Proposed solution:

This question comes from Tanner's manual Introduction to the Physics of Electrons in Solids, at the chapter dedicated to the application of the Fermi gas model. The Fermi energy is $\mathsf{E}_F$. The model used here is the free electron model, and the susceptibility is $$ \mathcal{\chi} = \frac{\bf \mathcal{M}}{\bf B} $$

Assuming that the density of states at the Fermi level $D(\mathsf{E}_F)$ is approximated to: $$D(\mathsf{E}_F) = \frac{3N}{2} \mathsf{E}_F $$ where $N$ is the total number of electronic states. It is demonstrated that: $$ \mathcal{\chi}=\frac{3n\mu_B^2}{2\mathsf{E}_F}$$ where $n$ is the number of conduction electrons per unit volume given to be $n=2.5 \times 10^{28}\,m^{-3}$, $\mu_B$ is the Bohr magneton given to be $\mu_B = 9.3 \times 10^{-24}\,J\cdot T^{-1}$, the Fermi energy for $\ce{Na}$ is $\mathsf{E}_F = 5.2 \times 10^{-19}\,J$, and therefore the susceptibility is $\mathcal{\chi}=6.2\times 10^1\, MKS$, which can be converted in SI by multiplication of $4\pi \times 10^{-7}$, so the result is $\mathcal{\chi} = 7.8\times 10^{-5} \, SI$

Now my question:

This result is absolutely not consistent with the literature that states $$\mathcal{\chi} = 1.3\times 10^{-5} \, SI$$

What is the problem with Tanner or (most certainly) with my calculus ?

  • 1
    $\begingroup$ Bohr magneton value is not correct and formula for free-electron susceptibility should be multiplied by susceptibility of vacuum. $\endgroup$
    – 10ppb
    Apr 19, 2022 at 11:57
  • 1
    $\begingroup$ @10ppb - The susceptibility of a vacuum is zero, so what we're really talking about here is the relative susceptibility (I believe) - I'm not sure there needs to be a term by which this is multiplied. I also see a calculated value of 3.1 eV (McQuarrie, Statistical Mechanics) for sodium, which is slightly off from what the OP has given. I believe the error lies as you indicated with the value for the Bohr magneton; and the number density given/computed. $\endgroup$ Apr 19, 2022 at 19:37
  • 1
    $\begingroup$ The formula given for chi is not a correct SI formula. It should be multiplied by mu_zero = 4*pi*10^-7 H/m, whatever you want to call it. $\endgroup$
    – 10ppb
    Apr 20, 2022 at 2:21
  • $\begingroup$ Many thanks @ToddMinehardt, I corrected the question according to your suggestions. There is still a factor of 4 between both results but this must be understandable from the model, ... and not from my calculation mistakes ... $\endgroup$ Apr 29, 2022 at 18:31


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