having lattice parameters 2.9 $A^0$ and density 7.87 $g/cm^3$. Atomic Weight of metal is 55.85, Avogadro number is $6.023 × 10^{23}/g-mole$. -

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: Dec 2012

Given:-

a = 2.9 $A^0$ = 2.9 × $10^-8$ cm

Atomic Weight, A = 55.85

Avogadro's number, $N_A = 6.023 ×10^{23}/g-mole$

ρ = 7.87 $g/cm^3$

To Find:-

Number of atoms per unit cell (n)

Solution:-

Using the Formula:-

$a^3 ρ = \frac{nA}{N_A}$

$∴(2.9 × 10^{-8})^3 × 7.87= \frac{n × 55.85}{6.023 × 10^{23}}$

$∴n = 2.069 ≈ 2$

As n = 2, the structure of the atom is BCC