Question

Molybdenum forms body-centered cubic crystals and at 20 oC the density is 10.3 g/mL. Calculate the distance between the centers of the nearest molybdenum atoms.

Answer 1

Answer:

ρ=

N

A

×a

3

Z×M

A

Z=2 (for BCC) , ρ=10.3g/cm

3

M

A

=95.94 gmol

−1

10.3=

6.023×10

23

×a

3

2×95.94

⇒a

3

=31.14×10

−24

cm

3

⇒a=3.146

A

˚

.

Answer 2

Answer:

The distance between the centers of the nearest molybdenum atoms is equal to 2.72Å .

Explanation:

The distance (d) between the centers of the nearest atoms in BCC unit cell will be given by:

$d=\frac{\sqrt{3} a}{2}$                                                                         ..................(1)

First we have to find, the edge length of unit cell:

Given, the density of the BCC unit cell, ρ = 10.3g/ml

The number of atoms in BCC unit cell of Mo, Z = 2

Molecular mass of molybdenum, M = 95.94 g/mol

We know the formula for density;

$\rho = \frac{Z*M}{ N_{A} * a^3}$                                                                      ..................(2)

Substitute the value of ρ, M, N and Z in equation (2);

$10.3=\frac{(2)*(95.94)}{(6.023*10^{23})*(a^3)}$

$a^3=31.14*10^{-24}cm^3$

$a=3.146$ Å

The nearest distance between two Mo atoms from equation (1);

$d=\frac{\sqrt{3}\;*(3.146) }{2}$

$d=2.72$ Å

Therefore, the nearest distance between two Mo atoms will be equal to 2.72Å .