# Free-electron Density.

1. Jan 6, 2009

### hhhmortal

1. The problem statement, all variables and given/known data

The movovalent metal Na (sodium) has the body-centred cubic structure with a unit cell side length of 4.23 x 10^-10 m

What is the free electron density of sodium?

2. Relevant equations

3. The attempt at a solution

I used:

n = (6.02 x 10^23)(density of sodium) / (atomic mass) = 2.54 x 10^28 m^-3

But isnt this the number of atoms per unit volume not free electron density?

Thanks.

2. Jan 6, 2009

### chaoseverlasting

Its given that the structure is a BCC type unit cell. The total number of atoms present in a BCC are 2. You are given the dimensions of this unit cell. Since its a cubic cell, you can find the volume enclosed by it.

Each cell contains two atoms, how many cells do you need to get Na (Avagardo Number) atoms?

Once you have that, you know the atomic weight of Na sodium atoms. This weight is equal to the weight of the protons+neutrons+electrons. You know the weight of one electron. You know how many free electrons are present in one sodium atom (valency). Hence you know how many electrons are present in Na sodium atoms.

You also know the total volume enclosed by Na atoms (which you calculated in the beginning). So now you have the total number of free electrons present in a given volume, and can now find the free electron density.

3. Jan 6, 2009

### nasu

You can do it an a simpler way:
There are 2 free electrons per a volume of (4.23x10^-10)^3.
n=2/(4.23x10^-10)^3 =2.6x10^28

4. Jan 6, 2009

### hhhmortal

Hi, I dont understand how I could get 'n' from this. I assumed that since its a monovalent metal it only has one valence electron therefore free electron density of sodium = density of sodium atoms. Hence I used the formula above to get the free electron density.

5. Aug 11, 2010

### mikehuihui

you are right