2020-05-22T01:39:52-04:00
Calculate the Fermi energy of sodium at 0 K if it has one free electron per
atom and density 970 kg/m3 and atomic weight 23.
1
2020-05-25T10:49:31-0400
n c = N ρ M = 6.02 ⋅ 1 0 26 ( 970 ) 23 = 2.5 ⋅ 1 0 28 m − 3 n_c=\frac{N\rho}{M}=\frac{6.02\cdot10^{26}(970)}{23}=2.5\cdot10^{28}\ m^{-3} n c = M Nρ = 23 6.02 ⋅ 1 0 26 ( 970 ) = 2.5 ⋅ 1 0 28 m − 3
E F ( 0 ) = h 2 2 m ( 3 n c 8 π ) 2 3 = ( 6.625 ⋅ 1 0 − 34 ) 2 2 ( 9.11 ⋅ 1 0 − 31 ) ( 3 ( 2.5 ⋅ 1 0 28 ) 8 π ) 2 3 E_F(0)=\frac{h^2}{2m}\left(\frac{3n_c}{8\pi}\right)^{\frac{2}{3}}=\frac{(6.625\cdot10^{-34})^2}{2(9.11\cdot10^{-31})}\left(\frac{3(2.5\cdot10^{28})}{8\pi}\right)^{\frac{2}{3}} E F ( 0 ) = 2 m h 2 ( 8 π 3 n c ) 3 2 = 2 ( 9.11 ⋅ 1 0 − 31 ) ( 6.625 ⋅ 1 0 − 34 ) 2 ( 8 π 3 ( 2.5 ⋅ 1 0 28 ) ) 3 2
E F ( 0 ) = 5.11 ⋅ 1 0 − 19 J = 3.1 e V E_F(0)=5.11\cdot10^{-19}J=3.1\ eV E F ( 0 ) = 5.11 ⋅ 1 0 − 19 J = 3.1 e V
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