Discussion
The energy of a particle in a three-dimensional box of equal side lengths is given as :
\[ \begin{aligned} E = &\frac{h^{2}}{8ma^{2}} \\ &\left(\frac{n^{2}}{x} + \frac{n^{2}}{y} + \frac{n^{2}}{z}\right) \end{aligned} \]
The degeneracy of energy for the levels nx ny nz = 1, 1, 1, 2, 1, 1 and 3, 2, 1 respectively are :
Answer: C
The degeneracies are 1, 3 and 6.
For a cubic box the bracket stands for the sum of squares of the three quantum numbers, nx2 + ny2 + nz2 (the paper prints the subscripts x, y, z beneath each n2). Every arrangement of the same three numbers gives the same energy, so the degeneracy is the number of distinct permutations:
- (1, 1, 1): sum 3; only one arrangement, so degeneracy 1.
- (2, 1, 1): sum 6; the 2 can be in any of three places, (2,1,1), (1,2,1), (1,1,2), so degeneracy 3.
- (3, 2, 1): sum 14; three different numbers can be arranged in 3! = 6 ways, so degeneracy 6.
No other set of positive integers has squares summing to 14 or 6, so there is no accidental degeneracy to add. 3, 4, 6 and 1, 3, 3 miscount the arrangements, and 1, 1, 1 treats every level as non-degenerate, which is true only for a box with three unequal sides.
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