A solid niobium block is held at in a cryostat

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Condensed Matter, Materials & Soft Matter · Origins of quantum theory

Thermal Physics, Ch Photons and phonons: blackbody radiation and the heat capacity of solids

Problem

A solid niobium block is held at 15.0 K15.0\,\mathrm{K} in a cryostat. Machined inside it is an evacuated cavity of volume 2.0 cm32.0\,\mathrm{cm^3} whose walls are at the block's temperature, connected to the outside by an aperture of area 0.50 mm20.50\,\mathrm{mm^2} that opens into a space cold enough that nothing comes back in. Niobium has a Debye temperature θD=275 K\theta_D = 275\,\mathrm{K}; ignore the contribution of the conduction electrons to its heat capacity. ((a)) Using the standing-wave state count in a box, find how many electromagnetic modes the cavity holds with frequency below 1.00 THz1.00\,\mathrm{THz}, counting both polarisations. ((b)) Find the mean number of photons ⟨n⟩\langle n\rangle in one such mode at 1.00 THz1.00\,\mathrm{THz}, and state in one sentence why the chemical potential to be used for the photons is zero. ((c)) Using the effusion argument (a flux 14cu\tfrac14 c u for quanta of speed cc and energy density uu), find the radiant power leaving through the aperture. ((d)) Find the molar heat capacity of the niobium at 15.0 K15.0\,\mathrm{K} from the Debye low-temperature law CV=(12π4/5)R(T/θD)3C_V = (12\pi^4/5)R(T/\theta_D)^3, and check that T≪θDT \ll \theta_D. Use kB=1.381×10−23 J K−1k_B = 1.381\times10^{-23}\,\mathrm{J\,K^{-1}}, h=6.626×10−34 J sh = 6.626\times10^{-34}\,\mathrm{J\,s}, c=2.998×108 m s−1c = 2.998\times10^{8}\,\mathrm{m\,s^{-1}}, R=8.314 J K−1 mol−1R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}, σSB=5.67×10−8 W m−2 K−4\sigma_{\rm SB} = 5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}.

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