Question

A given metal crystallizes out with a cubic structure having an edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom?
(a)- 80 pm
(b)- 108 pm
(c)- 40 pm
(d)- 127.6 pm

Answer 1

Hint: The number of atoms in the unit cell decides the type of structure of the solid. If the unit cell has 4 atoms then it is a face-centered cubic structure. The formula that can be used is $r=\dfrac{a}{2\sqrt{2}}$, where r is the radius of the atom and an edge length of the unit cell.

Complete answer:
Solids are particles that have structures called lattice and these lattices are made up of small cells known as the unit cell. All the atoms and molecules are arranged in the unit cell. The number of atoms or molecules in the unit cell decides the structure of the solid.
So, in the question, it is given that, in one unit cell there are four metal atoms. This means that it is a face-centered cubic structure because the unit cell has four atoms.
For the face-centered cubic structure, we can relate the radius of the atom to the edge of the unit cell as:
$r=\dfrac{a}{2\sqrt{2}}$
Where r is the radius of the atom and the edge length of the unit cell.
This formula can be simplified as:
$r=0.3535\text{ a}$
In the question the edge length is given 361 pm. So, applying this in the formula we get:
$r=0.3535 \times \text{ 361}$
$r=127.6$
So, the radius of the atom is 127.6 pm.

Therefore, the correct answer is an option (d)- 127.6 pm.

Note:
If the solid has simple structure then the formula is: $r=\dfrac{a}{2}$, and if the solid has body-centered structure then the formula is: $r=\dfrac{\sqrt{3}}{4}a$. In a simple structure, the number of atoms is 1 and in a body-centered structure, the number of atoms is 2.