STRUTTURA DELLA MATERIA 1 - 19 June 2017 - solutions

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  1. 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;
  2. 24350 eV (K); 3604 eV (LI); 3330 eV (LII); 3173 eV (LIII); the K shell; 42.30
  3. 74.23 pm; 1.942 10-26 J/K; 1.381 10-23 J/K
  4. εF = -5.259 eV; vF = 1.139×106 m/s; m*=2.326×10-30 kg
created: 12 Jul 2017
last modified: 27 Sep 2020
by Nicola Manini