 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-3/35c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

(2 π)^{d} / V

k

and including spin degeneracy gives

g (E) d E = \frac{g _{s} V S _{d - 1}}{( 2 π ) ^{d}} k^{d - 1} d k .

Using

k = (E / A)^{1/ α} /ℏ

gives the stated power with

B = \frac{g _{s} S _{d - 1}}{α ( 2 π ) ^{d} ℏ ^{d} A ^{d / α}} .

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