STRUTTURA DELLA MATERIA 1 - 19 June 2017 - solutions
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Define x=ε/(kB T).
(a) CV = N kB x²exp(x)/(1+exp(x))²;
(b) CV ≈ (kB/N) x²exp(x)/(1+exp(x)/N)².
Good agreement for exp(x) >> N,
alias T << ε/(kB log(N)).
For x=23.209,
(a) CV = 6.192 10-25 J/K;
CV = 6.192 10-19 J/K
(b) CV = 6.191 10-25 J/K;
CV = 8.721 10-23 J/K;
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24350 eV (K);
3604 eV (LI);
3330 eV (LII);
3173 eV (LIII);
the K shell; 42.30
- 74.23 pm; 1.942 10-26 J/K; 1.381 10-23 J/K
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εF = -5.259 eV;
vF = 1.139×106 m/s;
m*=2.326×10-30 kg