Temperature isn't defined for other distributions, such as that
of a gas of mono-energetic particles.
> Second, can anyone explain to me how an atom can absorb a photon? It
> seems to me that if an electron absorbs a photon, it goes to a higher
> level of energy but must come right back down reemiting a photon? No
> additionnal energy is acquired by the atom. By which process is it
> possible for an atom (alone, in a cristal structure, in a gas...) to
> absorb a photon's energy and to keep it in the form of vibrational
> energy, kinetic energy...
Old Man isn't inclined to write a book for Doodedski.
Get a book on quantum mechanics.
> Oh, and what happens if the photon hits the nucleus?
Completely negligible.
The nuclear cross-section is 10^(-10) times smaller than
that of atoms. Except for an overlap between x-rays and
gamma-rays, nuclei aren't excited by photons with atomic
excitation energies.
[Old Man]
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