Question

The number of unit cells present in 2 of gold (At. Wt. = 197) is 1.53 1021
The lattice of gold is
(a) Simple cubic
(b) FCC
(C) BCC
(d) HCP​

Answer 1

Answer:

Explanation:

∴ When number of atoms are 120.44197×1023 , number of unit cells are = 14×120.44197×1023 =30.11197×1023=1.53×1022 unit cells.

Answer 2

Answer:

Complete answer:

Firstly, let us discuss face centred cubic lattice (fcc).

A face-centred unit cell contains atoms at all the corners and at the centre of all the faces of the cubic unit cell. There are a total 8 corners atoms and 6 face-centre atoms in the cubic unit cell.

Now, contribution of an atom located at the corner in a fcc unit cell = 18 per unit cell.

Contribution of an atom located at the face-centre in a fcc unit cell = 12 per unit cell.

Thus, total number of atoms in the fcc unit cell = 8 corner atoms ×18 atom per unit cell + 6 face - centred atoms × 12 atom per unit cell

∴ Total number of atoms in a unit cell of fcc lattice = 1+3=4 atoms

Given that, atomic mass of gold = 197g.

Number of atoms in 1mole of gold i.e., 197 g of gold = NA=6.022×1023 atoms

∴ Number of atoms in 20 g of gold = 6.022×1023197×20=120.44197×1023 atoms

And, we calculated above that there are 4 atoms in 1 unit cell of fcc lattice.

∴ When number of atoms are 120.44197×1023 , number of unit cells are = 14×120.44197×1023 =30.11197×1023=1.53×1022 unit cells.

Hence, option (3) is correct.