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Photon collision with H2 molecule

  1. Jul 14, 2007 #1
    1. The problem statement, all variables and given/known data
    A photon of a given wavelength "lambda" collides with a hydrogen molecule that's initially at rest (photon is absorbed by H2). Assuming absorption event conserves momentum, find velocity and translational energy of H2 after the event.


    2. Relevant equations
    momentum of photon = h/wavelength
    momentum of object=mv
    kinetic energy=1/2mv^2


    3. The attempt at a solution
    For the velocity, since momentum is conserved, and H2 is initially at rest:
    h/wavelength=mv
    where left side of equation is for photon and right side of equation is for H2.
    So I plug in values to find "v". But I'm not sure of what to put in for "m". Is it just mass of two hydrogen atoms (which is really the mass of 2 protons) or is it the reduced mass of two hydrogen atoms or protons?

    (In general, I'm confused as to when to use sum of masses as opposed to reduced mass)

    And then for the kinetic energy, I guess I would just use 1/2mv^2... where "v" is velocity I find in the first part and "m" is... again, I'm not sure what "m" is.
     
  2. jcsd
  3. Jul 14, 2007 #2

    Astronuc

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    Staff: Mentor

    m would be the mass of the hydrogen molecule, which means 2 protons and 2 electrons.
     
  4. Jul 14, 2007 #3
    Oh... so when would you ever use the reduced mass?

    Also, can't I just ignore the 2 electrons' mass since they're about 4 orders of magnitude smaller than the proton?


    Also, just to verify... translational energy of a molecule is the same thing as the kinetic energy of a molecule??
     
    Last edited: Jul 14, 2007
  5. Jul 14, 2007 #4

    Astronuc

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    Staff: Mentor

    See this for a discussion on reduced mass.
    http://en.wikipedia.org/wiki/Reduced_mass

    and this on the two body problem - http://en.wikipedia.org/wiki/Two-body_problem

    I would expect that if the photon interacted with one H atom, and the result was a change in rotation of the molecule, then one would use the reduce mass in the appropriate equation.

    One could ignore the mass of the electrons and maintain reasonable approximation.
     
  6. Jul 14, 2007 #5
    Thanks for the help! I think that makes more sense.

    Last question, is translational energy the same thing as a kinetic energy? It would seem so since they are both the energy due to movement, but then I also read somewhere that kinetic energy is the sum of translational, rotational, and vibrational energy? I'm not sure... :(
     
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