The discussion revolves around calculating the free-electron density in sodium, which is a body-centered cubic (BCC) metal with a specified unit cell side length. Participants are exploring the relationship between atomic density and free electron density in the context of sodium's atomic structure.
There are multiple interpretations of how to calculate free electron density, with some participants providing methods based on the BCC structure and others questioning the assumptions made regarding valence electrons. A few participants have offered calculations, but consensus on the approach has not been reached.
Participants are navigating the complexities of atomic structure, valency, and the definitions of electron density versus atomic density. The discussion is influenced by the specifics of sodium's properties and the BCC unit cell configuration.
chaoseverlasting said: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.