91影视

From equation\(42.2\),the Fermi energy at absolute zero\({E_{F0}}\)as a function of electron concentration, the number of free electrons per unit volume,\(n\)is given by: \({E_{F0}} = \frac{{{3^{\frac{2}{3}}}{\pi ^{\frac{4}{3}}}{\hbar ^2}}}{{2m}}{n^{\frac{2}{3}}}\)

Given that the density of potassium is\[\rho = 851kg/{m^3}\]and the mass of a single potassium atom is\(m = 6.49 \times {10^{ - 26}}kg\).

First, we calculate the volume of one potassium atom as follows:

\(V = \frac{m}{\rho } = \frac{{6.49 \times {{10}^{ - 26}}}}{{851}} = 7.63 \times {10^{ - 29}}{m^3}\)

The reciprocal of this quantity is the number of atoms per unit volume,

\(\frac{N}{V} = \frac{1}{V} = \frac{1}{{7.63 \times {{10}^{ - 29}}}} = 1.31 \times {10^{28}}atom/{m^3}\)

Since each potassium atom contributes with one free electron to the lattice, theelectronconcentrationis:

\(\begin{aligned}{l}n = \left( {1.31 \times {{10}^{28}}atom/{m^3}} \right)\left( {1electron/atom} \right)\\n = 1.31 \times {10^{28}}electron/{m^3}\end{aligned}\)

Finally, we plug this value into the first equation, so we get theFermienergyof potassium at absolute temperature:

\({E_{F0}} = 3.25 \times {10^{ - 19}}J = 2.03eV\)