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
Atomic radius (r) = ?
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.
Answer:
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.