Final answer:
The kinetic energy U0 of a three-dimensional gas of N free electrons at 0 K is shown to be (3/5)NEF by integrating over all electron states up to the Fermi energy EF, acknowledging the Pauli exclusion principle and electron spin states.
Step-by-step explanation:
To demonstrate that the kinetic energy of a 3D gas of N free electrons at 0 K is U0 = (3/5)NEF, we must use concepts from quantum mechanics and statistical physics. At absolute zero, electrons populate states up to the Fermi energy (EF), which is the highest occupied energy level in a system of fermions at 0 K. Each energy level can be occupied by two electrons due to their spin, and in three dimensions, the energy states can be derived as being proportional to the square of quantum numbers (n1, n2, n3).
Starting by identifying the density of states g(E) and the energy of the system, we can integrate over all possible states up to the Fermi energy to obtain the total energy U. Considering the volume of a quarter of a sphere in quantum number space, where N corresponds to the number of states inside the sphere with radius R, the total number of particles is N = 2(1/8)(4/3)πR3. The factor 2 accounts for the two possible spin states for each electron.
The total energy for a Fermi gas at 0 K is then obtained by integration over all states up to the Fermi energy. Since kinetic energy, KE, for a particle is given by KE = (1/2)mv2 and the velocity is related to the momentum, which in turn is related to the energy levels in the quantum system, the result of the integration yields U = (3/5)NEF, where EF is the Fermi energy.