a.) (2 pts) Considering the normal state electronic entropy in the figure, what is the electronic specific heat of Al? S(T) = 0T (C/T’)dT’, where Celec/T = , so Snelec (Tc) = 1.6 mJ/molK = Tc, so =1.6mJ/molK/1.18 K = 1.36 mJ/molK2 (correct literature value) b.) (2 pts) We have discussed in class that, before the BCS theory, the superconducting electronic specific heat, C, was approximated as const.*T3. Plug this into the integral for Sscelec(Tc), what is the const.? 1.6 mJ/molK = 0Tc (C/T’)dT’ = const.*Tc3/3 => const.= 3*(1.6 mJ/molK)/1.183K3 =2.92 mJ/molK4 c.) (3 pts) What is the condensation energy of Al (keep 5 sig figs until the answer, in 3 sig figs)? Sscelec = T ; Snelec = 2.92 mJ/molK4 T3/3 U = 0Tc [1.36*T mJ/molK2 (2.92/3) mJ/molK4(T3)]dT = 1.36 *1.182/2 – in units of mJ/mol = 0.94683 0.47177 = 0.475 mJ/mol (vs literature – value of 0.44)
a.) (2 pts) Considering the normal state electronic entropy in the figure, what is the electronic specific heat of Al? S(T) = 0T (C/T’)dT’, where Celec/T = , so Snelec (Tc) = 1.6 mJ/molK = Tc, so =1.6mJ/molK/1.18 K = 1.36 mJ/molK2 (correct literature value) b.) (2 pts) We have discussed in class that, before the BCS theory, the superconducting electronic specific heat, C, was approximated as const.*T3. Plug this into the integral for Sscelec(Tc), what is the const.? 1.6 mJ/molK = 0Tc (C/T’)dT’ = const.*Tc3/3 => const.= 3*(1.6 mJ/molK)/1.183K3 =2.92 mJ/molK4 c.) (3 pts) What is the condensation energy of Al (keep 5 sig figs until the answer, in 3 sig figs)? Sscelec = T ; Snelec = 2.92 mJ/molK4 T3/3 U = 0Tc [1.36*T mJ/molK2 – (2.92/3) mJ/molK4(T3)]dT = 1.36 *1.182/2 in units of mJ/mol = 0.94683 – 0.47177 = 0.475 mJ/mol (vs literature value of 0.44)
Test%20IISpring2016solnsfinal.docx