Question

In a cubic lattice each edge of the unit cell is 400pm.Atomic weight of the element is 60 and its density is 6.25g/c.c.Avogadro number =6.023×1023.What will be the crystal lattice.

Answer 1

Let the number of atoms in a unit cell =x
mass of x-atoms i.e one unit cell = (60*X)/6 * 10^23
Volume of the unit cell = (edge length)3
⇒(400*10^−12*100)^3
⇒(400*10^−10)^3
⇒(4*10^−8cm)^3 = 64*10^−24cm3
Density = 6.25 = Mass of unit cell/Volume of unit cell
So, 6.25 = (60*X)/6.023*10^23 *64*10^−24
x = (6.25*6*64*10^−1)/60=4
Since the unit cell contains 4 atoms,so it is face centred cubic unit cell.

Answer 2

Answer:

Explanation:

Let the number of atoms in a unit cell =x

mass of x-atoms i.e one unit cell = (60*X)/6 * 10^23

Volume of the unit cell = (edge length)3

⇒(400*10^−12*100)^3

⇒(400*10^−10)^3

⇒(4*10^−8cm)^3 = 64*10^−24cm3

Density = 6.25 = Mass of unit cell/Volume of unit cell

So, 6.25 = (60*X)/6.023*10^23 *64*10^−24

x = (6.25*6*64*10^−1)/60=4

Since the unit cell contains 4 atoms,so it is face centred cubic unit cell.