Chapter 9: 56E (page 407)

(a) From equation (9.34) and the Fermi-Dirac distribution given in Exercise 53, obtain an expression for E_F(T), the Fermi temperature for a collection of fermion oscillators, (b) Show that ${E_{F}}_{o} = ε$ . (c) Plot E_F(T)versus $\frac{k_{B} T}{ε}$ from 0to $\frac{k_{B} T}{6} = 1.5$ . (d) By what percent does the Fermi energy drop from its maximum T=0value when k_BTrises to 25%of $ε$ ?

Short Answer

Expert verified

a) The expression for E_F(T) is $k_{B} Tln (e^{\frac{E}{k_{B} T}} - 1)$ .

b) It is proved that data-custom-editor="chemistry" $E_{F 0} = E$ .

c) Graph is shown in the solution part.

d) The percent drop in Fermi energy is $0.462 %$ .

Step by step solution

Formula used

The expression for Fermi Dirac distribution is given by,

$\frac{1}{e^{\frac{(E - E_{F})}{k_{B} T + 1}}} = \frac{1}{\frac{e^{\frac{E}{k_{B} T}}}{\frac{E}{e^{\frac{E}{B^{T} T}}} + 1}}$

The expression for drop in Fermi energy is

$∆ = (\frac{E_{F 0} - E_{F}}{E_{F 0}}) \times 100 %$

Calculate the expression for Fermi Dirac distribution

The expression for E_F(T) is calculated as,

$\frac{1}{e^{\frac{(E - E_{F})}{k_{B} T + 1}}} = \frac{1}{\frac{e^{\frac{E}{k_{B} T}}}{\frac{E}{e_{B} T}} + 1} e^{\frac{(E - E_{F})}{k_{B} T}} = \frac{e^{\frac{E}{k_{B} T}}}{e^{\frac{E}{k_{B} T}} - 1} \ln [e^{\frac{(E - E_{F})}{k_{B} T}}] = \ln [\frac{e^{\frac{E}{k_{B} T}}}{e^{\frac{E}{k_{B} T}} - 1}] \frac{(E - E_{F})}{k_{B} T} = \frac{E}{k_{B} T} - \ln [e^{\frac{E}{k_{B} T}} - 1] \frac{E}{k_{B} T} - \frac{E_{F}}{k_{B} T} = \frac{E}{k_{B} T} - \ln [e^{\frac{E}{k_{B} T}} - 1] E_{F} = k_{B} Tln [e^{\frac{E}{k_{B} T}} - 1]$

Calculate the expression for EF0

The expression for E_F0 is calculated as,

$\begin{array}{rcl} E_{F} & = & k_{B} Tln [e^{\frac{E}{k_{B} T}} - 1] \\ E_{F 0} & = & k_{B} Tln [e^{\frac{E}{k_{B} T}}] \\ = & k_{B} T (\frac{E}{k_{B} T}) \\ = & E \end{array}$

The graph of EF(T) versus kBTm

The graph is

Calculate the percent drop in Fermi energy

The expression for drop in Fermi energy is calculated as,

$\begin{array}{rcl} ∆ & = & (\frac{E_{F 0} - E_{F}}{E_{F 0}}) \times 100 % \\ = & (\frac{E - E_{F}}{E_{F}}) \times 100 % \\ = & (1 - \frac{E}{E_{F}}) \times 100 % \\ = & \{1 - [\frac{k_{B} T}{E}] \ln (e^{\frac{E}{k_{B} T}} - 1)\} \times 100 % \\ ∆ & = & \{1 - [\frac{k_{B} T}{E}] \ln (e^{\frac{E}{k_{B} T}} - 1)\} \times 100 % \\ = & \{1 - [0.25] \ln [e^{\frac{1}{0.25}} - 1]\} \times 100 % \\ = & 0.462 % \end{array}$