Let's solve the problem in **three steps** as requested: --- ### **Step 1: Calculate the Number Density (n) of Potassium Atoms** Given: - Density (\(\rho\)) = \(860 \, \text{kg/m}^3\) - Atomic mass of K (\(M\)) = \(39.1 \, \text{amu}\) - \(1 \, \text{amu} = 1.66 \times 10^{-27} \, \text{kg}\) First, convert atomic mass to kg: \[ M = 39.1 \times 1.66 \times 10^{-27} \approx 6.49 \times 10^{-26} \, \text{kg} \] Number of atoms per unit volume (\(n\)): \[ n = \frac{\rho}{M} = \frac{860}{6.49 \times 10^{-26}} \approx 1.33 \times 10^{28} \, \text{atoms/m}^3 \] **Explanation:** We calculated the number density of potassium atoms, which is needed for the Fermi energy calculation. --- ### **Step 2: Calculate the Fermi Energy (\(E_F\))** Fermi energy formula: \[ E_F = \frac{\hbar^2}{2m} (3\pi^2 n)^{2/3} \] Where: - \(m\) = mass of electron \(= 9.11 \times 10^{-31} \, \text{kg}\) - \(\hbar = 1.055 \times 10^{-34} \, \text{J s}\) - \(n = 1.33 \times 10^{28} \, \text{m}^{-3}\) Plug values in: \[ E_F = \frac{(1.055 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31}} (3\pi^2 \times 1.33 \times 10^{28})^{2/3} \] Calculate \(3\pi^2 n\): \[ 3\pi^2 n = 3 \times (\pi^2) \times 1.33 \times 10^{28} \approx 3 \times 9.87 \times 1.33 \times 10^{28} \approx 39.36 \times 10^{28} \] Now, \((3\pi^2 n)^{2/3}\): \[ (39.36 \times 10^{28})^{2/3} \approx (3.936 \times 10^{29})^{2/3} \approx 2.52 \times 10^{20} \] Now plug back: \[ E_F = \frac{(1.055 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31}} \times 2.52 \times 10^{20} \] \[ (1.055 \times 10^{-34})^2 = 1.113 \times 10^{-68} \] \[ \frac{1.113 \times 10^{-68}}{2 \times 9.11 \times 10^{-31}} = \frac{1.113 \times 10^{-68}}{1.822 \times 10^{-30}} \approx 6.11 \times 10^{-39} \] \[ E_F = 6.11 \times 10^{-39} \times 2.52 \times 10^{20} \approx 1.54 \times 10^{-18} \, \text{J} \] Convert to eV (\(1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}\)): \[ E_F = \frac{1.54 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 9.6 \, \text{eV} \] **Explanation:** We used the Fermi energy formula for a free electron gas and found the Fermi energy of potassium. --- ### **Step 3: Calculate the Compressibility (\(\kappa\))** Given formula: \[ \kappa = \frac{3}{2} \cdot \frac{1}{n E_F} \] Plugging values: \[ \kappa = \frac{3}{2} \cdot \frac{1}{1.33 \times 10^{28} \times 1.54 \times 10^{-18}} \] \[ = \frac{1.5}{2.05 \times 10^{10}} \] \[ = 7.32 \times 10^{-11} \, \text{N}^{-1}\text{m}^2 \] **Explanation:** We applied the formula for the compressibility of a Fermi gas using the number density and Fermi energy we calculated. --- ## **Final Answers** - **Fermi Energy (\(E_F\)) of potassium:** \(\boxed{9.6 \, \text{eV}}\) - **Compressibility (\(\kappa\)) of potassium at 0K:** \(\boxed{7.32 \times 10^{-11} \, \text{N}^{-1}\text{m}^2}\) Let me know if you want to see the compressibility calculation with the value in the problem statement!

The solution primarily relies on concepts from free electron theory and quantum mechanics. First, the number density of electrons is determined from the atomic density of potassium atoms, assuming one free electron per atom. Next, the Fermi energy is calculated using the free electron model formula, which relates the electron density to the highest occupied energy level at absolute zero. Finally, the compressibility is derived from the inverse of the Fermi energy and number density, reflecting how the electron gas responds to pressure changes. These concepts collectively connect microscopic atomic properties to macroscopic electronic behavior in metals.