Question

A bcc lattice is made up of hollow spheres of B. Spheres of solid A are present in hollow spheres of B. The radius of A is half of the radius of B. The ratio of total volume of spheres of B unoccupied by A in a unit cell and volume of unit cell is $A\times \dfrac { \pi \sqrt { 3 } }{ 64 }$. Find the value of A.

Answer 1

In this question, a BCC unit cell is given. For a BCC unit cell, the total number of the spheres that belong exclusively to the unit cell is equal to 2 since $\dfrac { 1 }{ 8 }$ of the total number of spheres present at the corners along with the sphere present at the body centre belong exclusively to the unit cell. The total number of spheres present at the corners is 8.

Complete answer:
In a BCC unit cell, the spheres are present at the corners as well as at the centre. Due to the sphere present at the centre, the spheres present at the corner do not touch each other. But, the sphere present at the body centre touches all the spheres present at the corners of the unit cell.
Let the radius of the spheres B be R and the radius of sphere A be r.
Therefore $r=\dfrac { R }{ 2 }$
The total number of spheres B present in the unit cell will be 2. Similarly the total number of spheres A present in the unit cell will also be 2.
In order to find out the volume of spheres B not occupied by the spheres A we need to subtract the total volume of the spheres A from the total volume of spheres B.

Volume of spheres of B un-occupied by A=$\left( 2\times \dfrac { 4 }{ 3 } \pi { R }^{ 3 } \right) -\left( 2\times \dfrac { 4 }{ 3 } \pi { r }^{ 3 } \right)$

Since $r=\dfrac { R }{ 2 }$, Therefore
Volume of spheres of B un-occupied by A=$\left( 2\times \dfrac { 4 }{ 3 } \pi { R }^{ 3 } \right) -\left( 2\times \dfrac { 4 }{ 3 } \pi \dfrac { { R }^{ 3 } }{ 8 } \right)$

Volume of squares of B un-occupied by A=$2\times \dfrac { 4 }{ 3 } \pi { R }^{ 3 }\times \dfrac { 7 }{ 8 }$
This is a BCC unit cell, therefore, we will use the relation: $a\sqrt { 3 } = 4R$
Where ‘a’ is the side of the unit cell.

Therefore, $a=\dfrac { 4R }{ \sqrt { 3 } }$

In the question it is stated that $\dfrac { Volume\quad of\quad spheres\quad B\quad unoccupied }{ Total\quad volume\quad of\quad the\quad unit\quad cell } =\dfrac { \pi \sqrt { 3 } \times A }{ 64 }$

Therefore
$\dfrac { \dfrac { 8\times \pi { R }^{ 3 }\times 7 }{ 3\times 8 } }{ \dfrac { { 4 }^{ 3 }{ R }^{ 3 } }{ { \sqrt { 3 } }^{ 3 } } } =\dfrac { \pi \sqrt { 3 } \times A }{ 64 }$
So, $\dfrac { 7\times \sqrt { 3 } \pi }{ 64 } =\dfrac { \pi \sqrt { 3 } \times A }{ 64 }$
Therefore A=7.

Hence the correct answer is 7.

Note: In a BCC unit cell, the spheres present at the corners do not touch each other. Therefore the sum of the radius of the sphere is not equal to the side of the unit cell. But, the sphere present at the body centre touches all the spheres present at the corners. Therefore the length of the body diagonal will be equal to four times the radius of the sphere present in the unit cell.