LIST-I (Energy of a particle in a box of length L)
LIST-II (Degeneracy of the states)
A.
\( \frac{14h^2}{8mL^2} \)
I.
1
B.
\( \frac{11h^2}{8mL^2} \)
II.
3
C.
\( \frac{3h^2}{8mL^2} \)
III.
6
Choose the correct answer from the options given below:
Show Hint
To find the degeneracy for a 3D particle in a box, you are looking for the number of ways you can sum the squares of three positive integers to get a specific number. Systematically test combinations of small integers (1, 2, 3, 4, ...) to find the sets that work.
Step 1: Recall the energy formula for a particle in a 3D cubic box.
The energy levels for a particle of mass \(m\) in a cubic box of side length \(L\) are given by:
\[
E = \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2)
\]
where \(n_x, n_y, n_z\) are positive integers (1, 2, 3, ...). Degeneracy is the number of different combinations of \((n_x, n_y, n_z)\) that yield the same energy value.
Step 2: Calculate the degeneracy for each energy level.
- A. Energy \(14h^2/(8mL^2)\):
We need to find the number of integer sets \((n_x, n_y, n_z)\) such that \(n_x^2 + n_y^2 + n_z^2 = 14\).
By inspection, \(1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14\). The distinct sets are permutations of (1, 2, 3). The number of permutations of three distinct numbers is \(3! = 6\). The states are (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1). The degeneracy is 6. So, A \(\rightarrow\) III.
- B. Energy \(11h^2/(8mL^2)\):
We need \(n_x^2 + n_y^2 + n_z^2 = 11\).
By inspection, \(1^2 + 1^2 + 3^2 = 1 + 1 + 9 = 11\). The distinct sets are permutations of (1, 1, 3). The number of permutations is \(\frac{3!}{2!} = 3\). The states are (1,1,3), (1,3,1), (3,1,1). The degeneracy is 3. So, B \(\rightarrow\) II.
- C. Energy \(3h^2/(8mL^2)\):
We need \(n_x^2 + n_y^2 + n_z^2 = 3\).
The only possible combination is \(1^2 + 1^2 + 1^2 = 3\). The state is (1,1,1). There is only one such state. The degeneracy is 1. So, C \(\rightarrow\) I.
Step 3: Formulate the correct matching sequence.
The correct matches are: A \(\rightarrow\) III, B \(\rightarrow\) II, and C \(\rightarrow\) I. This corresponds to option (3).
