Question

An atom has fcc crystal whose density is 10g m^-3 and cell edge is 100 pm. How many atoms are present in its 100g?

Answer 1

Total No. of atoms = \(\frac{Z \times M}{a^3 d}\)

where,

Z = No. of atoms in a unit cell 

M =  Mass of metal

a = Edge of unit cell

d = Density of metal

Here, \(Z_{fcc} = 8 \times \frac 18 + 6 \times \frac 12 = 4\)

Substituting the values in the above relation,

No. of atoms = \(\frac{4 \times 100 g}{(100 \times 10^{-12}m)^3 \times 10 g m^{-3}}\) = = 4 × 10³¹ atoms