Metals — Question 6

Question 6

Pauli exclusion suppresses the electronic heat capacity

For a degenerate free-electron gas use $C_e=(\pi^2/2)Nk_B(T/T_F)$, where $T_F=E_F/k_B$. Take $E_F=5.00\,\mathrm{eV}$ and $k_B=8.617\times10^{-5}\,\mathrm{eV/K}$.

1. At 300 K find $T_F$ and the ratio to the classical value $3Nk_B/2$.

2. Integrate the heat capacity from zero temperature to find the thermal energy increment and express it as a fraction of $NE_F$.

Question 6 - Solution

Pauli exclusion suppresses the electronic heat capacity

(a) Method and calculation

$$\boxed{T_F=5.00/(8.617\times10^{-5})=5.80\times10^4\,\mathrm K}.$$$$\boxed{\frac{C_e}{3Nk_B/2}=\frac{\pi^2}{3}\frac{300}{T_F}=.0170}.$$

(b) Interpretation and further calculation

$$\Delta U=\int_0^TC_e(T\prime)dT\prime=\frac{\pi^2}{4}\frac{Nk_BT^2}{T_F}.$$$$\boxed{\Delta U/(NE_F)=\frac{\pi^2}{4}(T/T_F)^2=6.60\times10^{-5}}.$$

Physical check

Only states within an energy scale kBT of the Fermi surface can change occupation appreciably. The low-temperature expansion requires T much less than TF.