What is the correct equation for calculating free-electron density in sodium?

[hhhmortal]

Homework Statement

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?

Homework Equations

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 isn't this the number of atoms per unit volume not free electron density?


Thanks.

 

[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 N_a (Avagardo Number) atoms?

Once you have that, you know the atomic weight of N_a 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 N_a sodium atoms.

You also know the total volume enclosed by N_a 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.

 

[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

 

[hhhmortal]

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 N_a (Avagardo Number) atoms?

Once you have that, you know the atomic weight of N_a 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 N_a sodium atoms.

You also know the total volume enclosed by N_a 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.

Hi, I don't 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.

 

[mikehuihui]

you are right