Question 13

Solution

(ii) Face centered unit cell |
End-centered unit cell |

It contains one particle present at the centre of its each face, besides the ones that are at its corners. |
In this unit cell, one constituent particle is present at the centre of any two opposite faces besides the ones present at its corners. |

Question 14

Solution

(i) Primitive cubic unit cell has atoms only at its corner. Each atom at a corner is shared between eight adjacent unit cells . four unit cells in the same layer and four unit cells of the upper (or lower) layer Therefore only 1/8th of an atom (molecule or ion) actually belongs to a particular unit cell.

since each cubic unit cell has 8 atoms on its corners, the total number of atoms in one unit cell is $\mathbf{8}\mathbf{\times}\mathbf{1}\mathbf{/}\mathbf{8}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$ atom.

(ii) The atom at the body centre wholly belongs to the unit cell in which it is present. Thus in a body-centered cubic (bcc) unit cell:

(i) 8 corners${\times}{1}{/}{8}$ per corner atom =${\mathbf{8}}{\mathbf{\times}}{\mathbf{1}}{\mathbf{/}}{\mathbf{8}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{1}}{\mathbf{}}{\mathbf{atom}}$

(ii) 1 body centre atom =${1}{\times}{1}{=}{1}$

therefore total number of atom per unit cell =2 atom.

Question 15

Solution

The co-ordination number is 4.

Question 16

Solution

Number of closed packed particeles =$0.5\times 6.022\times {10}^{23}=3.011\times {10}^{23}$

therefore number of octhahedral voids = 3.011$\times {10}^{23}$

and number oftetrahedral voids =$2\times 3.011\times {10}^{23}=6.022\times {10}^{23}$

therefore total number of voids=$3.011\times {10}^{23}+6.022\times {10}^{23}=9.033\times {10}^{23}$.

therefore number of octhahedral voids = 3.011$\times {10}^{23}$

and number oftetrahedral voids =$2\times 3.011\times {10}^{23}=6.022\times {10}^{23}$

therefore total number of voids=$3.011\times {10}^{23}+6.022\times {10}^{23}=9.033\times {10}^{23}$.