Question

A crystal structure has BCC structure and its lattice constant is $3.6{A^o}$ then what is its atomic radius?
A. $3.6{A^o}$
B. $1.8{A^o}$
C. $1.27{A^o}$
D. $1.56{A^o}$

Answer 1

Find the relation between the lattice constant and the atomic radius for a BCC structure. This can be done by using the packing factor and the number of atoms present in one crystal with BCC structure. In a BCC structure there is one atom on every single corner and one in the center.

Complete answer:
In a BCC structure, there are eight atoms at eight different corners of the crystal and there is one single atom in the middle. These eight atoms on the corners are shared by other crystals as well so there is only $1/8$ th of an atom in the corner.
We know that the lattice constant is the length of the side of the crystal and is given as $a$
Now the diagonal of this cubic crystal structure would be $\sqrt 3 a$
In BCC structure, the diagonal passes through one central atom and two corner atoms and therefore the diagonal should also be equal to $4r$ considering $r$ as the radius of our atom.
Equating these two: $\sqrt 3 a = 4r$ and $r = \dfrac{{\sqrt 3 a}}{4}$
Now we substitute the given value of lattice constant $r = \dfrac{{\sqrt 3 \times 3.6}}{4}{A^\circ }$
By solving, we get $1.5588{A^\circ }$ which is approximately $1.56{A^\circ }$

**So the correct answer is option D. $1.56{A^o}$

Note:**
The relation between the lattice constant and atomic radius is different for different structural arrangements. It is important to remember the atomic arrangement for different structures to be able to differentiate between them and calculate the correct relation between lattice constant and atomic radius.