Speed of atoms ejected from an oven

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The average speed of atoms ejected from an oven can be estimated using the Maxwell-Boltzmann distribution, resulting in the formula for average speed as v = √(8kT/πm). This contrasts with the RMS speed derived from the thermodynamic expression E = 3/2 kT, which gives v = √(3kT/m) and represents the average of the squared speeds. The distinction arises because the RMS speed accounts for all directions of particle motion, leading to an average velocity of zero in a gas. In a Stern-Gerlach experiment, where atoms are collimated, the average speed derived from the Maxwell-Boltzmann distribution is more relevant. Understanding these differences is crucial for accurately predicting the behavior of atoms in experimental setups.
Jolb
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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 v=\sqrt{\frac{3kT}{m}}
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:
\langle v \rangle = \int_0^{\infty} v \, f(v) \, dv= \sqrt { \frac{8kT}{\pi m}}
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|.
 
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The second one.
 
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.
 
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.
 
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