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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.

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  • $\begingroup$ As a hint: N is fixed and the energy is a function of temperature and volume E=E(T,V) and that will change your total differential. $\endgroup$ – Ignacio Vergara Kausel Nov 3 '13 at 20:45
  • $\begingroup$ That was my initial assumption. But doesn't $N$ vary with temperature? $\endgroup$ – user28823 Nov 3 '13 at 20:59
  • $\begingroup$ Sodium is a metal, I would say it's constant. Still, somehow you need to have something depending on temperature to be able to replace the volume expansion coefficient. $\endgroup$ – Ignacio Vergara Kausel Nov 3 '13 at 21:01
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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).

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