 Solution

Assume three-dimensional isotropic gapless excitations, zero chemical potential and one independent branch. Momentum-state counting givesSubstituting $x=ϵ/(k_{B}T)$ into the Bose energy integral shows $U/V\propto T^{1+3/n}$ and $C/V\propto T^{3/n}$. The observed exponent therefore gives $n=2$.

For the quadratic-boson heat-capacity coefficient, the required Bose integral isThe equality follows by expanding $(e^x-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-rx}$ and integrating term by term. HenceThe supplied units make its $C$ a heat capacity per volume. Equating its coefficient to $220\mathrm{J}{\mathrm{K}}^{-5/2}{\mathrm{m}}^{-3}$ givesEquivalently $ϵ=p^2/(2m_{\ast })$ has $m_{\ast }=1.09\times 10^{-28}\mathrm{kg}$. For $q_b$ identical branches, the measured coefficient fixes $q_b/{\alpha }^{3/2}$, so instead $\alpha =(4.58\times 10^{27}\mathrm{k}{\mathrm{g}}^{-1})q_b^{2/3}$. The one-branch convention is necessary for the quoted numerical value. The supplied integral with $x^4$ is not the moment arising for the exponent determined here.