The electrons' contribution to the heat capacity of a metal is small and goes to 0 as...

The electrons' contribution to the heat capacity of a metal is small and goes to 0 as T→ 0. We might try to calculate it via the total internal energy,  but it is one of those integrals impossible to do in closed form, and N(E)FD is the culprit. Still, we can explain why the heat capacity should go to zero and obtain a rough value, (a) Starting with N(E)FD expressed as in equation (9-34), show that the slope dN(E)FD/dE at E = EF is -1/(4kBT). (b) Based on part (a), the accompanying figure is a good approximation to N(E)FD when T is small. In a normal gas, such as air, when T is raised a little, all molecules, on average, gain a little energy, proportional to kBT. Thus, the internal energy U increases linearly with T, and the heat capacity, , is roughly constant. Argue on the basis of the figure that in thin fermion gas, as the temperature increases from 0 to a small value T, while some particles gain energy of roughly kBT, not all do, and the number doing so is also roughly proportional to T. What effect does this have on the heat capacity? (c) Viewing the total energy increase as simply ∆U = (number of particles whose energy increases) X (energy change per particle) and assuming the density of states is simply a constant D over the entire range of particle energies, show that the heat capacity under these lowest-temperature conditions should be proportional to (kBR/EF)T. (Trying to be more precise is not really worthwhile, for the proportionality constant is subject to several corrections from effects we ignore.)