Question

A hypothetical metal has the BCC crystal structure, a density of 7.24 g/cm3, and an atomic weight of 48.9 gmol. a) Sketch a BCC unit. b) cell determine the atomic radius of this metal. c) calculate the side length of the BCC unit cell

Answer 1

Answer :

(a) The sketch of BCC unit cell is shown below.

(b) The atomic radius of this metal is, $1.221\times 10^{-8}cm$

(c) The side length of the BCC unit cell is, $2.82\times 10^{-8}cm$

Solution : Given,

Number of atom in unit cell of BCC (Z) = 2

Atomic mass (M) = 48.9 g/mole

Density = $7.24g/cm^3$

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

First we have to calculate the edge length of unit cell.

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$      .............(1)

where,

$\rho$ = density  = $7.24g/cm^3$

Z = number of atom in unit cell  = 2

M = atomic mass  = 48.9 g/mole

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

a = edge length of unit cell  or the side length of the BCC unit cell =?

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

$7.24g/cm^3=\frac{2\times (48.9g/mol)}{(6.022\times 10^{23}mol^{-1}) \times (a)^3}$

$a=2.82\times 10^{-8}cm$

The edge length of unit cell is, $2.82\times 10^{-8}cm$

Now we have to determine the atomic radius of this metal.

Formula used :

$a=\frac{4r}{\sqrt{3}}$

where,

a = edge length of unit cell

r = nearest neighbor distance  or atomic radius of metal

Now put all the given values in this formula, we get :

$2.82\times 10^{-8}cm=\frac{4\times r}{\sqrt{3}}$

$r=1.221\times 10^{-8}cm$

The atomic radius of this metal is, $1.221\times 10^{-8}cm$