91影视

We have been given that the density of states and the chemical potential for a two-dimensional Fermi gas which we have found earlier in Problem $7.28$is$g\left(蔚\right)=\frac{N}{{蔚}_F}$.

We need to find the heat capacity of this gas in the limit $kT鈮�{蔚}_F$and show that the heat capacity has the expected behavior when $kT鈮�{蔚}_F$. Also we need to sketch the heat capacity as a function of temperature.

The total energy is given by the following integral:

$U={鈭�}_0^{鈭�}蔚g\left(蔚\right){\stackrel{炉}{n}}_{FD}\left(蔚\right)d蔚$ (Let this equation be (localid="1650103209259" $1$))

We have found the energy density earlier in Problem $7.28$such as

localid="1650102270353" $g\left(蔚\right)=\frac{N}{{蔚}_F}$

Substitute the value of $g\left(蔚\right)$in equation (localid="1650103517684" $1$), we get:

$U=\frac{N}{{蔚}_F}{鈭�}_0^{鈭�}蔚{\stackrel{炉}{n}}_{FD}\left(蔚\right)d蔚$

Integrating by parts, we get:

$U=\frac{N}{{蔚}_F}\frac{{蔚}^2}{2}{\stackrel{炉}{n}}_{FD}{|}_0^{鈭�}-\frac{N}{{蔚}_F}{鈭�}_0^{鈭�}\frac{{蔚}^2}{2}\frac{鈭�{\stackrel{炉}{n}}_{FD}}{鈭�蔚}d蔚$ (Let this equation be ($2$))

If we substitute $蔚=0$, the integral will become zero due to the dependence of the term on ${蔚}^2$and the first term vanishes.

The equation ($2$) reduces to :

$U=-\frac{N}{{蔚}_F}{鈭�}_0^{鈭�}\frac{{蔚}^2}{2}\frac{鈭�{\stackrel{炉}{n}}_{FD}}{鈭�蔚}d蔚$ (Let this equation be ($3$))

We know that ${\stackrel{炉}{n}}_{FD}=\frac{1}{e^{{\displaystyle \raisebox{1ex}{$(e-渭)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}$

Taking derivative with respect to $蔚$, we get:

localid="1650103912932" $\frac{鈭�{\stackrel{炉}{n}}_{FD}}{鈭�蔚}=\frac{鈭�\left[{\displaystyle \frac{1}{e^{{\displaystyle \raisebox{1ex}{$(蔚-渭)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}}\right]}{鈭�蔚}$

$\frac{鈭�{\stackrel{炉}{n}}_{FD}}{鈭�蔚}=-\frac{1}{kT}\frac{e^{{\displaystyle \raisebox{1ex}{$(蔚-渭)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}}{{(e^{{\displaystyle \raisebox{1ex}{$(蔚-渭)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1)}^2}$

Let $\raisebox{1ex}{$(蔚-渭)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.=x$, then we can write as:

$dx=\frac{d蔚}{kT}$

We also need to change the boundaries of integration, so:

$蔚鈫�鈭�$ $x鈫�鈭�$

$蔚鈫�0$ $x鈫�-\frac{渭}{kT}$

As $kT鈮�渭$, we can put $-鈭�$as the lower limit of integration.

The integral in equation ($3$) will become:

$U=\frac{N}{2{蔚}_F}{鈭�}_{-鈭�}^{鈭�}{蔚}^2\frac{e^x}{{(e^x+1)}^2}dx$ (Let this equation be (localid="1650114310341" $4$))

Now expanding ${蔚}^2$about $渭$using Taylor series, we get:

$$\begin{gathered} {蔚}^2={渭}^2+2渭(蔚-渭)x+(蔚-渭)x^2 \\ {蔚}^2={渭}^2+2渭\left(kTx\right)+{\left(kTx\right)}^2 \end{gathered}$$

By substituting the value of in equation ($4$), we get three integrals say $I_1,I_2,I_3$

The equation ($4$) will become:

where,

$$\begin{gathered} I_1={渭}^2{鈭�}_{-鈭�}^{鈭�}\frac{e^x}{{(e^x+1)}^2}dx \\ {I}_2=2kT渭{鈭�}_{-鈭�}^{鈭�}\frac{x{e}^x}{{(e^x+1)}^2}dx \\ {I}_3={\left(kT\right)}^2{鈭�}_{-鈭�}^{鈭�}\frac{x^2{e}^x}{{(e^x+1)}^2}dx \end{gathered}$$

Simplifying the integrals $I_1,I_2,I_3$, we get:

$I_1={渭}^2$

The second integral is odd integral and the limits of integration are from to , so this integral is simply zero.

$I_2=0$

$I_3={\left(kT\right)}^2\left[\frac{蟺2}{3}\right]=\frac{{\left(kT蟺\right)}^2}{3}$

Substitute the values of in equation ($5$), we get:

$U=\frac{N}{2{蔚}_F}\left[{渭}^2+\frac{{\left(kT蟺\right)}^2}{3}\right]$

Let $渭={蔚}_F$, we get:

$$\begin{gathered} U=\frac{N}{2{蔚}_F}\left[{蔚}_F^2+\frac{{\left(kT蟺\right)}^2}{3}\right] \\ U=\frac{N{蔚}_F}{2}+\frac{N{\left(kT蟺\right)}^2}{6{蔚}_F} \end{gathered}$$

The heat capacity is the partial derivative of with respect to the temperature, that is:

$$\begin{gathered} C_V=\frac{鈭�U}{鈭�T} \\ {C}_V=\frac{鈭�\left[{\displaystyle \frac{N{蔚}_F}{2}}+{\displaystyle \frac{N{\left(kT蟺\right)}^2}{6{蔚}_F}}\right]}{鈭�T} \\ {C}_V=\frac{N{\left(k蟺\right)}^{2}T}{3{蔚}_F} \end{gathered}$$

The graph of the heat capacity as a function of temperature is linear, where the slope of the line is $Slope=\frac{N{\left(k蟺\right)}^2}{3{蔚}_F}$.