Crystal Physics - Solved Problems

1. Copper has FCC structure and its lattice parameter is $3.6Å$ . Find the atomic radius.

Given data

Lattice parameter of copper $(a)=3.6Å$

Solution:

Atomic radius of copper

$\begin{aligned} ' \mathrm{r} & =\frac{a \sqrt{2}}{4} \\ & =\frac{3.6 \times 10^{-10} \times \sqrt{2}}{4} \end{aligned}$

= 1.273 x 10^-10m

r = 1.273 Å

2. Silver has FCC structure and its atomic radius is $1.414Å$ . Find the spacing of (110) plane.

Given data

atomic radius r = 1.414Å

r = 1.414 x 10^-10m.

Solution:-

For FCC, lattice constant $a=\frac{4 r}{\sqrt{2}}$

Therefore,

$\begin{aligned} & a=\frac{4 \times 1.414 \times 10^{-10}}{\sqrt{2}} \\ & a=4 \times 10^{-10} \mathrm{~m} \end{aligned}$

We know in the case of cubic system

$\left.\begin{array}{l}\text { The inplanar spacing between } \\ \text { planes of Miller indices }(h k l)\end{array}\right\} d=\frac{a}{\sqrt{h^2+k^2+\ell^2}}$

$\left.\begin{array}{l}\text { Therefore the spacing of the } \\ \text { plane with Miller indices }(110)\end{array}\right\} d=\frac{4 \times 10^{-10}}{\sqrt{1^2+1^2+0}}$

d = 2.828 x 10^-10 m

3. Iron is BCC with atomic radius 0.123 $Å$ . Find the lattice constant and also the volume of the unit cell.

Given data

r = 0.123 Å = 0.123 x 10^-10m

Solution:-

For BCC Lattice,

$\begin{aligned} & a=\frac{4 r}{\sqrt{3}} \\ & =\frac{4 \times 0.123 \times 10^{-10}}{\sqrt{3}} \end{aligned}$

$a=0.284 \times 10^{-10} \mathrm{~m}$

∴ Volume of the unit cell $(\mathrm{V})=a^3=\left(0.284 \times 10^{-10}\right)^3$

=2.2906×10^-32m³

4. The interplanar distance between the planes of a crystal is 2.9 $Å$ . It is found that first order Bragg’s reflection occurs at an angle of 9.5° What is the wavelength of x- rays?

Given data

d = 2.9

Å

= 2.9×10^-10m

θ = 9.5°

n=1

Solution:-

From Bragg’s law nλ = 2d sinθ

Therefore,

$\begin{aligned} & \lambda=\frac{2 d \sin \theta}{n}=\frac{2 \times 2.9 \times 10^{-10} \times \sin 9.5^{\circ}}{1} \\ \end{aligned}$

λ=0.957 × 10^-10m

λ=0.957 Å

5. The Bragg’s angle corresponding to the first order reflection from (110) plane is a crystal is $32°$ When X- rays of wavelength of $1.7Å$ is used. Find the interatomic spacing in a cubic lattice.

Given data

(hkl) = (110)

Therefore

d₁₁₀ = a/√2

λ = 1.7Å =1.7 × 10^-10m

θ = 32°

Also n = 1

Solution:- For a Cubic lattice, the lattice constant ‘a’, and interplanar spacing ‘a’ for a plane of Miller indice ( $hkl)$ is given by

$d_{h k l}=\frac{a}{\left(h^2+k^2+\ell^2\right)^{1 / 2}}$

Subsituting the above values in Bragg’s Law,

2d sinθ = nλ

$\begin{gathered} 2 d \sin \theta=n \lambda \\ 2 \times \frac{a}{\sqrt{2}} \sin 32^{\circ}=1.7 \times 10^{-10} \end{gathered}$

Therefore

$\begin{aligned} & a=\frac{\sqrt{2} \times 1.7 \times 10^{-10}}{2 \sin 32} \\ & a=2.268 \times 10^{-10} \mathrm{~m} \end{aligned}$

6. Copper has FCC structure and atomic radius $1.278 A°$ . The atomic weight of copper is 63.54 . Avagadro’s number is $6.023\times10 26$ $kg$ . mole ^$-1$. Find the lattice parameter and density of copper.

Given data

atomic radius r = 1.278Å = 1.278 x 10^-10m

Atomic weight of copper M = 63.54

Avagadro’s number N = 6.023 x 10²⁶ kg.mole^-1

Solution :-

In FCC No of atoms per unit cell $n = 4$

Lattice constant $\begin{aligned} & a=\frac{4 r}{\sqrt{2}} \\ & a=\frac{4 \times 1.278 \times 10^{-10}}{\sqrt{2}} \\ & a=3.615 \times 10^{-10} \mathrm{~m} \end{aligned}$

We know the density of copper

$\begin{aligned} & \rho=\frac{n M}{N a^3} \\ & =\frac{4 \times 63.54}{\left(3.615 \times 10^{-10}\right)^3 \times 6.023 \times 10^{26}} \\ & =\frac{254.16}{0.0285} \\ & \rho=8.932 \times 10^3 \mathrm{Kgm}^{-3} \end{aligned}$

7. An element has HCP structure. If the radius ‘r’ of the atom is $1.605 A°$ . Find the volume of the unit cell.

Given data

r = 1.605A° = 1.605 × 10^-10m.

Solution :-

For HCP structure,

$\begin{aligned} & \frac{\mathrm{c}}{\mathrm{a}}=\sqrt{\frac{8}{3}} \\ & \frac{\mathrm{c}}{2 \mathrm{r}}=\sqrt{\frac{8}{3}} \\ & \mathrm{c}=\sqrt{\frac{8}{3}} \times 2 \times 1.605 \times 10^{-10} \\ & \mathrm{c}=1.632 \times 2 \times 1.605 \times 10^{-10} \\ & \mathrm{c}=5.24 \times 10^{-10} \mathrm{~m} \end{aligned}$

we know,

$\frac{\mathrm{c}}{\mathrm{a}}=1.633$

∴ Lattice constant $(\mathrm{a}) \quad=\frac{\mathrm{c}}{1.633}$

$=\frac{5.24 \times 10^{-10}}{1.633}$

a = 3.21 × 10^-10m

∴ Volume of the unit cell $(\mathrm{v})=\frac{3 \sqrt{3} \mathrm{a}^2 \mathrm{c}}{2}$

$\begin{aligned} & =\frac{3 \times 1.732 \times\left(3.21 \times 10^{-10}\right)^2 \times 5.24 \times 10^{-10}}{2} \\ & =\frac{2.8 \times 10^{-28}}{2} \\ & v=1.4 \times 10^{-28} \mathrm{~m}^3 \end{aligned}$

8. Calculate the interplanar spacing for a (311) plane in a simple cubic lattice whose Lattice constant is $2.109\times10 -10 m$ .

Given data

(h k l) = (311)

a = 2.109 × 10^-10m

Solution :-

$\text { Interplanar distance }(\mathrm{d}) \quad=\frac{\mathrm{a}}{\sqrt{\mathrm{h}^2+\mathrm{k}^2+\mathrm{l}^2}}$

$\begin{aligned} & =\frac{2.109 \times 10^{-10}}{\sqrt{3^2+1^2+1^2}} \\ & =\frac{2.109 \times 10^{-10}}{3.317} \\ & =0.6358 \times 10^{-10} \\ & \mathrm{~d}=0.6358 \mathrm{~A}^{\circ} \end{aligned}$

9. Sodium is a BCC crystal. Its density is $9.6\times10 2 kgm -3$ and atomic weight is 23 . Calculate the lattice constant for sodium crystal.

Given data

e = 9.6 × 10² kgm^-3

M = 23

For BCC

Number of atoms per unit cell (n) = 2

Solution :-

Density $\begin{aligned} \mathrm{e} & =\frac{\mathrm{nM}}{\mathrm{Na}^3} \\ \Rightarrow \mathrm{a}^3 & =\frac{\mathrm{nM}}{\mathrm{Ne}} \\ \mathrm{a} & =\left[\frac{2 \times 23}{6.023 \times 10^{26} \times 9.6 \times 10^2}\right]^{\frac{1}{3}} \\ \mathrm{a} & =4.3 \times 10^{-10} \mathrm{~m} . \\ \mathrm{a} & =4.3 \mathrm{~A}^{\circ} \end{aligned}$