Why Does the Kinetic Energy Formula for Ag Atoms Use 2kT Instead of (3/2)kT?

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tanaygupta2000
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It says that for Ag atoms, from Kinetic Theory, the velocity vx of an atom of mass M is evaluated by setting
(1/2)M(vx)^2 = 2kT​
This is my confusion. What I have studied is that it should be equal to (3/2)kT instead of 2kT.
 
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That's wrong anyway. This gives the average (root-mean-squared) speed of atoms in a cavity at rest. The correct formula for that is
$$\langle E_{\text{kin}} \rangle=\frac{m}{2} \langle \vec{v}^2 \rangle=\frac{3}{2} k_{\text{B}} T$$
for an ideal gas (for the Ag vapour in the SGE it's safe to assume it to be an ideal gas).

For the average velocity of the Ag atom to leave the oven you have to calculate the current out of the cavity in a given direction. This gives an additional factor ##v##. Don't worry, even Stern himself made this mistake when in 1920 he measured the molecular speed with this molecular beam method for the first time, and then he wrote an erratum thanking Einstein for pointing out this mistake to him :-).
 
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