 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-1/section-a/2/solution

K = L^{2} / (2 I)

. For a three-dimensional linear rigid rotor, angular momentum operator replaces the squared magnitude by

ℏ^{2} l (l + 1)

, where

l = 0, 1, 2, \dots

. Hence

E_{l} = \frac{ℏ ^{2}}{2 I} l (l + 1), \frac{ℏ ^{2}}{I} = 1.1121 \times 1 0^{- 21} J = 6.941 \times 1 0^{- 3} eV .

The three lowest distinct energy levels are therefore

E_{0} = 0, E_{1} = 6.94 \times 1 0^{- 3} eV, E_{2} = 2.08 \times 1 0^{- 2} eV .

Their magnetic quantum degeneracy values are

1, 3, 5

. A single moment of inertia specifies this rigid rotor model; an arbitrary asymmetric body would require its other principal moments.

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