Question

The vacant space in BCC unit cell is:
(A) 23%
(B) 32%
(C) 26%
(D) 48%

Answer 1

BCC unit cell contains eight atoms at the corner of the cube and one atom at the center of the unit cell. We can find % vacant space by following the equation.
% Vacant space = 100 - % Volume occupied by atoms in unit cell

Complete answer:
Here, they have asked us to calculate the vacant or empty space in the BCC unit cell.
- BCC stands for body centered cubic lattice. It is one of the specific arrangements of atoms in crystalline state. In the BCC unit cell, eight atoms occupy the corners of the cube. There is one atom at the center of the cube as well. Now, we will find the total number of atoms present in the unit cell.
- We know that there are eight atoms at the corners of the cube but those atoms are shared by other eight unit cells that are in the vicinity. The atom at the center of the lattice is not shared by any other unit cell. So, we can say that the total number of atoms present in one BCC unit cell is $\dfrac{1}{8} \times 8 + 1 = 1 + 1 = 2$ atoms.
- Thus, we obtained that the total atoms present in one BCC unit cell are 2.
- Now, we will first find the packing efficiency by calculating the volume and then we will find the % vacant space.
- Suppose that the length of the unit cell is a. So, we need to find the radius of the atom in order to find volume.
For a face diagonal, we can write that its length d = $\sqrt {{{(a)}^2} + {{(a)}^2}} = \sqrt 2 a$
Now, to find the distance between two edges, we can write that its distance b = $\sqrt {{{(a)}^2} + {{(\sqrt 2 a)}^2}} = \sqrt 3 a$
Now, we know that this distance is equal to four times the radius. So, we can write that
$\sqrt 3 a = 4r$
Where r is the radius
$r = \dfrac{{\sqrt 3 }}{4}a$
Now, we can say that the volume of the 2 spheres (atoms), will be $2 \times \dfrac{4}{3}\pi {r^3}$
Now, the volume of the unit cell (cube) will be ${a^3} = \dfrac{4}{{\sqrt 3 }}{r^3}$
Now, we can write the packing efficiency for the BCC cell as
${\text{Packing efficiency = }}\dfrac{{{\text{Volume occupied by two spheres}} \times 100}}{{{\text{Volume of unit cell}}}}$
So,
${\text{Packing efficiency = }}\dfrac{{2 \times \dfrac{4}{{\sqrt 3 }}\pi {r^3} \times 100}}{{\dfrac{4}{{\sqrt 3 }}\pi {r^3}}} = \dfrac{{\dfrac{8}{3}\pi {r^3} \times 100}}{{\dfrac{{64}}{{(3\sqrt 3 ){r^3}}}}} = 68\%$
Thus, we can say that the % volume occupied by atoms in a unit cell is 68%.
So, % Vacant space = 100 - % volume occupied by atoms
% Vacant space = 100 – 68 = 32%

Therefore, the correct answer is (B).

Note:
Do not get confused with FCC lattice in which the atoms occupy the corners of the unit cell as well as the centers of the faces of the unit cell. Remember that in a BCC unit cell, the atoms at the corners of the cube are equally shared by eight similar unit cells.