Question

Question:
For a metal that has the body-centered cubic crystal structure, calculate the atomic radius if the metal has a density of 7.25 g/cm3 and an atomic weight of 50.99 g/mol.

Atomic Radius
This problem is related to the concept of solid states. The smallest part of a crystal is the unit cell. This concept can explain the crystal structure of different compounds. The Atomic radius is proportional to the value of edge length of unit cell.

Answer 1

Answer:

The atomic radius is $1.238\times 10^{-8}cm$

.Explanation:

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

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

a = edge length of unit cell

We have :

M = 50.99 g/mol

a =?

Z = 2 ( BCC)

$\rho =7.25 g/cm^3$

On substituting all the given values , we will get the value of 'a'.

$7.25 g/cm3=\frac{2\times 50.99 g/mol}{6.022\times 10^{23} mol^{-1}\times (a)^{3}}$

$a = 2.86\times 10^{-8} cm$

The unit cell edge length is $4.3\times 10^{-8} m$

For BCC: radius is related to edge length by :

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

r = 0.433a

$r=0.433\times 2.86\times 10^{-8} cm=1.238\times 10^{-8} cm$

The atomic radius is $1.238\times 10^{-8} cm$.