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Effective mass for metals

Physics Asked by Ragon Loso on March 14, 2021

How can I calculate effective mass for of the electrons in $Al$ and $Ca$? I did real from some sources and they were mentioning that it could be evaluated from density of states but how? Also is there any table that shows for all metals?

2 Answers

Effective mass is calculated from the dispersion relation in a material. I can give a simplified explanation assuming your minimum is at $k=0$ and the dispersion relation is isotropic. The concept is that free electrons have a dispersion of $$E = frac{hbar^2 k^2}{2m}$$

Now in a solid, at the bottom of the band (small $E$) the dependence of $E$ on $k$ will still be quadratic, but with a different factor than for free electrons. We push that additional factor into the mass and call it the effective mass $m^*$. However, depending on the energy region you are interested in, this quadratic dependence might not hold anymore, so some care must be taken.

To get this effective mass, we take the second derivative of the energy with respect to $k$, which for free electrons would be $$ left.frac{partial^2 E}{partial k^2}right|_{k=0} = frac{hbar^2}{m}$$ Therefore for low energies, we set $$ m^* = frac{hbar^2}{left.frac{partial^2 E}{partial k^2}right|_{k=0}} $$

If your material is anisotropic, this expression will give you different masses in different directions, and you need to take the derivative with respect to $vec{k}$ in the direction you are interested in.

In general, all of this depends on where in the band structure you are, and possibly on the direction you are looking in. Therefore there is no blanket constant $m^*$ for any given metal.

Answered by noah on March 14, 2021

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You can do a DFT calculation for bulk Al and Ca, then fit the band structure obtained using the above formula to obtain the effective mass.

Answered by Yuan Fang on March 14, 2021

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