Let's solve the question **step by step**: --- ### **Given:** Show that the de Broglie wavelength of an electron with **energy equal to Fermi energy** is \[ \lambda = \left(\frac{8\pi}{3n}\right)^{1/3} \] where \( n \) is the number density of electrons. --- ## **Step 1: Fermi Energy Expression** For a free electron gas (in a metal, for example), the Fermi energy \( E_F \) is given by: \[ E_F = \frac{\hbar^2 k_F^2}{2m} \] where: - \(\hbar\) is the reduced Planck’s constant, - \(k_F\) is the Fermi wave number, - \(m\) is the electron mass. --- ## **Step 2: Number Density and Fermi Wave Number** The number density \( n \) (number of electrons per unit volume) is related to \( k_F \) by: \[ n = \frac{N}{V} = \frac{g}{(2\pi)^3} \cdot \frac{4}{3}\pi k_F^3 \] where \( g = 2 \) (spin degeneracy for electrons). So, \[ n = \frac{2}{(2\pi)^3} \cdot \frac{4}{3}\pi k_F^3 = \frac{1}{3\pi^2} k_F^3 \] \[ k_F^3 = 3\pi^2 n \] \[ k_F = (3\pi^2 n)^{1/3} \] --- ## **Step 3: de Broglie Wavelength at Fermi Energy** The de Broglie wavelength is: \[ \lambda = \frac{2\pi}{k} \] At the Fermi energy, \( k = k_F \): \[ \lambda_F = \frac{2\pi}{k_F} \] --- ## **Step 4: Substitute \( k_F \) in terms of \( n \):** \[ \lambda_F = \frac{2\pi}{(3\pi^2 n)^{1/3}} \] --- ## **Step 5: Simplify the Expression** Let’s simplify: \[ \lambda_F = 2\pi \cdot (3\pi^2 n)^{-1/3} \] \[ = 2\pi \cdot 3^{-1/3} \cdot (\pi^2)^{-1/3} \cdot n^{-1/3} \] \[ = 2\pi \cdot 3^{-1/3} \cdot \pi^{-2/3} \cdot n^{-1/3} \] \[ = 2 \cdot 3^{-1/3} \cdot \pi^{1 - 2/3} \cdot n^{-1/3} \] \[ = 2 \cdot 3^{-1/3} \cdot \pi^{1/3} \cdot n^{-1/3} \] Now, write \( 2 = (8)^{1/3} \): \[ = (8)^{1/3} \cdot 3^{-1/3} \cdot \pi^{1/3} \cdot n^{-1/3} \] \[ = (8\pi/3n)^{1/3} \] --- ## **Final Answer** \[ \boxed{\lambda_F = \left( \frac{8\pi}{3n} \right)^{1/3}} \] --- ### **Summary of Steps:** 1. Wrote Fermi energy and Fermi wave number expressions. 2. Related number density \( n \) to \( k_F \). 3. Used de Broglie formula \( \lambda = 2\pi / k \). 4. Substituted \( k_F \) and simplified to get the required result. Let me know if you'd like a further breakdown or explanation!