Question

In face-centered cubic (fcc) crystal lattice, the edge length of the unit cell is \[{\text{400 pm}}\]. Find the diameter of the greatest sphere which can be fitted into the interstitial void without distortion of the lattice.

Answer 1

Hint:Let us use the relationship \[{\text{a}}\sqrt 2 = 4r\] between the edge length and atomic radius to calculate the atomic radius. Use the relationship \[a = 2\left( {r + R} \right)\] between the edge length, the atomic radius and the radius of an atom present in the octahedral void.

Complete step-by-step solution:
Let ‘a’ represent the unit cell edge length. The edge length of the unit cell is \[{\text{400 pm}}\] .
\[{\text{a = 400 pm}}\]
Let ‘r’ be the radius of the atom that is present at either corner or face center.
We can write the following relationship between the edge length and the atomic radius for a face-centered cubic (fcc) crystal lattice,
\[{\text{a}}\sqrt 2 = 4r\]
Rearrange above expression in terms of the atomic radius:
\[r = \dfrac{{{\text{a}}\sqrt 2 }}{4}\]
Substitute \[{\text{400 pm}}\] for ‘a’ in the above expression
\[\Rightarrow r = \dfrac{{{\text{400 pm}} \times \sqrt 2 }}{4}\]
\[\Rightarrow r = {\text{100 pm}} \times 1.414\]
\[\Rightarrow r = {\text{141}}{\text{.4 pm}}\]
Hence, the radius of the atom that is present at either corner or face center is \[{\text{141}}{\text{.4 pm}}\] .
Let ‘R’ be the radius of an atom present in the octahedral void in face-centered cubic (fcc) crystal lattice. The following relationship holds true.
\[a = 2\left( {r + R} \right)\]
On Rearrange the above expression
\[\Rightarrow a = 2r + 2R\]
\[\Rightarrow a - 2r = 2R\] … …(1)
But \[D = 2R\] … …(2)
Here, D is the diameter of the sphere.
On Substituting equation (2) in equation (1).
\[D = a - 2r\]… …(3)
Let us Substitute \[{\text{400 pm}}\] for ‘a’ and \[{\text{141}}{\text{.4 pm}}\]for ‘r’ in the equation (3).
\[\Rightarrow D = {\text{400 pm}} - 2\left( {{\text{141}}{\text{.4 pm}}} \right)\]
\[\Rightarrow D = {\text{400 pm}} - {\text{282}}{\text{.8 pm}}\]
\[\Rightarrow D = {\text{117}}{\text{.2 pm}}\]
Hence, the diameter of the greatest sphere which can be fitted into the interstitial void without distortion of the lattice is \[{\text{117}}{\text{.2 pm}}\].

Note:In face-centered cubic (fcc) crystal lattice, each corner is occupied by an atom and each face center is occupied by an atom. In an octahedral void, the atom present has coordination number of six.