91Ó°ÊÓ

We have been given that,

$A=L^2$

In two dimensional case,

$N=2{∫}_0^{n_{max}}{∫}_0^{\mathrm{π}/2}ndnd\mathrm{θ}=\frac{\mathrm{π}}{2}{n}_{max}^2$

We know that,

$$\begin{gathered} {∈}_F=\frac{h^2{n}_{max}^2}{8m{L}^2}=\frac{h^2}{8mA}×\frac{2N}{\mathrm{Ï€}} \\ {∈}_F=\frac{h^2}{4\mathrm{Ï€³¾´¡}}N \end{gathered}$$

Now, we can find the total energy $U$,

$$\begin{gathered} U=2{∫}_0^{n_{max}}{∫}_0^{\raisebox{1ex}{$\mathrm{Ï€}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}ndnd\mathrm{θ}=\frac{\mathrm{Ï€}}{2}{n}_{max}^2 \\ U={∫}_0^{{∈}_F}∈\left[\mathrm{Ï€}\sqrt{\frac{8mA}{h^2}}\sqrt{∈}\sqrt{\frac{8\mathrm{mA}}{{\mathrm{h}}^2}}\frac{1}{2\sqrt{∈}}\right]d∈ \\ U={∫}_0^{{∈}_F}∈\left[\frac{4\mathrm{Ï€³¾´¡}}{h^2}\right]d∈ \\ U={∫}_0^{{∈}_F}∈\frac{N}{{∈}_F}d∈ \\ U=\frac{1}{2}N{∈}_F \end{gathered}$$

So, we can easily get average energy of the particles is $\frac{U}{N}=\frac{{∈}_F}{2}$

We need to find out the density of states.

From part (a),

Density of states is,

$g\left(∈\right)=\frac{N}{{∈}_F}$

That is independent of$∈$

We need to find out behavior of chemical properties

Since the density of state is constant, the behavior of the chemical potential is the same as that of the Fermi-Dirac distribution. As we know that Fermi-Dirac distribution has some symmetrical property but when the energy difference from $\mathrm{μ}$is higher than the chemical potential, it gets broken because the occupancy is zero when $∈<0$. So, the electrons with energy less than the Fermi energy are excited to a higher level than the Fermi energy with broken symmetry, which means that the number of electron loss with energy $∈<{∈}_F$is lower than the number of electron gain with energy $∈>{∈}_F$.

To be sensible with the difference in no. of electrons, chemical potential should move i.e. Fermi-Dirac distribution has to move totally. As the number of electrons with energy $∈>{∈}_F$is larger than $∈<{∈}_F$, so Fermi-Dirac distribution has to move on the left side with fixed ${∈}_F$, which means that the chemical potential decreases. When the temperature slightly grows up, then the difference in the number of electrons also grows up. So, the chemical potential will decrease continuously and at some point, it will have a negative value.

We need to find out value of$\mathrm{μ}$

We need to find out that the chemical properties of this system is the same as that of an ordinary ideal gas