The susceptibility of a simple metal has a contribution ##\chi_{c.c}## from the conduction electrons and a contribution ##\chi_{ion}## from the diamagnetic response of the closed-shell core electrons.

Taking the conduction electron susceptibility to be given by the free electron values of the Pauli paramagnetic and Landau diamagnetic susceptibilities, show that:

$$ \frac{\chi_{ion}}{\chi_{c.c}} = -\frac{1}{3} \frac{Z_c}{Z_v}\langle (k_F r)^2 \rangle $$

where ##Z_v## is the valence, ##Z_c## is the number of core electrons, and ##<r^2>## is the mean square ionic radius defined in (31.26).

2. Relevant equations

$$(31.26) <r^2> = \frac{1}{Z_i} \sum <0|r_i^2 |0>$$

$$\chi^{molar} = -Z_i (e^2/(\hbar c))^2 \frac{N_A a_0^3}{6}\langle (r/a_0)^2 \rangle$$

$$\chi_{pauli} = \bigg(\frac{\alpha}{2\pi}\bigg)^2 (a_0k_F)$$

$$\chi_{Landau} = -1/3 \chi_{Pauli}$$

3. The attempt at a solution

I thought that ##\chi_{molar}=\chi_{ion}## and that ##\chi_{c.c} = \chi_{Landau}+\chi_{Pauli} = 2/3 \chi_{Pauli}##.

But when I divide between the two susceptibilities I don't get the right factors, has someone already done this exercise from Ashcroft and Mermin?

I tried searching google for a solution but to a veil.