Question

A given metal crystallizes out with a cubic structure having an edge length of $361pm$. If there are four metal atoms in one unit cell, what is the radius of one atom?
A.$80pm$
B.$108pm$
C.$40pm$
D.$127.6pm$

Answer 1

To solve this numerical one must have the basic knowledge of the solid states. When the metal crystallizes out with a cubic structure we need to determine the lattice first. Once we come to know about the structure we can easily derive the relation between the radius and edge length of the atom.
Formula Used:
The relation between the radius and edge length for FCC.
$4r = \sqrt 2 a$
Where, $r$ is the radius of the atom,
$a$ is the edge length of the cubic structure.

Complete step by step answer:
First, we will try to determine the lattice of a cubic structure. For that, we will read the question properly and we have given in the question that there are four metal atoms in one unit cell. Therefore, $Z = 4$ where $Z$ represents the number of atoms in a unit cell. $Z = 4$ is for the face-Centered cubic lattice (fcc).
Now we have determined the lattice of a cubic structure that is Face centered cubic lattice. So, we will use the relation between radius and edge length for fcc. We have, $a = 361pm$ , we will substitute it in the relation of radius and edge length.
$4r = \sqrt 2 a$ $- \left( 1 \right)$
Now substitute $a = 361pm$, in the above equation. We get,
$\Rightarrow 4r = \sqrt 2 \times 361pm$
$\Rightarrow r = \dfrac{{\sqrt 2 \times 361}}{4}$
$\Rightarrow r = 127.6pm$
From the above calculation, we get the radius of one atom as $r = 127.6pm$.

Therefore, the correct option is (D).

Note:
We can derive the relation between the radius of atom and edge length of the lattice by considering a cubic FCC lattice with atoms at the faces and the vertex.
Due to the different arrangements of atoms in a lattice the radius of an atom is different for different lattices.