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Question 5.23: The bulk modulus of a substance is the ratio of a small decr...

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

B=-V \frac{d P}{d V} .

Show that B = (5/3)P, in the free electron gas model, and use your result in Problem 5.21(d) to estimate the bulk modulus of copper. Comment: The observed value is 13.4 \times 10^{10} N / 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.

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P=\frac{\left(3 \pi^{2}\right)^{2 / 3} \hbar^{2}}{5 m}\left(\frac{N q}{V}\right)^{5 / 3}=A V^{-5 / 3} \Rightarrow B=-V \frac{d P}{d V}=-V A\left(\frac{-5}{3}\right) V^{-5 / 3-1}=\frac{5}{3} A V^{-5 / 3}=\frac{5}{3} P .

For copper, B=\frac{5}{3}\left(3.84 \times 10^{10} N / m ^{2}\right)=6.4 \times 10^{10} N / m ^{2} .

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