Find the number of free electrons per one sodium atom at `T=0` if the Fermi level is equal to `E_(F)= 3.07eV` and the density of sodium is `0.97 g//cm^(3)`.

Explanation

We calculate the concentration `n` of electron in the `Na` metal from
`E_(max)=E_(F)=( ħ^(2))/(2m)(3 pi^(2)n)^(2//3)`
we get from `E_(F)= 3.07eV`
`n= 2.447xx10^(22) per c.c`
From this we get the number of electrons per one `Na` atom as
`(n)/(rho).(M)/(N_(A))`
where `rho=` density of `Na, M=` molar weight in gm of `Na,N_(A)=` Avagadro number we get
`0.963` electrons per one `Na` atom.