When the attractive forces of the ions in a crystal are taken into account, the allowed electron energies are no longer given by the simple formula 7.36; instead, the allowed energies are grouped into bands, separated by gaps where there are no allowed energies. In a conductor the Fermi energy lies within one of the bands; in this section we have treated the electrons in this band as "free" particles confined to a fixed volume. In an insulator, on the other hand, the Fermi energy lies within a gap, so that at T = 0 the band below the gap is completely occupied while the band above the gap is unoccupied. Because there are no empty states close in energy to those that are occupied, the electrons are "stuck in place" and the material does not conduct electricity. A semiconductor is an insulator in which the gap is narrow enough for a few electrons to jump across it at room temperature. Figure 7 .17 shows the density of states in the vicinity of the Fermi energy for an idealized semiconductor, and defines some terminology and notation to be used in this problem.

(a) As a first approximation, let us model the density of states near the bottom of the conduction band using the same function as for a free Fermi gas, with an appropriate zero-point: $g\left(ϵ\right)=g0\sqrt{ϵ-{ϵ}_c}$, where go is the same constant as in equation 7.51. Let us also model the density of states near the top

Figure 7.17. The periodic potential of a crystal lattice results in a densityof-states function consisting of "bands" (with many states) and "gaps" (with no states). For an insulator or a semiconductor, the Fermi energy lies in the middle of a gap so that at T = 0, the "valence band" is completely full while the-"conduction band" is completely empty. of the valence band as a mirror image of this function. Explain why, in this approximation, the chemical potential must always lie precisely in the middle of the gap, regardless of temperature.

(b) Normally the width of the gap is much greater than kT. Working in this limit, derive an expression for the number of conduction electrons per unit volume, in terms of the temperature and the width of the gap.

(c) For silicon near room temperature, the gap between the valence and conduction bands is approximately 1.11 eV. Roughly how many conduction electrons are there in a cubic centimeter of silicon at room temperature? How does this compare to the number of conduction electrons in a similar amount of copper?

( d) Explain why a semiconductor conducts electricity much better at higher temperatures. Back up your explanation with some numbers. (Ordinary conductors like copper, on the other hand, conduct better at low temperatures.) (e) Very roughly, how wide would the gap between the valence and conduction bands have to be in order to consider a material an insulator rather than a semiconductor?

Step by step solution

01

part(a) Step 1:given information

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

02

part(a) Step 2: Simplify

energy distribution is given by:

${\overline{n}}_{\mathrm{FD}}=\frac{1}{e^{\left(ϵ-{ϵ}_F\right)/kT}+1}g\left(ϵ\right)=g_0\sqrt{ϵ-ϵ}$

$N_C\approx g_0{\int }_{{ϵ}_C}^{\infty }\sqrt{ϵ-{ϵ}_C}{e}^{-\left(ϵ-{ϵ}_C\right)/kT}{e}^{-\left({ϵ}_C-{ϵ}_F\right)/kT}dϵ$

03

part(b) Step 1: Given information

we have been given that at zero temperature there will be some electrons in the conductors

04

part(b) Step 2: Explanation

$\text{a point of}ϵ=\mu \text{, so we can conclude that the probability of state at}ϵ\text{being occupied is equals the probability of state}$

05

part(c) step 1: given information

we have been given that$v_Q={\left(\frac{{\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)}^2}{2\pi \left(9.11\times {10}^{-31}\mathrm{kg}\right)\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)(300\mathrm{K})}\right)}^{3/2}$

06

part(c) Step 2: Simplify

at room temperature

$\frac{N_C}{V}=\frac{2}{8.00\times {10}^{-26}{\mathrm{m}}^3}{e}^{-\left(1.778\times {10}^{-19}\mathrm{J}\right)/\left(2\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)(300\mathrm{K})\right)}$

electrons in copper is:

$\frac{N_C}{V}=8.47\times {10}^{28}{\mathrm{m}}^{-3}$

07

part(d) Step 1: given information

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

08

part(d) step 2: simplify

The terms we get:

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

therefore volume is

$\frac{N_C}{V}=\frac{2}{6.328\times {10}^{-26}{\mathrm{m}}^3}{e}^{-\left(1.778\times {10}^{-19}\mathrm{J}\right)/\left(2\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)(350\mathrm{K})\right)}$

09

Part(e) Step 1: Given information

The number of electrons per unit volume

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

10

Part(e) Step 2:Simplify

The width of the gap is measured

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

$e^{-\Delta ϵ/(0.052\mathrm{eV})}=8.00\times {10}^{-26}$

$\Delta ϵ=3.0\mathrm{eV}$

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