Q. 7.34 - An Introduction to Thermal Physics

Question

In a real semiconductor, the density of states at the bottom of the conduction band will differ from the model used in the previous problem by a numerical factor, which can be small or large depending on the material. Let us, therefore, write for the conduction band $g\left(ϵ\right)=g_{0c}\sqrt{ϵ-{ϵ}_c}$ where $g_{0c}$is a new normalization constant that differs from $g_0$ by some fudge factor. Similarly, write $g\left(\in \right)$ at the top of the valence band in terms of a new normalization constant $g_{0v}$.

(a) Explain why, if $g_{0v}\ne g_{0c}$, the chemical potential will now vary with temperature. When will it increase, and when will it decrease?

(b) Write down an expression for the number of conduction electrons, in terms of $T, \mu , {\in }_c and g_{0c}$ Simplify this expression as much as possible, assuming ${ϵ}_c-\mu \gg kT$.

(c) An empty state in the valence band is called a hole. In analogy to part (b), write down an expression for the number of holes, and simplify it in the limit $\mu -{ϵ}_v\gg kT$.

(d) Combine the results of parts (b) and (c) to find an expression for the chemical potential as a function of temperature. 

(e) For silicon, $\raisebox{1ex}{$g_{0c}$}\!\left/ \!\raisebox{-1ex}{$g_{0 }=1.09 and {\displaystyle \raisebox{1ex}{$g_{0v}$}\!\left/ \!\raisebox{-1ex}{$g_0=0.{44}^{*}.$}\right.}$}\right.$Calculate the shift inµ for silicon at room temperature. 

Step-by-Step Solution

Answer

  (a) The chemical potential will vary$N_C=g_{0c}{\int }_{{ϵ}_C}^{\infty }\frac{\sqrt{ϵ-{ϵ}_C}}{e^{\left(ϵ-{ϵ}_F\right)/kT}+1}dϵ$

(b) The number of conduction electrons$N_C\approx g_{0c}\frac{\sqrt{\pi }}{2}(kT{)}^{3/2}{e}^{-\left({ϵ}_C-\mu \right)/kT}$

(c) The valence bond is$N_v=g_{0v}\frac{\sqrt{\pi }}{2}{e}^{-\left(\mu -{ϵ}_v\right)/kT}(kT{)}^{3/2}$

(d) The chemical potential is$\mu =\frac{{ϵ}_C+{ϵ}_v}{2}-\frac{kT}{2}\ln \left(\frac{g_{0c}}{g_{0v}}\right)$

(e) The shift for silicon is${\mu }_{\text{shift }}=-0.0118\mathrm{eV}$

Part(a) Step 1: Given information

We have been given that $g\left(ϵ\right)=g_{0c}\sqrt{ϵ-{ϵ}_C}$

Part(a) Step 2: Given information

The solution is as follows:

$N_C={\int }_{{ϵ}_C}^{\infty }g\left(ϵ\right){\overline{n}}_{\mathrm{FD}}dϵ$

$N_C=g_{0c}{\int }_{{ϵ}_C}^{\infty }\frac{\sqrt{ϵ-{ϵ}_C}}{e^{\left(ϵ-{ϵ}_F\right)/kT}+1}dϵ$

Part(b) Step 1: Given information

We have been given that $\frac{\sqrt{\pi }}{2}$

Part(b) Step 2: Simplify

The numbers of electrons in the conduction

${\int }_0^{\infty }\sqrt{x}{e}^{-x}dx=\frac{\sqrt{\pi }}{2}$

$N_C\approx g_{0c}\frac{\sqrt{\pi }}{2}(kT{)}^{3/2}{e}^{-\left({ϵ}_C-\mu \right)/kT}$

Part(c) Step 1: GIven information

We have been given that  $\frac{N_C}{V}$

Part(c) Step 2: Simplify

The solution is as follows:

${\overline{n}}_{\mathrm{FD}}=\frac{1}{e^{(ϵ-\mu )/kT}+1} g\left(ϵ\right)=g_{0v}\sqrt{{ϵ}_v-ϵ}$

$N_v=g_{0v}\frac{\sqrt{\pi }}{2}{e}^{-\left(\mu -{ϵ}_v\right)/kT}(kT{)}^{3/2}$

Part (d) Step 1: Given information

We have been given $N\propto T^{3/2}{e}^{-c/T}$

Part(d) Step 2:Simplify

The temperature increases

$v_Q={\left(\frac{{\left(6.626\times {10}^{-34} \mathrm{J}·\mathrm{s}\right)}^2}{2\pi \left(9.11\times {10}^{-31} \mathrm{kg}\right)\left(1.38\times {10}^{-23} \mathrm{J}/\mathrm{K}\right)(350 \mathrm{K})}\right)}^{3/2}$

$=6.328\times {10}^{-26}{ \mathrm{m}}^3$

Step 1: Given information

We have been given that $\frac{N_C}{V}$

Step 2: Simplify

The steps are given below

$\frac{N_C}{V}=2{ \mathrm{m}}^{-3}=\frac{2}{8.00\times {10}^{-26}{ \mathrm{m}}^3}{e}^{-\Delta ϵ/(0.052\mathrm{eV})}$

$\Delta ϵ=3.0\mathrm{eV}$