Specific heat — UPSC Physics Optional CSE 1990 Paper I

Thermal energy of a solid is given by the relation E = \int_0^{ u_m} \frac{h\nu g(\nu)d\nu}{e^{h\nu/k_B T} - 1} where u_m = k_B \theta_D/h, \theta_D being the Debye temperature. Given g(\nu) = 6N h^2 \nu(k_B \theta_D)^2, deduce the expression for E for T \ll \theta_D and discuss the temperature variation of specific heat for T \ll \theta_D. \left( \int_0^\infty \frac{x^3 dx}{e^x - 1} = 2.404 \right)