Verifica del 04 febbraio 2003 - solution

1. initial state: ⁵S₂; final states: ⁵P_{1,2 and 3}; number of lines: 9, 12 (NOT 13!!), 15
2. d=7.42×10^-11m; E_zero=236.6 meV; C_V(150 K)=1.68×10^-13k_B C_V(800 K)=0.049 k_B
3. To express the Fermi energy in terms of the number density of fermions of 1 spin kind only, a simple way is to note that for fermions of 2 spin kinds ε_F(n)=a n^2/3= a 2^2/3 (n_up)^2/3, where a is a suitable constant. Then the required maximum density n_max is given by equating 2 µ_B B = a 2^2/3 (n_max)^2/3. One finds: n_max=32 3^-1 π (m_e µ_B B)^3/2 h^-3 = 3.16×10²² m^-3
4. C_V^el=R 0.0001095 T/K; C_V^ph= R 3.153×10^-6 (T/K)³; C_V^el = C_V^ph at T=5.89 K