As we learned in class if a material’s crystal structure is known a theoretical density, ????, can be computed from a tiny fundamental unit using the formula ???? = ???????? ????c???????? Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its theoretical density with its experimentally measured density of 7.87 g/cm^3.

Respuesta :

Answer : Yes, theoretical density can be computed from a tiny fundamental unit using the formula [tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex].

Explanation :

Nearest neighbor distance, r = [tex]0.124nm=1.24\times 10^{-8}cm[/tex] [tex](1nm=10^{-7}cm)[/tex]

Atomic mass (M) = 55.85 g/mol

Avogadro's number [tex](N_{A})=6.022\times 10^{23} mol^{-1}[/tex]

For BCC = Z = 2

Given density = [tex]7.87g/cm^3[/tex]

First we have to calculate the cubing of edge length of unit cell for BCC crystal lattice.

For BCC lattice : [tex]a^3=(\frac{4r}{\sqrt{3}})^3=(\frac{4\times 1.24\times 10^{-8}cm}{\sqrt{3}})^3=2.35\times 10^{-23}cm^3[/tex]

Now we have to calculate the density of unit cell for BCC crystal lattice.

Formula used :  

[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex] .............(1)

where,

[tex]\rho[/tex] = density

Z = number of atom in unit cell (for BCC = 2)

M = atomic mass

[tex](N_{A})[/tex] = Avogadro's number  

a = edge length of unit cell

Now put all the values in above formula (1), we get

[tex]\rho=\frac{2\times (55.85g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (2.35\times 10^{-23}Cm^3)}=7.89g/Cm^{3}[/tex]

From this information we conclude that, the given density is approximately equal to the given density.

Yes, theoretical density can be computed from a tiny fundamental unit using the formula [tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex].