Susceptibility of a simple metal (Problem 31.6 in Ashcroft's
1. The problem statement, all variables and given/known data
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 closedshell 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
$$(31.26) <r^2> = \frac{1}{Z_i} \sum <0r_i^2 0>$$
$$\chi^{molar} = Z_i (e^2/(\hbar c))^2 \frac{N_A a_0^3}{6}\langle (r/a_0)^2 \rangle$$
$$\chi_{pauli} = \bigg(\frac{\alpha}{2\pi}\bigg)^2 (a_0k_F)$$
$$\chi_{Landau} = 1/3 \chi_{Pauli}$$
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.
