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1) $ \begin{align*} n& = \int_{E_c}^\infty D(E)f(E)dE\\ &=\int_{E_c}^\infty\frac{2(E - E_c)}{\pi\hslash^2V_F^2}\cdot\frac{1}{1+e^{(E-E_F)/kT}}dE \end{align*} $

 Let; 

$ \begin{align*} \eta &=\frac{E-E_c}{kT}\:\:\:\:\:\:\therefore dE = kTd\eta\\ \eta_c &=\frac{E_F-E_c}{kT} \end{align*} $

 $  \begin{align*} n& = \frac{2}{\pi\hslash^2V_F^2}\cdot(kT)^2\int_0^\infty\frac{\eta d\eta}{1+e^{\eta-\eta_c}}\\ &=\frac{2(kT)^2}{\pi\hslash^2V_F^2}\cdot\cancelto{1!}{\Gamma 2}\cdot F_1(\eta_c)\\ &=\frac{2(kT)^2}{\pi\hslash^2V_F^2} F_1(\eta_c)\\ \end{align*}  $

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2)

At $ T = 0\:\:\:\:\:f(E) = 1 $ for $ E\le E_F $

$ \begin{align*} \therefore n &=\int_{E_c}^{E_F}D(E)dE\\ &=\int_{E_c}^{E_F}\frac{2(E-E_c)}{\pi\hslash^2V_F^2}dE\\ &=\frac{2}{\pi\hslash^2V_F^2}\cdot\frac{(E-E_c)^2}{2}\bigg\vert_{E_c}^{E_F}\\ &=\frac{(E_F-E_c)^2}{\pi\hslash^2V_F^2} \end{align*} $

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3)

For Maxwell Boltzmann Statistics

$ F_1(\eta_c)\to e^{\eta_c} $

if

$ \eta_c\le-3 $

$ E_F-E_c\le-3kT $

$ E_c-E_F\ge3kT $

$ \therefore n = \frac{2(kT)^2}{\pi\hslash^2V_F^2}\cdot e^{(E_F-E_c)/kT} $

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4)

$ \bar{u} = \frac{\int_{E_c}^\infty D(E)f(E)(E-E_c)dE}{\int_{E_c}^\infty D(E)f(E)dE} $

from (1);

$   \begin{align*} \text{Denominator} &= \frac{2(kT)^2}{\pi\hslash^2V_F^2}F_1(\eta_c)\\ \text{Numerator} &= \int_{E_c}^\infty\frac{2(E-E_c)^2}{\pi\hslash^2V_F^2}\cdot\frac{1}{1+e^{(E-E_F)/kT}}dE\\ &=\frac{2}{\pi\hslash^2V_F^2}(kT)^3\int_0^\infty\frac{\eta^2d\eta}{1+e^{\eta-\eta_c}}\\ &=\frac{2(kT)^3}{\pi\hslash^2V_F^2}\cdot\cancelto{2!}{\Gamma 3}\cdot F_2(\eta_c)\\ &=\frac{4(kT)^3}{\pi\hslash^2V_F^2} F_2(\eta_c)  \end{align*}   $

$ \therefore\bar{u} = 2kT\frac{F_2(\eta_c)}{F_1(\eta_c)} $

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5) At $ T=0 $

$ \bar{u} = \frac{\int_{E_c}^E D(E)(E-E_c)dE}{\int_{E_c}^{E_F} D(E)dE} $

$   \begin{align*} n &= D(E)f(E)\\ &=\frac{\int(E-E_c)D(E)f(E)}{\int D(E)f(E)dE}  \end{align*}  $

$   \begin{align*} \text{Denominator} &= \frac{(E_F-E_c)^2}{\pi\hslash^2V_F^2}\\ \text{Numerator} &= \int_{E_c}^{E_F}\frac{2(E-E_c)^2}{\pi\hslash^2V_F^2}dE\\ &=\frac{2}{\pi\hslash^2V_F}\cdot\frac{(E-E_c)^3}{3}\bigg\vert_{E_c}^{E_F}\\ &=\frac{2(E_F-E_c)^3}{3\pi\hslash^2V_F}  \end{align*}   $

$ \therefore\bar{u} = \frac{2}{3}(E_F - E_c) $

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6) For Maxwell-Boltzmann statistics At $ T=0 $

$ F_1(\eta_c) = F_2(\eta_c)\to e^{\eta_c} $

$ \therefore \bar{u} = 2kT. $

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