Which of the following relationship helps to calcuate the denstity of the crystal?

Options

A. `d=(a^(3)N_(A))/(ZM)`
B. `d=(ZM)/(a^(3)N_(A))`
C. `d=(Za^(3))/(MN_(A))`
D. `d=(a^(3)M)/(ZN_(A))`

Answer

2

Explanation

We can determine the structure and dimension of a unit cell by diffraction methods. Once we know the unit cell dimensions and the structure of a crystal ,some interesting calculations and can be made . for instance ,suppose we find that a metallic solid has a face entered cubic unit celll with all atoms at lattic points.and we determine the edge length of the unit cell.
From this unit cell dimension we can calculate the volume of the unit cell.Then knowing the density of the metal we can the unit cell .Then knowing the density of the metal ,we can calculate the mass of the atoms in the unit cell. Beacuse we know that the unit cell is face-centered cubic with all atoms at lattice points and that such a unit cell has four atoms ,we can obtain the mass of an individus atoms . This determination of the mass of an individual atom . This determination of the mass of a single atom was one of the first accurate determination of Avogardo's number . If a is taken is cm and `M` is molar mass in `g mol^(-1)` then
`d=(ZM)/(a^(3)N_(A))gcm^(-3)`
In terms of `SI` units `M` is in `kg mol^(-1)` a is in meteres then
`d=(ZM)/(a^(3)N_(A))kgcm^(-3)`