Chapter 38, Problem 32P

(a)

To determine

Proof of $P=\frac{2}{5}\frac{N{E}_F}{V}=C{V}^{-5/3}$ .

Answer to Problem 32P

The proof of $P=\frac{2}{5}\frac{N{E}_F}{V}=C{V}^{-5/3}$ has been determined.

Explanation of Solution

Given:

The monatomic ideal-gas relation is,

  $PV=\frac{2}{3}N{E}_{\text{av}}$   ..... (1)

The Fermi energy from equation 38-22b is,

  $E_F=\left(0.3646\text{eV}\cdot {\text{nm}}^{\text{2}}\right){\left(\frac{N}{V}\right)}^{2/3}$ …... (2)

The average energy from equation 38-23 is,

  $E_{\text{av}}=\frac{3}{5}{E}_F$   ..... (3)

Calculation:

From equation (1) and (3),

  $\begin{array}{c}PV=\frac{2}{3}N\left(\frac{3}{5}{E}_F\right)\\ PV=\frac{2}{5}N{E}_F\\ P=\frac{2}{5}\frac{N{E}_F}{V}\end{array}$   ..... (4)

From equation (4) and (2),

  $\begin{array}{c}P=\frac{2}{5}\frac{N}{V}\left(0.3646\text{eV}\cdot {\text{nm}}^{\text{2}}\right){\left(\frac{N}{V}\right)}^{2/3}\\ =\left(0.3646\text{eV}\cdot {\text{nm}}^{\text{2}}\right)\frac{2}{5}{\left(\frac{N}{V}\right)}^{5/3}\\ =\left(0.3646\text{eV}\cdot {\text{nm}}^{\text{2}}\right)\frac{2}{5}{N}^{5/3}{V}^{-5/3}\\ =C{V}^{-5/3}\end{array}$   ..... (5)

Hence, equation $P=\frac{2}{5}\frac{N{E}_F}{V}=C{V}^{-5/3}$ is proved.

Conclusion:

Therefore, the proof of $P=\frac{2}{5}\frac{N{E}_F}{V}=C{V}^{-5/3}$ has been determined.

(b)

To determine

Proof of $B=\frac{5}{3}P=\frac{2}{3}\frac{N{E}_F}{V}$ .

Answer to Problem 32P

The proof of $B=\frac{5}{3}P=\frac{2}{3}\frac{N{E}_F}{V}$ has been determined.

Explanation of Solution

Given:

The expression of bulk modulus is,

  $B=-V\frac{\partial P}{\partial V}$   ..... (6)

From equation (5) and (6),

  $\begin{array}{c}B=-V\frac{\partial \left(C{V}^{-5/3}\right)}{\partial V}\\ =-V\left(-\frac{5}{3}C{V}^{-8/3}\right)\\ =\frac{5}{3}C{V}^{-5/3}\\ =\frac{5}{3}P\left(\because P=C{V}^{-5/3}\right)\end{array}$

Substitute value of $P$ from equation (4) in above equation,

  $\begin{array}{c}B=\frac{5}{3}\left(\frac{2}{5}\frac{N{E}_F}{V}\right)\\ =\frac{2}{3}\frac{N{E}_F}{V}\end{array}$   ..... (7)

Conclusion:

Therefore, the proof of $B=\frac{5}{3}P=\frac{2}{3}\frac{N{E}_F}{V}$ has been determined.

(c)

To determine

The bulk modulus for copper.

Answer to Problem 32P

The bulk modulus for copper is $63.6\times {10}^9\text{N}/{\text{m}}^{\text{2}}$ .

Explanation of Solution

Given:

The value of $N/V$ for Copper from Table 38-1 is $N/V=84.7\text{electrons}/{\text{nm}}^{\text{3}}$

The Fermi energy for copper from Table 38-1 is $E_F=7.03\text{eV}$

Calculation:

From equation (7), the bulk modulus for copper is calculated as,

  $\begin{array}{c}B=\frac{2}{3}\left(84.7\text{electrons}/{\text{nm}}^{\text{3}}\right)\left(7.03\text{eV}\right)\\ =\frac{2}{3}\left(84.7\times {10}^{27}\text{electrons}/{\text{m}}^{\text{3}}\right)\left(7.03\text{eV}\right)\left(\frac{1.602\times {10}^{-19}\text{N}\cdot \text{m}}{1\text{eV}}\right)\\ =63.593\times {10}^9\text{N}/{\text{m}}^{\text{2}}\\ \approx 63.6\times {10}^9\text{N}/{\text{m}}^{\text{2}}\end{array}$

Comparison of $63.6\times {10}^9\text{N}/{\text{m}}^{\text{2}}$ with $140\times {10}^9\text{N}/{\text{m}}^{\text{2}}$ is,

  $\frac{140\times {10}^9\text{N}/{\text{m}}^{\text{2}}}{63.6\times {10}^9\text{N}/{\text{m}}^{\text{2}}}=2.2$

Conclusion:

Therefore, the bulk modulus for copper is $63.6\times {10}^9\text{N}/{\text{m}}^{\text{2}}$ .