Question

A particle of mass \( m \) is placed in a three-dimensional cubic box of side \( a \). What is the degeneracy of its energy level with energy \( \frac{14 \hbar^2 \pi^2}{2m a^2} \)?

Hint:

In a three-dimensional box, the degeneracy of an energy level is the number of different combinations of quantum numbers that yield the same energy.

Answer 1

Step 1: Understanding the energy levels.
The energy levels for a particle in a three-dimensional box are given by: \[ E_n = \frac{n_x^2 + n_y^2 + n_z^2 \hbar^2 \pi^2}{2m a^2} \] where \( n_x, n_y, n_z \) are positive integers, and the energy is quantized. The energy given is \( \frac{14 \hbar^2 \pi^2}{2m a^2} \), so we need to find the combination of \( n_x, n_y, n_z \) that satisfies this energy.
Step 2: Determine the quantum numbers.
We need to find the values of \( n_x, n_y, n_z \) such that: \[ n_x^2 + n_y^2 + n_z^2 = 14 \] The possible combinations of \( (n_x, n_y, n_z) \) that satisfy this are: \[ (3, 1, 2), (2, 3, 1), (1, 3, 2), \text{ and permutations.} \] Thus, there are 6 combinations.
Step 3: Conclusion.
Thus, the degeneracy of this energy level is 6.