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?
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 6.023×1026 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 2.109×10−10 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 9.6×102 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}
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