# What is the paramagnetic susceptibility of sodium at around room temperature?

## 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 ?

• Bohr magneton value is not correct and formula for free-electron susceptibility should be multiplied by susceptibility of vacuum. Apr 19, 2022 at 11:57
• @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. Apr 19, 2022 at 19:37
• 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. Apr 20, 2022 at 2:21
• 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 ... Apr 29, 2022 at 18:31