CALC When the pressure $p$ on a material increases by an amount $\Delta p,$ the volume of the material will change from $V$ to $V+\Delta V,$ where $\Delta V$ is negative. The bulk modulus $B$ of the material is defined to be the ratio of the pressure change $\Delta p$ to the absolute value $|\Delta V / V|$ of the fractional volume change. The greater the bulk modulus, the greater the pressure increase required for a given fractional volume change, and the more incompressible the material (see Section 11.4$)$ . since $\Delta V<0$ , the bulk modulus can be written as $B=-\Delta p /\left(\Delta V / V_{6}\right) .$ In the limit that the pressure and volume changes are very small, this becomes $$B=-V \frac{d p}{d V}$$ (a) Use the result of Problem 42.55 to show that the bulk modulus

for a system of $N$ free electrons in a volume $V$ at low temperatures is $B=\frac{5}{3} p .$ (Hint: The quantity $p$ in the expression $B=$ $-V(d p / d V)$ is the external pressure on the system. Can you explain why this is equal to the internal pressure of the system itself, as found in Problem 42.55$\%$ (b) Evaluate the bulk modulus for the electrons in copper, which has a free-electron concentration of $8.45 \times 10^{28} \mathrm{m}^{-3}$ . Express your result in pascals. (c) The actual bulk modulus of copper is $1.4 \times 10^{11} \mathrm{Pa}$ . Based on your result in

part (b), what fraction of this is due to the free electrons in copper? (This result shows that the free electrons in a metal play a major role in making the metal resistant to compression.) What do you

think is responsible for the remaining fraction of the bulk modulus?