The density of copper is$8.96\mathrm{g}/{\mathrm{cm}}^3,$and its atomic weight is$63.5\mathrm{g}/\mathrm{mole}$

(a) Calculate the Fermi energy for copper (Equation 5.43). Assume d = 1, and give your answer in electron volts.

$E_F=\frac{{ħ}^2}{2m}{\left(3{\mathrm{\rho \pi }}^2\right)}^{2/3}$ (5.43).

(b) What is the corresponding electron velocity? Hint: Set$E_F=\left(1/2\right)m{v}^2$Is it safe to assume the electrons in copper are nonrelativistic?

(c) At what temperature would the characteristic thermal energyrole="math" localid="1656065555994" $(k_{B}T,where{k}_{B}kB$is the Boltzmann constant and T is the Kelvin temperature) equal the Fermi energy, for copper? Comment: This is called the Fermi temperature,$T_F$

. As long as the actual temperature is substantially below the Fermi temperature, the material can be regarded as “cold,” with most of the electrons in the lowest accessible state. Since the melting point of copper is 1356 K, solid copper is always cold.

(d) Calculate the degeneracy pressure (Equation 5.46) of copper, in the electron gas model.

$P=\frac{2}{3}\frac{E_{tot}}{V}=\frac{2}{3}\frac{{ħ}^2{k}_F^5}{10{\pi }^{2}m}=\frac{{\left(3{\pi }^2\right)}^{2/3}{ħ}^2}{5m}{\mathrm{\rho }}^{5/3}$

Step by step solution

01

Given data

density of copper d =$8.96gm/cm3$

atomic weightof copper M =$\mathrm{m}=63.5gm/mol$

02

(a) Calculating the Fermi energy for copper

$$\begin{gathered} E_F=\frac{ħ}{2m}{\left(3{\mathrm{\rho \pi }}^2\right)}^{2/3}.\mathrm{\rho }=\frac{Nq}{V}=\frac{N}{V}=\frac{\mathrm{atoms}}{\mathrm{mole}}\times \frac{\mathrm{moles}}{\mathrm{gm}}\times \frac{\mathrm{gm}}{\mathrm{volume}}=\frac{{\mathrm{N}}_{\mathrm{A}}}{\mathrm{M}}.\mathrm{d} \\ M=atomicmass=63.5gm/mol,d=density=8.96gm/cm3. \\ \mathrm{\rho }=\frac{\left(6.02\times {10}^{23}\right)\left(8.96gm/c{m}^3\right)}{\left(63.5gm\right)}=8.49\times {10}^{22}/c{m}^3=8.49\times {10}^{28}/m^3 \\ {E}_F=\frac{\left(1.055\times {10}^{-34}J.s\right)\left(6.58\times {10}^{-16}eV.s\right)}{\left(2\right)\left(9.109\times {10}^{-31}kg\right)}{\left(3{\pi }^{2}8.49\times {10}^{28}/m^3\right)}^{2/3}=7.04eV. \end{gathered}$$

03

Step3:(b) Corresponding electron velocity

$7.04eV=\frac{1}{2}\left(0.511\times {10}^{6}eV/c^2\right)v^2\Rightarrow \frac{v^2}{c^2}=\frac{14.08}{.511\times {10}^6}=2.76\times {10}^{-5}\Rightarrow \frac{v}{c}=5.25\times {10}^{-3}$

so it’s nonrelativistic.$v=\left(5.25\times {10}^{-3}\right)\times \left(3\times {10}^8\right)=1.57\times {10}^{6}m/s$

04

Step4:(c) Temperature

$T=\frac{7.04eV}{8.62\times {10}^{-5}eV/K}=8.17\times {10}^{4}K$

05

Step5:(d) Calculating the Degeneracy of copper

$P=\frac{{\left(3{\pi }^2\right)}^{2/3}{ħ}^2}{5m}{\mathrm{\rho }}^{5/3}=\frac{{\left(3{\mathrm{\pi }}^2\right)}^{2/3}{\left(1.055\times {10}^{-34}\right)}^2}{5\left(9.109\times {10}^{-31}\right)}{\left(8.49\times {10}^{28}\right)}^{5/3}N/m^2=3.84\times {10}^{10}N/m^2$

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