The Solid State

• Question 161

## A metal (at. mass = 50) has a bcc crystal structure. The density of the metal is 5.96 g cm–3. Find the volume of its unit cell?

Solution
Solution:
We have given that
Atomic mass of the metal = 50g
bcc unit cell, Z = 2
Density of metal = 5.96 g/cm
3
Therefore, volume of the unit,

Question 162

## Calculate the density of silver which crystallizes in a face centred cubic lattice with unit cell length 0.4086 nm (At. mass of Ag = 107.88)

Solution

Solution:
We have given that

Unit cell length,
a = 0.40806 nm  = 4.086 x 10-10 m

If fcc lattice the number of atoms per unit cell,
i.e. Z = 4

M for Ag = 107.88 g mol-1

Density of Ag, d =

Thus the density of silver is 1.051 x 104 kgm-3

Question 163

## An element crystallizes in a structure having a fcc unit cell of an edge 200 pm. Calculate its density if 200 g of this element contains 24 x 1023 atoms.

Solution

Solution:
We have given that
Edge length of the unit cell
= 200 pm = 200 x 10–10 cm

Vol. of the unit cell
= (200 x 10–10 cm)= 8 x 10–24 cm3
Therefore, volume of the substance
= Vol. of unit cell x Vol. of 1 unit cell.

Since the element has a fcc unit cell, number of atoms per unit cell = 4.
Total number of atoms = Atoms/unit cell x number of unit cells.
= 24 x 1023atoms = 4

atoms/unit cell x No. of unit cells.

∴   No. of unit cells =

∴  Volume of the substance

Now,

Question 164

## An element X with an atomic mass of 60 g/mol has density of 6.23 g/cm–3. If the edge length of its cubic unit cell is 400 pm, identify the type of cubic unit cell. Calculate the radius of an atom of this element.

Solution
Solution:
we have given that
Density, d = 6.23 g/cm3
a = 400 pm
M = 60g/mol-1

Volume  = (a3)

$\mathrm{Z}=\frac{\mathrm{d}×{\mathrm{a}}^{3}×{\mathrm{N}}_{\mathrm{A}}}{\mathrm{M}}$

Hence, the type of cubic unit cell is FCC

So radius of element is 141.4 pm