having the lattice 5 parameter $2.9A^0$ and density 7.87 gm/cm2. Atomic weight of metal is 55.85. Avogadro number is $6.023 × 10^{23} /gm mole$. - Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: May 2015

Given:

$a = 2.9 A = 2.9 × 10^{-10}m$

$Density (ρ ) = 7.87 gm/cm2 = 7.87 × 10^{-1}m$

M=55.85

$N_a = 6.023 × 10^{23} gm/mole = 6.023 × 10^{26} kg/Nm$

To find: n = ?

soln: $ρ = \frac{Mn}{N_Aa^3}$

$n = \frac{ρ × N_A × a^3}{M} $

$n = \frac{7.87 × 10^{-1} × 6.023 × 10^{26} × (2.9 × 10^{-10})^3}{55.85}$

$n = 20.699 × 10^{-5}$

$n = 0.207 × 10^{-3}$