Question

At 1000K, the fraction of the conduction electrons in a metal that have energies greater than the Fermi energy is equal to the area under the curve of Fig. 41-8bbeyond ${\mathit{E}}_{\mathbf{F}}$ divided by the area under the entire curve. It is difficult to find these areas by direct integration. However, an approximation to this fraction at any temperature T is $\mathit{f}\mathit{r}\mathit{a}\mathit{c}\mathbf{=}\frac{\mathbf{3}\mathbf{k}\mathbf{T}}{\mathbf{2}{\mathbf{E}}_{\mathbf{F}}}$.

Note that frac = 0 for T = 0 K, just as we would expect. What is this fraction for copper at (a) 300 K and (b) 1000 K? For copper ${\mathit{E}}_{\mathbf{F}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{}\mathit{e}\mathit{V}$. (c) Check your answers by numerical integration using Eq. 41-7.

Short answer

a) The value of the fraction for copper at $T=300\mathrm{K}\mathrm{is}5.5\times {10}^{-3}$.

b) The value of the fraction for copper at $T=1000K\mathrm{is}1.8\times {10}^{-2}$.

c) The answers for the above temperature values T = 300 K and T = 1000 K are 0.00385 and 0.0129 respectively by numerical integration.

Step-by-step solution

The given data

a) The given temperature values are: T = 300 K andb T = 1000 K

b) Fermi energy for copper,$E_{\mathrm{F}}=7\mathrm{eV}$

c) The given approximation to the fraction at any temperature,

$frac=\frac{3kT}{2E_{\mathrm{F}}},\mathrm{where}\mathrm{k}=8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K}$

Understanding the concept of fraction of conduction electrons

The fraction of the conduction electrons can be found by calculating the area under the curve of the graph between the number density and the energy of the level. The fraction of the conduction electrons having energy greater than Fermi level is equal to the ratio of the area under the curve above the Fermi level to that of the total area under the curve.

Formula:

The fraction to be calculated by numerical integration is-

$$\begin{gathered} frac=\frac{{\int }_{E_{\mathrm{F}}}^{\infty }\sqrt{E}/\left(e^{\left(E-E_{\mathrm{F}}\right)kT}+1\right)dE}{{\int }_0^{\infty }\sqrt{E}/\left(e^{\left(E-E_{\mathrm{F}}\right)kT}+1\right)dE}.............................\left(1\right) \\ \mathrm{where}\mathrm{k}=8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K} \end{gathered}$$

a) Calculation of the fraction at T = 300 K

Using the given data and the temperature value T = 300K in the given equation of fraction, we obtain the fraction of conduction electrons at T = 300Kas follows:

$$\begin{gathered} frac=\frac{3\left(8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)\left(300\mathrm{K}\right)}{2\left(7.0\mathrm{eV}\right)} \\ =5.5\times {10}^{-3} \end{gathered}$$

Hence, the value of the fraction is $5.5\times {10}^{-3}$.

b) Calculation of the fraction at T = 1000K

Using the given data and the temperature value T = 1000 K in the given equation of fraction, we obtain the fraction of conduction electrons at T = 1000 K as follows:

$$\begin{gathered} frac=\frac{3\left(8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)\left(1000\mathrm{K}\right)}{2\left(7.0\mathrm{eV}\right)} \\ =1.8\times {10}^{-2} \end{gathered}$$

Hence, the value of the fraction is $1.8\times {10}^{-2}$.

Calculation of the above values by numerical integration

Using the math software package MAPLE that has built-in numerical integration routines. Setting up the ratios of the integrals of Eq. 41-7 and canceling the common factors, we obtain the equation from equation (1)

$$\begin{gathered} frac=\frac{{\int }_{E_{\mathrm{F}}}^{\infty }\sqrt{E}/\left(e^{\left(E-E_{\mathrm{F}}\right)kT}+1\right)dE}{{\int }_0^{\infty }\sqrt{E}/\left(e^{\left(E-E_{\mathrm{F}}\right)kT}+1\right)dE} \\ \mathrm{where}\mathrm{k}=8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K} \end{gathered}$$

Now, we have the Fermi energy value of copper and hence evaluating this for and T = 1000 K, we find the value of the fractions as frac = 0.00385 and frac = 0.0129 respectively.