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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 241
E7D.13(b) �e energy levels for a 2D rectangular box, side lengths L1, L2 are
En1 ,n2 =
h2
8m
(n
2
1
L21
+ n
2
2
L22
)
where n1 and n2 are integers greater than or equal to 1.
For the speci�c case where L1 = L, L2 = 2L,
En1 ,n2 =
h2
8m
(n
2
1
L2
+ n22
(2L)2
) = h2
8mL2
(n21 +
n22
4
)
�e energy of the state with n1 = 2, n2 = 8 is then
E2,2 =
h2
8mL2
(22 + 8
2
4
) = h2
8mL2
(20)
By inspection, the �rst term in the sum is equal to 4 and the second is equal to
16, it can be arranged for these to be swapped.�is requires n21 = 16 and so n1 =
4 and n22/4 = 4, and so n2 = 4. Hence, the state (4, 4) is degenerate with (2, 8) .
�e question notes that degeneracy frequently accompanies symmetry, and
suggests that one might be surprised to �nd degeneracy in a box with unequal
lengths. Symmetry is a matter of degree. �is box is less symmetric than a
square box, but it ismore symmetric than boxes whose sides have a non-integer
or irrational ratio. Every state of a square box except those with n1 = n2 is
degenerate (with the state that has n1 and n2 reversed). Only a few states in this
rectangular box are degenerate. In this system, a state (n1 , n2) is degenerate
with a state (n2/2, 2n1) as long as the latter state (a) exists (that is n2/2 must
be an integer) and (b) is distinct from (n1 , n2). A box with incommensurable
sides, say, L and
√
2L, would have no degenerate levels .
E7D.14(b) �e energy levels of a cubic box are given by [7D.13b–267]
En1 ,n2 ,n3 =
h2
8mL2
(n21 + n22 + n23)
where n1 , n2 , n3 are integers greater than or equal to 1. Hence the lowest energy
state is that with n1 = n2 = n3 = 1, with energy
E(1,1,1) =
h2
8mL2
(12 + 12 + 12) = h2
8mL2
(3)
and so the energy of the level with energy 14/3 times that of the lowest is
14h2/8mL2, whichwill be produced by states for which n21+n22+n23 = 14.�ere
are six stateswith this energy, (n1 , n2 , n3) = (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1),
(3, 1, 2), (3, 2, 1), and so the degeneracy is 6 .
E7D.15(b) �e transmission probability [7D.20a–269] depends on the energy of the tun-
nelling particle relative to the barrier height (ε = E/V0 = (1.5 eV)/(2.0 eV)