The degeneracy of the state having energy \(\rm\frac{27 h^2}{8mL^2}\) for a particle in a 3-D cubic box of length L is

Concept:

In a three-dimensional cubic box, the energy of a particle is given by the equation:

\(E = h²(n₁² + n₂² + n₃²) / 8mL²\)

where: E is the energy, h is Planck's constant, n₁, n₂, and n₃ are the quantum numbers associated with the particle (they can be any positive integer), m is the mass of the particle, and L is the length of the box.

Explanation:

You're given that the energy E is 27h²/ 8mL² . Setting this equal to the energy equation, we get:

\(27h² / 8mL² = h²(n₁² + n₂² + n₃²) / 8mL²\)

Solving for n₁² + n₂² + n₃², we find that it equals 27. This means that the sum of the squares of the three quantum numbers equals 27.

The possible sets of quantum numbers (n₁, n₂, n₃) that satisfy this equation are (3, 3, 3), (1, 1, 5), (1, 5, 1), and (5, 1, 1). Each set of quantum numbers corresponds to a different state of the particle, so there are four states that have this energy.

Conclusion:-

Therefore, the degeneracy of the state with energy \(27h² / 8mL²\) is 4