91影视

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 $g\left(系\right)=g_{0c}\sqrt{系-{系}_c}$where $g_{0c}$is a new normalization constant that differs from $g_0$by some fudge factor. Similarly, write $g\left(鈭�\right)$at the top of the valence band in terms of a new normalization constant $g_{0v}$.

(a) Explain why, if $g_{0v}鈮�g_{0c}$, 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 $T,渭,{鈭�}_{c}and{g}_{0c}$Simplify this expression as much as possible, assuming ${系}_c-渭鈮�kT$.

(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 $渭-{系}_v鈮�kT$.

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

Step by step solution

01

Part(a) Step 1: Given information

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

02

Part(a) Step 2: Given information

The solution is as follows:

$N_C={鈭�}_{{系}_C}^{鈭�}g\left(系\right){\stackrel{炉}{n}}_{\mathrm{FD}}d系$

$N_C=g_{0c}{鈭�}_{{系}_C}^{鈭�}\frac{\sqrt{系-{系}_C}}{e^{\left(系-{系}_F\right)/kT}+1}d系$

03

Part(b) Step 1: Given information

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

04

Part(b) Step 2: Simplify

The numbers of electrons in the conduction

${鈭�}_0^{鈭�}\sqrt{x}{e}^{-x}dx=\frac{\sqrt{蟺}}{2}$

$N_C鈮�g_{0c}\frac{\sqrt{蟺}}{2}(kT{)}^{3/2}{e}^{-\left({系}_C-渭\right)/kT}$

05

Part(c) Step 1: GIven information

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

06

Part(c) Step 2: Simplify

The solution is as follows:

${\stackrel{炉}{n}}_{\mathrm{FD}}=\frac{1}{e^{(系-渭)/kT}+1}g\left(系\right)=g_{0v}\sqrt{{系}_v-系}$

$N_v=g_{0v}\frac{\sqrt{蟺}}{2}{e}^{-\left(渭-{系}_v\right)/kT}(kT{)}^{3/2}$

07

Part (d) Step 1: Given information

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

08

Part(d) Step 2:Simplify

The temperature increases

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

$=6.328脳{10}^{-26}{\mathrm{m}}^3$

09

Given information

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

10

Simplify

The steps are given below

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

$螖系=3.0\mathrm{eV}$

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!