91影视

We have to find that evaluation the$N$integral ($7.53$) for the case$\frac{kT}{{蔚}_0}$and $渭=0$, and checking the answer is consistent with the graph shown above.

To find no. of particles,

$$\begin{gathered} N={鈭�}_0^{鈭�}g\left(蔚\right)\frac{1}{e^{{\displaystyle \raisebox{1ex}{$\left(蔚-渭\right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}d蔚 \\ g\left(蔚\right)=\frac{3N}{2{{蔚}_F}^{3/2}}\sqrt{蔚} \end{gathered}$$

putting these things, $$\begin{gathered} N={鈭�}_0^{鈭�}\frac{3N}{{2{蔚}_F}^{3/2}}\sqrt{蔚}\frac{1}{e^{{\displaystyle \raisebox{1ex}{$\left(蔚-渭\right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}d蔚 \\ 1={鈭�}_0^{鈭�}\frac{3}{{2{蔚}_F}^{3/2}}\sqrt{蔚}\frac{1}{e^{{\displaystyle \raisebox{1ex}{$\left(蔚-渭\right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}d蔚 \end{gathered}$$

We assume , $t=\frac{kT}{{蔚}_F},c=\frac{渭}{{蔚}_F},x=\frac{蔚}{{蔚}_F}鈫�dx=\frac{d蔚}{{蔚}_F}$

We get , $$\begin{gathered} 1=\frac{3}{2}{鈭�}_0^{鈭�}\frac{\sqrt{x}}{e^{\left(x-c\right)/t}+1}dx \\ 1=\frac{3}{2}{鈭�}_0^{鈭�}\frac{\sqrt{x}}{e^{x/t}+1}dx \end{gathered}$$

using Matlab , RHS integral can be calculated , it comes out to be $1.017$, so $渭$is not equals to 0 when ${蔚}_F=kT$.

We need to vary$渭$holding $T$fixed, to get the desired value,$N$, for values of $kT/{蔚}_F$ranging from $0.1$up to $2$,and plotting the results to reproduce Figure $7.15$

Here, for specific value of $c$and , we well find integral , also temperature will be fixed, so we assume the value of$t=1$. We will find the value of $c$,so that we get correct result of integral which is $1$,to find this we will use Matlab. With respect to the two variables,$x$ and$c$, equation 1 will be defined, we will integrate with respect to $x$from $0$and $鈭�$ , so here function will be in variable only, we will calculate the value of $c$ , the interval will be from $-1$to $0.1$ ,

So, $c=-0.0215$.

Then , We will find constant with different temperatures value, here t will be changed and same instructions will be done . The interval of solutions, will also be changed from $-1$to $2$. We will change$t$from $0.1$to $2.0$. The following graph will be obtained.

We are asked to check that the heat capacity has the expected behavior at both low and high temperatures.

By plotting data in Excel, we will add trend line to actually see the equation which fits into data ,

Here, the equation between $t$and $c$from graph can be seen ,

$$\begin{gathered} c=-\left(08469\right)t^2-\left(0.2019\right)x+1.0412 \\ U={鈭�}_0^{鈭�}蔚g\left(蔚\right)\frac{1}{e^{{\displaystyle \raisebox{1ex}{$\left(蔚-渭\right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}d蔚 \end{gathered}$$

We can find the value of energy density here as;

$$\begin{gathered} g\left(蔚\right)=\frac{3N}{{2{蔚}_F}^{3/2}}\sqrt{蔚} \\ t=\frac{kT}{{蔚}_F},c=\frac{渭}{{蔚}_F},x=\frac{蔚}{{蔚}_F}鈫�dx=\frac{d蔚}{{蔚}_F} \end{gathered}$$

Finally , $\frac{U}{N_{{蔚}_F}}=\frac{3}{2}{鈭�}_0^{鈭�}\frac{x^{3/2}}{e^{{\displaystyle \raisebox{1ex}{$\left(x-c\right)$}\!\left/ \!\raisebox{-1ex}{$t$}\right.}}+1}dx$

We can find heat capacity by doing partial derivative of total energy with respect to to temperature.