Question

Thebulk modulus of a substance is the ratio of a small decrease in pressure to the resulting fractional increase in volume:

$\mathit{B}\mathbf{=}\mathbf{-}\mathit{V}\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}{\mathbf{V}}^{\mathbf{.}}}$

Show that$\mathit{B}\mathbf{=}\left(5/3\right)\mathit{P}$, in the free electron gas model, and use your result in Problem 5.16(d) to estimate the bulk modulus of copper. Comment: The observed value is $13.4\times {10}^{10}\mathrm{N}/{\mathrm{m}}^2$, but don’t expect perfect agreement—after all, we’re neglecting all electron–nucleus and electron–electron forces! Actually, it is rather surprising that this calculation comes as close as it does.

Short answer

$\begin{array}{rcl}\mathit{B}& \mathbf{=}& \frac{\mathbf{5}}{\mathbf{3}}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{84}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathbf{N}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}\mathbf{)}\\ & \mathbf{=}& \mathbf{6}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathit{N}\mathbf{/}{\mathit{m}}^{\mathbf{2}}\end{array}$

Step-by-step solution

Definition of bulk modulus of a substance

The volumetric stress to volumetric strain ratio for any given material is known as the bulk modulus.

Showing that B = (5/3) P in the free electron gas model

We need to find bulk modulus in the free electron gas model. And after that, we need to estimate the bulk modulus of copper from the previous problem.

$p=\frac{2E_{tot}}{3V}=\frac{2}{3}\frac{{\hslash }^2{k}_F^5}{5m}{\rho }^{5/3}$

$P=\frac{{\left(3{\pi }^2\right)}^{2/3}{\hslash }^2}{5m}{\rho }^{5/3}=A{V}^{-5/3},A=\frac{{\hslash }^2{\left(3{\pi }^2\right)}^{2/3}{\left(Nq\right)}^{5/3}}{5m}.$

Bulk modulus is equal to:

role="math" localid="1658144523557" $\begin{array}{rcl}B& =& -V\frac{dP}{dV}\\ & =& -V\left(-\frac{5}{3}A{V}^{-8/3}\right)=\frac{5}{3}A{V}^{-5/3}\\ & =& \frac{5}{3}P\\ B& =& \frac{5}{3}P\end{array}$

Degeneracy pressure of copper was $P=3.84\times {10}^{10} \text{Pa}$. So Bulk modulus, according to the formula is,For copper,

$\begin{array}{rcl}B& =& \frac{5}{3}\left(3.84\times {10}^{10} N/m^2\right)\\ & =& 6.4\times {10}^{10} N/m^2.\\ & & \end{array}$