Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Speed of atoms ejected from an oven

  1. Aug 1, 2012 #1
    My question is this: what is the average speed of atoms released from an oven at some temperature T? For example, in a Stern-Gerlach experiment, hydrogen atoms are emitted from an oven and collimated into a beam by passing them through a slit (and then sent into an inhomogenous magnetic field, but I don't really care about these details for this particular question).

    The reason I am confused about this is because my gut instinct would be to estimate this using the old thermodynamic expression for an ideal gas: E = 3/2 kT = 1/2 mv^2 which would imply [tex]v=\sqrt{\frac{3kT}{m}}[/tex]
    However, upon cracking open my thermodynamics textbook, it seems as though this is the RMS speed of atoms in a gas, rather than the boring-old average speed. The boring old average speed is found by calculating the expectation value of speed in the Maxwell-Boltzmann distribution, as such:
    [tex]\langle v \rangle = \int_0^{\infty} v \, f(v) \, dv= \sqrt { \frac{8kT}{\pi m}} [/tex]
    They are very close: the factor of 3 just changes to 8/∏. But which one should I use in a Stern-Gerlach question? Which one would actually be observed in a collimated beam of atoms emitted from an oven at temperature T?

    Edit: Just to clarify, even though I am using the symbol v, which is usually reserved for the velocity, here I am referring to the speed, which is equal to the magnitude of the velocity vector. That is to say, v=|v|.
    Last edited: Aug 1, 2012
  2. jcsd
  3. Aug 2, 2012 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    The second one.
  4. Aug 2, 2012 #3
    The second one does seem more obvious, but why does my "gut instinct" derivation fail? It seems to me that if we know some expression for the average kinetic energy (E=3/2 kT), the expression should give the average speed, not the RMS speed. Why does that derivation give RMS rather than the correct answer?

    Also, references would be appreciated.
  5. Aug 4, 2012 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    Depends what you are deriving and how you go about it.

    The rms velocity is what you get in the first one because the particles in the gas are equally likely to be heading an any direction - giving an average velocity of zero. You can see why this is not useful.

    Put a hole in the walls though, and you are selecting for those particles that are headed in a subset of possible directions.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook