# Solid State

## NCERT Solution (part 3)

Question – 1.11 – Silver crystallizes in fcc lattice. If edge length of the cell is 4.07 xx 10– 8 cm and density is 10.5 g cm– 3. Calculate the atomic mass of silver.

Answer: Given, edge length (a) =4.07xx10^(-8)

Density (d) = 10.5 g cm-3

For fcc number of atoms per unit cell (z) = 4

We know that Atomic Mass (M) =(da^3N_A)/z

Thus, M = (10.5gcm^3xx4.07xx10^(-8)xx6.033xx10^(23)text(mol)^(-1))/4

Or, M = (10.5gxx67.419xx10^(-24)xx6.022xx10^(23)text(mol)^(-1))/4

Or, M = (10.5gxx67.419xx10^(-1)xx6.022text(mol)^(-1))/4

Or, M =(426.2970789)/4g\text(mol)^(-1)

Or, M=106.574g\text(mol)^(-1)=107 g\text(mol)^(-1)

Question – 1.12 - A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?

Answer: Given, Atoms of Q are at the corners of the cube and P at the body-centre.

So, number of atoms Q in one unit cell =8xx1/8=1

Number of atoms of P in one unit cell = 1

So, ratio of P and Q atoms = P : Q = 1 : 1

So, formula of given compound = PQ

Since it is bcc

Hence, coordination number of P and Q = 9

Question – 1.13 - Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93 u.

Answer: Given , density (d) = 8.55 g cm-3

Atomic Mass (M) = 93 u = 93 g mol-1

We know, Avogadro Number NA = 6.022 xx 1023mol-1

Since, given lattice is bcc

Hence, number of atoms per unit cell (z) = 2

We know that, d=(zM)/(a^3N_A)

Or, 8.55 g cm-3 = (2xx93g\text(mol)^(-1))/(a^3xx6.022xx10^(23)text(mol)^(-1))

Or, a^3=(2xx93g)/(8.55g\cm^(-3)xx6.022xx10^(23))

Or, a^3=(186)/(51.4881xx10^(23)\cm^3

Or, a^3=3.6124xx10^(-23)cm^3

Or, a^3=36.124xx10^(-24)cm^3

Or, a=3.3057xx10^(-8)cm

Thus atomic radius of niobium = 14.31 nm

Question – 1.14 - If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

In the given figure, let an sphere having centre ‘O’ is fitted in the octahedral void.

As given, radius of the sphere fitted in the octahedral void = r

And radius of the atoms in close packing = R

Here, ∠AOD = 90°

In ΔAOD, DA^2=OA^2+OD^2

Or, (R+R)^2=(R+r)^2+(R+r)^2

Or, 4R^2=2(R+r)^2

Or, sqrt2R=R+r

Or, r=sqrt2R-R

Or, r=R(sqrt2-1)

Or, r=R(1.414-1)

Or, r=0.414R

This is the required relation between r and R.