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Valence/conduction electron density of a classical gas

  • Thread starter iLIKEstuff
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  • #1
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sorry if this is in the wrong section. the forum rules say a textbook related issue goes here.

i'm reading the ashcroft and mermin solid state physics book and referring to page 4.

in the context of drude theory, the book states that the valence/conduction electron densities for metals are typically a thousand times greater than those of a classical gas at normal temps and pressures.

but i'm confused, how would you calculate such a density for a classical gas. i'm assuming when they say "classical gas" i'm assuming that it is a gas that obeys the ideal gas law. so how would you calculate the CONDUCTION ELECTRON density for air?

(the units they are using for this density is "conduction electrons per cubic centimeter")

thanks guys.
 

Answers and Replies

  • #2
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Hmm, good question. I know that for a conduction electron in a metal, a 3D infinite square well is a very good approximation. You can use the fermi energy of the electron to see that the number of states is very large, and that the points in "n-space" (or "phase space") form a near continuum in the limit of large n. So the total number of states will look like a spherical surface because n^2=nx^2+ny^2+nz^2. Since you only want the first quadrant, the rest is nonphysical, you can come up with:

[tex]N(E)=\frac{1}{8}(\frac{4}{3}\pi n^3)[/tex]

Then the density of states would just be dN(e)/dE.

If you can apply the same infinite sq. well. approx. to the gas case, you could make a similar argument for the gas and then compare.
 
  • #3
Dick
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I'm going out on a limb here, but I think you are reading to much into this. I think the only content here is that liquids and solids are about a thousand times denser than gas. So if you ionize a gas the conduction electrons should be about a thousand times less dense than in a conductive liquid or solid. I'm guessing this is true because it's only on page 4. So it probably doesn't have any hugely technical intent.
 

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