 Solution

ID: past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2013/paper-1/1/c/solution

Use a three-dimensional ideal gas at the same number density n for both isotopes. An unpolarized helium-3 gas has spin degeneracy two, hence

k_{B} T_{F} = \frac{ℏ ^{2}}{2 m _{3}} (3 π^{2} n)^{2/3} .

Helium-4 has a single spin state, and the ideal Bose-gas saturation condition gives

k_{B} T_{B} = \frac{2 π ℏ ^{2}}{m _{4}} [\frac{n}{ζ ( 3/2 )}]^{2/3} .

Eliminating n, using

m_{3} / m_{4} ≃ 3/4

and

ζ (3/2) = 2.612

, yields

\frac{T _{B}}{T _{F}} = \frac{4 π}{( 3 π ^{2} ) ^{2/3} ζ ( 3/2 ) ^{2/3}} \frac{m _{3}}{m _{4}} ≃ 0.519, T_{B} ≃ 2.6 K .

The fermionic spin degeneracy must be included. This is an ideal-gas estimate; strong interactions in liquid helium prevent identifying it with an exact experimental transition temperature.

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