Question

Derive expression for the density of state of the electron gas contained in the "cube box" of the metal with the edge of 3a, where a is lattice constant.

Answer 1

The density of states (DOS) for the electron gas in the cube box of the metal with an edge of (3a) is given by
`\(D(E) = (1)/(4 \pi^2) \left((2m)/(\hbar^2)\right)^(3/2) √(E)\)`, where (D(E)) is the density of states per unit volume per unit energy.

Step-by-step explanation:

In the expression for the density of states, (D(E)), several key factors contribute to the overall understanding of the electron gas in the cube box. The term
`\((1)/(4 \pi^2)\)` arises from the normalization factor, ensuring the probability density integrates to unity over all states. The term
`\(\left((2m)/(\hbar^2)\right)^(3/2)\)` incorporates the mass (m) and reduced Planck's constant
`(\(\hbar\))`, which are fundamental constants in quantum mechanics. Additionally, the
`\(√(E)\)`term represents the dependence of the density of states on the square root of the energy, illustrating the quantum nature of the electron gas.

This expression is derived from solving the Schrödinger equation for a particle in a box and applying quantum mechanics principles. The cube box of edge (3a) introduces the lattice constant (a), reflecting the periodic nature of the crystal lattice. The resulting DOS expression provides a valuable insight into the distribution of energy states available to electrons in the metal, crucial for understanding electronic properties and behaviors in condensed matter physics.

In summary, the density of states expression encapsulates the quantum characteristics of electrons in the cube box, considering the confinement within the crystal lattice. The parameters in the formula highlight the foundational constants in quantum mechanics, emphasizing the intricate relationship between energy and available states in the electron gas.