Sodium has a volume expansion coefficient of $15 * 10^{-5} K^{-1}$. Calculate the percentage change in the fermi energy as the temperature is raised from $T = 0K$ to $T = 300K$.
My attempt at the solution is below:
Volume expansion coefficient is given by $\alpha _v = \frac{1}{V}\left( \frac{\partial V}{\partial T} \right)$. Then just using the definition of the fermi energy I get:
$$E_f = \frac{ \hbar ^2 k_f^2}{2m} = \frac{ \hbar ^2 \left( 3 \pi^2 \frac{N}{V} \right)^{2/3}}{2m}$$
Where $N$ is the number of electrons that can contribute to conduction (They are in the conduction band I think, but their energy must be less than $E_f$ because they are inside the fermi surface so I am unsure about that.)
And $V$ is some volume containing the electrons. Then it follows that:
$$dE_f = \left( \frac{\partial E_f}{\partial N} \right) dN + \left( \frac{\partial E_f}{\partial V} \right) dV$$
$$=\frac{ \hbar ^2 \left( 3 \pi^2 \frac{1}{V} \right)^{2/3}}{2m}N^{-1/3}dN + \frac{ \hbar ^2 \left( 3 \pi^2 N \right)^{2/3}}{2m}V^{-5/2}dV$$
Now $dV$ I know from the thermal expansion coefficient, but am I supposed to evaluate this new term? I had the idea that each Na atom contributes 1 conduction electron, but from there I don't know how many atoms or what else. Also I find it strange that I am asked to evaluate the fermi energy at a non-zero temperature.
Best Answer
The energy is a function of temperature and volume, thus: $$E_F=E_F(T,V)$$
The total differential of the energy is given by
$$dE_F = \left( \frac{\partial E_F}{\partial T} \right) dT + \left( \frac{\partial E_f}{\partial V} \right) dV$$
Now, dividing by $dT$:
$$\frac{dE_F}{dT} = \left( \frac{\partial E_F}{\partial T} \right) + \left( \frac{\partial E_f}{\partial V} \right)\frac{\partial V}{\partial T}$$
Here, in the second term, we recognize that $\frac{\partial V}{\partial T} = V\alpha_v$.
Now you can calculate the differentials and replace stuff. Keep in mind that the percentage change is being asked, that will cancel a lot of the constants (so do it early and don't carry them around).