 Solution

A nonzero superconducting energy gap suppresses the population of thermally excited quasiparticles. When $k_{B}T\ll \Delta$, their Boltzmann factor produces an approximately activated electronic heat capacity,where the prefactor $A(T)$ varies algebraically rather than exponentially. For the rough estimate requested, neglect its change between the two quoted temperatures. Taking a ratio eliminates its unknown magnitude:ThereforeThe corresponding excitation energy is $\Delta \simeq 2.19\times 10^{-22}\mathrm{J}\simeq 1.36\mathrm{meV}$. The units $\mathrm{mJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ of the two heat-capacity readings cancel in their ratio.

This is a heat-capacity activation estimate, not an exact determination of a gap from two data points. More generally, if $C_e=A{T}^p{e}^{-\Delta /(k_{B}T)}$, the same measurements giveFor example, the low-temperature BCS theory form for a fully gapped isotropic superconductor has $p=-3/2$ and would give about $19.5\mathrm{K}$. The stated approximate data and constant-gap assumption support the simpler $16\mathrm{K}$ estimate; they do not specify a detailed density of states or heat-capacity prefactor. Here $\Delta$ means the minimum quasiparticle excitation energy, rather than automatically twice that value.