91影视

We have been given that the energy integral $7.54$is$U鈥�={鈭�}_0^{鈭�}蔚g\left(蔚\right){\stackrel{炉}{n}}_{FD}\left(蔚\right)d蔚$.

The Equation $7.67$, $U=\frac{3}{5}N\frac{{渭}^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{{蔚}_F}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}+\frac{3{蟺}^2}{8}N\frac{{\left(kT\right)}^2}{{蔚}_F}+.....$and equation $7.68$, $U=\frac{3}{5}N{蔚}_F+\frac{{蟺}^2}{4}N\frac{{\left(kT\right)}^2}{{蔚}_F}+.....$are given.

We need to Carry out the Sommerfeld expansion for the energy integral $7.54$to obtain equation $7.67$. Then we have to plug in the expansion for $渭$to obtain the final answer, equation $7.68$.

The given total energy integral $7.54$is

localid="1650054588139" $U={鈭�}_0^{鈭�}蔚g\left(蔚\right){\stackrel{炉}{n}}_{FD}\left(蔚\right)d蔚$

localid="1650054613027" $U=g_0{鈭�}_0^{鈭�}{蔚}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\stackrel{炉}{n}}_{FD}\left(蔚\right)d蔚$

Integrating by parts, we get:

$U=\frac{2}{5}{g}_0{蔚}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\stackrel{炉}{n}}_{FD}{|}_0^{鈭�}-\frac{2}{5}{g}_0{鈭�}_0^{鈭�}{蔚}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{鈭�{\stackrel{炉}{n}}_{FD}}{鈭�蔚}d蔚$ (Let this equation be ($1$))

If we substitute $蔚=0$, the integral will become zero$\left(0\right)$due to the dependence of the term on ${蔚}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$and the first term vanishes. If we substitute with $鈭�$, the exponential term in the denominator will grow faster than ${蔚}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

So the equation ($1$) reduces to :

$U=\frac{2}{5}{g}_0{鈭�}_0^{鈭�}{蔚}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{鈭�{\stackrel{炉}{n}}_{FD}}{鈭�蔚}d蔚$ (Let this equation be ($2$))

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

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

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

On simplifying,

we get$\frac{鈭�{\stackrel{炉}{n}}_{FD}}{鈭�蔚}=-\frac{1}{kT}\frac{e^{{\displaystyle \raisebox{1ex}{$(蔚-渭)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}}{{(e^{{\displaystyle \raisebox{1ex}{$(蔚-渭)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1)}^2}$ (Let this equation be ($3$))

Let $\raisebox{1ex}{$(蔚-渭)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.=x$, the equation ($3$) becomes :

$dx=\frac{d蔚}{kT}$

Substitute $dx,x$and $\frac{鈭�{\stackrel{炉}{n}}_{FD}}{鈭�蔚}$in equation ($2$), we get

$U=\frac{2}{5}{g}_0{鈭�}_0^{鈭�}{蔚}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{kT}\frac{e^x}{{(e^x+1)}^2}kTdx$

$U=\frac{2}{5}{g}_0{鈭�}_0^{鈭�}{e}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{e^x}{{(e^x+1)}^2}dx$ (Let this equation be ($4$))

As we need to change the integration boundaries also, so :

$$\begin{gathered} 蔚鈫�鈭�x鈫�鈭� \\ 蔚鈫�0x鈫�-\frac{渭}{kT} \end{gathered}$$

As $kT鈮�渭$, so we can put the lower limit of integral in equation ($4$) as $-鈭�$,

The integral in equation ($4$) becomes :

$U=\frac{2}{5}{g}_0{鈭�}_{-鈭�}^{鈭�}{蔚}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{e^x}{{(e^x+1)}^2}dx$ (Let this equation be ($5$))

Now expand the term ${蔚}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$using Taylor series about $渭$,

${蔚}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={渭}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5}{2}(蔚-渭){渭}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{15}{8}{(蔚-渭)}^2{渭}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+...$

Substitute $蔚-渭=kTx$,

${蔚}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={渭}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5}{2}\left(kTx\right){渭}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{15}{8}{\left(kTx\right)}^2{渭}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+...$

Substitute the value of ${蔚}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$in equation ($5$), we get three integrals say $I_1,I_2,I_3$,we get:

$U=\frac{2}{5}{g}_0(I_1+I_2+I_3)$ (Let this equation be ($6$))

where,

$I_1={渭}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{鈭�}_{-鈭�}^{鈭�}\frac{e^x}{{(e^x+1)}^2}dx$

$I_2=\frac{5}{2}kT{渭}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{鈭�}_{-鈭�}^{鈭�}\frac{x{e}^x}{{(e^x+1)}^2}dx$

$I_3=\frac{15}{8}{\left(kT\right)}^2{渭}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{鈭�}_{-鈭�}^{鈭�}\frac{x^2{e}^x}{{(e^x+1)}^2}dx$

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

$I_1={渭}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

The second integral $I_2$is odd integral, the integration of integral is from $-鈭�$to $鈭�$, so the integral $I_2$is directly zero .

$I_2=0$

$I_3=\frac{15}{8}{\left(kT\right)}^2{渭}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[\frac{蟺2}{3}\right]=\frac{5{蟺}^2}{8}{\left(kT\right)}^2{渭}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

By Substituting the values of the three integrals $I_1,I_2,I_3$in equation ($6$), we get

$$\begin{gathered} U鈮�\frac{2}{5}{g}_0\left[{渭}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{5{蟺}^2}{8}{\left(kT\right)}^2{渭}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right] \\ U鈮�\frac{2}{5}{g}_0{渭}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+g_0\frac{{蟺}^2}{4}{\left(kT\right)}^2{渭}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Substitute $g_0=\frac{3}{2}\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{${蔚}_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$}\right.$, we get:

$U鈮�\frac{3}{5}\frac{N}{{蔚}_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{渭}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{N}{{蔚}_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\frac{3{蟺}^2}{8}{\left(kT\right)}^2{渭}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Set $渭={蔚}_F$in second term, we get

$U鈮�\frac{3}{5}\frac{N}{{蔚}_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{渭}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{N}{{蔚}_F}\frac{3{蟺}^2}{8}{\left(kT\right)}^2$ (Let this equation be ($7$))

The equation ($7.66$) is given by:

${渭}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={蔚}_F^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[1-\frac{{蟺}^2}{12}{\left(\frac{kT}{{蔚}_F}\right)}^2\right]}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

By expanding the terms in the brackets, we get:

${渭}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={蔚}_F^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[1-\frac{5{蟺}^2}{24}{\left(\frac{kT}{{蔚}_F}\right)}^2\right]$

Substitute the value of ${渭}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$in equation ($7$), we get:

$U鈮�\frac{3}{5}\frac{N}{{蔚}_F^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{蔚}_F^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[1-\frac{5{蟺}^2}{24}{\left(\frac{kT}{{蔚}_F}\right)}^2\right]+\frac{N}{{蔚}_F}\frac{3{蟺}^2}{8}{\left(kT\right)}^2$

$U鈮�\frac{3}{5}N{蔚}_F\left[1-\frac{5{蟺}^2}{24}{\left(\frac{kT}{{蔚}_F}\right)}^2\right]+\frac{N}{{蔚}_F}\frac{3{蟺}^2}{8}{\left(kT\right)}^2$

$U鈮�\frac{3}{5}N{蔚}_F-\frac{1}{8}N{蟺}^2{蔚}_F{\left(\frac{kT}{{蔚}_F}\right)}^2+\frac{N}{{蔚}_F}\frac{3{蟺}^2}{8}{\left(kT\right)}^2$

$U鈮�\frac{3}{5}N{蔚}_F-\frac{N}{{蔚}_F}\frac{{蟺}^2}{8}{\left(kT\right)}^2+\frac{N}{{蔚}_F}\frac{3{蟺}^2}{8}{\left(kT\right)}^2$

$U鈮�\frac{3}{5}N{蔚}_F+\frac{N}{{蔚}_F}\frac{{蟺}^2}{{蔚}_F}\frac{{蟺}^2}{4}{\left(kT\right)}^2$