An element ‘A’ has the face-centred cubic structure with the edge length equal to \[361{\text{pm}}\] . The apparent radius of atom ‘A’ is:
A.$127.6{\text{pm}}$
B.$180.5{\text{pm}}$
C.$160.5{\text{pm}}$
D.$64{\text{pm}}$

Hint: In the face-centred cubic structure, there are 4 unit cells. $\dfrac{1}{8}$ of 8 atoms from the 8 corners of the cube and $\dfrac{1}{2}$ of 6 atoms from the 6 sides of the cubes.
Formula Used: The radius of the atom,
$r = \dfrac{1}{{2\sqrt 2 }}a$
where a is the length of the cube.

Complete step by step solution:
Let the unit cell edge of a face-centred cubic structure = ‘a’ and length the length of the diagonal = b. if we consider the surface. lf any cube, it has 4 vertices so we can consider the surface to be composed of two triangles inverted over each other. From this consideration, let the diagonal of the surface of the cube be the hypotenuse of the triangle and hence, from Pythagoras theorem
\[{\left( {{\text{the length of the hypotenuse}}} \right)^2} = {\text{ }}{\left( {{\text{the length of the base}}} \right)^2} + {\text{ }}{\left( {{\text{the length of the height}}} \right)^2}\]
For a cube, both the height and the base of the triangle will be equal to the edge of the cube = a.
Therefore, the diagonal = sq. rt. of the hypotenuse = $\sqrt 2 a$.
If the radius of one atom/ion of the unit cell = r, then for the diagonal of the cube we have,
$4r = \sqrt 2 a$
Or,
$r = \dfrac{1}{{2\sqrt 2 }}a$
Therefore, the radius of the atom ‘A’:
$r = \dfrac{1}{{2\sqrt 2 }} \times 361 = 127.6{\text{ pm}}$

So, the correct answer is option A.

Note:
A face centred cubic unit cell consists of atoms at all the corners and at all the centres of all the faces of a cube. It has been seen that atoms arranged in this manner have the most efficient packing and occupy the highest amount of space in the lattice.