Question

Consider a particle trapped in a three-dimensional potential well such that U(x, y, z) = 0 for 0 ≤ x ≤ a, 0 ≤ y ≤ a, 0 ≤ z≤a and U(x, y, z) = ∞ everywhere else. The degeneracy of the 5^th excited state is

• A. 1
• B. 3
• C. 6
• D. 9

Answer 1

The Correct Option is C

The problem describes a quantum mechanical particle in a three-dimensional infinite potential well (also known as a "particle in a box"). The potential well is defined such that the potential energy \(U(x, y, z)\) is zero inside a cube of side \(a\) and infinite outside.

The energy levels of a particle in a three-dimensional box are given by the formula:

\(E_{n_x, n_y, n_z} = \frac{h^2}{8m a^2}(n_x^2 + n_y^2 + n_z^2)\)

where \(n_x\), \(n_y\), and \(n_z\) are quantum numbers corresponding to the dimensions \(x\), \(y\), and \(z\), respectively, and they can take positive integer values (1, 2, 3, ...).

The ground state corresponds to \(n_x = 1, n_y = 1, n_z = 1\), and the energy for higher states increases with increasing values of \(n_x\), \(n_y\), or \(n_z\).

The degeneracy of an energy level refers to the number of different combinations of \((n_x, n_y, n_z)\) that result in the same energy level. To find the degeneracy, we need to consider all possible permutations of the quantum numbers that give the same sum \(n_x^2 + n_y^2 + n_z^2\).

Let us calculate the energy states:

1. First excited state: \(n_x, n_y, n_z\) can be (1, 1, 2) and permutations.
2. Second excited state: \(n_x, n_y, n_z\) can be (1, 2, 2) and permutations.
3. Third excited state: \(n_x, n_y, n_z\) can be (2, 2, 2).
4. Fourth excited state: \(n_x, n_y, n_z\) can be (1, 1, 3) and permutations.
5. Fifth excited state: Consider possible combinations:

For the 5th excited state, suitable combinations of quantum numbers are (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1).

Therefore, the number of permutations that lead to this energy level is 6.

Thus, the degeneracy of the 5th excited state is 6.