If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

 

 

 

Question 1.10) Calculate the efficiency of packing in case of a metal crystal for (i) simple cubic (ii) body-centred cubi

Question 1.10) Calculate the efficiency of packing in case of a metal crystal for

(i) simple cubic

(ii) body-centred cubic

(iii) face-centred cubic (with the assumptions that atoms are touching each other).

Answer:

 

(i) Simple cubic

In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.

Let the edge length of the cube be ‘a’ and the radius of each particle be r.

So, we can write:

a = 2r

Now, volume of the cubic unit cell $=a^3$

$={\left(2r\right)}^3$

$=8r^3$

We know that the number of particles per unit cell is 1.

Therefore, volume of the occupied unit cell

$=\frac{4}{3}\pi r^3$

 

packing efficiency =$\frac{Volume of the one particle}{Volume of cubic unit cell}\times 100\%$

$\frac{{\displaystyle \frac{4}{3}}\pi r^3}{8r^3}\times 100\%$

$=\frac{1}{6}\pi \times 100\%$

$=\frac{1}{6}\mathit{\times }\frac{\mathit{22}}{\mathit{7}}\times 100\%$

=52.4%

 

(ii) Body-centred cubic

It can be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged.

From ΔFED

$b^2=a^2+a^2$

${\mathrm{b}}^2=2{\mathrm{a}}^2$

$b=\sqrt{2a}$

 

  from ΔAFD

${\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2$

${\mathrm{c}}^2={\mathrm{a}}^2+2{\mathrm{a}}^2$

${\mathrm{c}}^2=3{\mathrm{a}}^2$

$\mathrm{c}=\sqrt{3\mathrm{a}}$

 

Let the radius of the atom be r.

Length of the body diagonal

$c= 4\pi$

$\sqrt{3a}=4r$

$\mathrm{a}=\frac{4\mathrm{r}}{\sqrt{3}}$

$\mathrm{r}=\frac{\sqrt{3\mathrm{a}}}{4}$

$a^3={\left(\frac{4r}{\sqrt{3}}\right)}^3$

 

A body-centred cubic lattice contains 2 atoms.

So, volume of the occupied cubic lattice

$=2\mathrm{\pi }\frac{4}{3}{\mathrm{r}}^3$

$=\frac{8}{3}{\mathrm{\pi r}}^3$

Packing efficiency $\frac{\mathrm{Volume} \mathrm{occupied} \mathrm{by} \mathrm{two} \mathrm{spheres} \mathrm{in} \mathrm{the} \mathrm{unit} \mathrm{cell}}{\mathrm{Total} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{unit} \mathrm{cell}}\times 100\%$

$=\frac{{\displaystyle \frac{8}{3}}{\mathrm{\pi r}}^3}{{\left({\displaystyle \frac{4}{\sqrt{3}}}\mathrm{r}\right)}^3}\times 100\%$

$=\frac{{\displaystyle \frac{8}{3}}{\mathrm{\pi r}}^3}{{\left({\displaystyle \frac{64}{\sqrt[3]{3}}}{\mathrm{r}}^3\right)}^{}}\times 100\%$

 

= 68%

 

(iii) Face-centred cubic

Let the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b.

From ΔABC

$A{C}^2=B{C}^2+A{B}^2$

$b^2=a^2+a^2$

$b^2=2a^2$

$\mathrm{b}=\sqrt{2\mathrm{a}}$

 

Let r be the radius of the atom.

from the figure, it can be observed that

$\mathrm{b}=4\mathrm{r}$

$\sqrt{2\mathrm{a}}=4r$

$\mathrm{a}=2\sqrt{2\mathrm{r}}$

Volume of the cube

$a3={\left(\sqrt[2]{2r}\right)}^3$

We know that the number of atoms per unit cell is 4.

Volume of the occupied unit cell

$=4\pi \frac{4}{3}{r}^3$

Packing efficiency = $\frac{Volume occupied by four spheres in the unit cell}{Total volume of the unit cell}\times 100\%$

$=\frac{4\pi \frac{4}{3}{r}^3}{{\left(\sqrt[2]{2r}\right)}^3}\times 100\%$

$=\frac{\frac{16}{3}\mathrm{\pi }{r}^3}{\sqrt[16]{2r^3}}\times 100\%$

=74%

 

 

 

Question 1.7) How will you distinguish between the following pairs of terms

 

Question 1.7) How will you distinguish between the following pairs of terms:

(i) Hexagonal close-packing and cubic close-packing?

(ii) Crystal lattice and unit cell?

(iii) Tetrahedral void and octahedral void?

Answer :

(a) Cubic close packing:-

When a third layer is placed over the second layer in a manner that the octahedral voids are covered by the spheres, a layer different from the first

(A) and second

(B) is obtained.

If we continue packing in this manner we get the cubic close packing.Hexagonal close packing: When the third layer is placed over the second layer in a way that the tetrahedral voids are covered by the spheres, a 3D close packing is produced where spheres in each third or alternate layers are vertically aligned. If we continue packing in this order we get hexagonal close packing.

(b) Unit cell:-

It is the smallest 3D dimensional portion of a complete space lattice, which when repeated over and over again in different directions from the crystal lattice.

Crystal lattice :-

it is a regular orientation of particles of a crystal in a 3D space.

(c) Octahedral void :-

it is a void surrounded by 6 spheres.

Tetrahedral void :-

it is a void surrounded by 4 spheres.