 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/35b/b/solution

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

The number of momentum states in a shell is

\frac{g _{s} V}{( 2 π ℏ ) ^{d}} S_{d - 1} p^{d - 1} d p .

With

E = p^{2} / (2 m)

this gives

g (E) = g_{s} B V E^{(d - 2) /2}, B = \frac{S _{d - 1} m ( 2 m ) ^{(d - 2) /2}}{( 2 π ℏ ) ^{d}} .

At zero temperature all states up to

E_{F}

are occupied, hence

N = \frac{2 g _{s} B V}{d} E_{F}^{d /2}, E_{F} = (\frac{d N}{2 g _{s} B V})^{2/ d} .

The energy is

U = d N E_{F} / (d + 2)

. Since a nonrelativistic ideal gas in

d

p V = 2 U / d

p V = \frac{2}{d + 2} N E_{F} .

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