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Doppler broadening in a gas

  1. Feb 28, 2006 #1
    Hi,

    A very hot gas is enclosed in an oven with a small window. The gas molecules emits radiation at a characteristic wavelength. I assume that because of the thermal motion of the molecules the emitted wavelengths will form a spectrum of some kind (Doppler broadening.) I am trying to derive an expression for this spectrum, but I don't know how... Any hints?

    Thanks.
     
  2. jcsd
  3. Feb 28, 2006 #2

    Astronuc

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    Staff: Mentor

    Yes. There would be no Doppler broadening for molecules moving parallel to the plane of the window. Molecules moving away would have red shift, and molecules moving toward the window would have a blue shift. The higher the temperature, the greater the range of velocities.

    Maxwell-Boltzmann distribution - http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html#c2
     
  4. Feb 28, 2006 #3

    Physics Monkey

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    Like Astronuc said, you know the Doppler shift as a function of atomic velocity, and you know the distribution for atomic velocities. It's basically a change of variables.
     
  5. Mar 5, 2006 #4
    I tried to perform the calculation of the spectrum.
    I got the spectrum dP/df (power emitted each frequency unit):

    dP/df = Pt >< dv/df >< fr(v)

    Pt is the total power emitted by the gas, fr(v) is the Maxwell distribution regarding the radial component of speed (component in observer's direction), and v(f) is the inverse of f(v) which gives the doppler frequency as a function of the radial velocity.
     
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