Question

A metal crystallizes in a face centered cubic structure. If the edge length of its unit cell is $\text{'}\mathrm{a}\text{'}$, the closest approach between two atoms in metallic crystal will be:

• A. $\sqrt{2}\mathrm{a}$
• B. $\frac{a}{\sqrt{2}}$
• C. $2a$
• D.

Answer 1

The correct option is B

$\frac{a}{\sqrt{2}}$

Explanation for the correct option:

(B)) $\frac{\mathbf{a}}{\sqrt{\mathbf{2}}}$

1. For face centered cubic structure, the closest approach between two atoms in metallic crystal will be equal to half of the length of the face diagonal.
2. It is represented as: $\frac{1}{2}\times \mathrm{length}\mathrm{of}\mathrm{face}\mathrm{diagonal}$ ……………$\left(1\right)$
3. Length of face diagonal for FCC is given by: $\sqrt{2}\mathrm{a}$ and the edge length of unit cell is $\text{'}\mathrm{a}\text{'}$.
4. Putting the values for length of face diagonal and edge length in equation $\left(1\right)$ we get: $\frac{1}{2}\times \sqrt{2}\mathrm{a}=\frac{\sqrt{2}\mathrm{a}}{2}=\frac{\mathrm{a}}{\sqrt{2}}$

Explanation for incorrect options:

(A) $\sqrt{\mathbf{2}}\mathbf{a}$

1. For FCC the closest approach between two atoms in metallic crystal is half of the length of the face diagonal.
2. $\sqrt{2}\mathrm{a}$ does not represent the half of the length of the face diagonal.

(C) $\mathbf{2}\mathbf{a}$

1. For FCC the closest approach between two atoms in metallic crystal is half of the length of the face diagonal.
2. $2\mathrm{a}$ does not represent the half of the length of the face diagonal.

(D) $\mathbf{2}\sqrt{\mathbf{2}}\mathit{a}$

1. For FCC the closest approach between two atoms in metallic crystal is half of the length of the face diagonal.
2. $2\sqrt{2}\mathrm{a}$ does not represent the half of the length of the face diagonal.

Hence option B is correct, for FCC the closest approach between two atoms in metallic crystal will be $\frac{\mathrm{a}}{\sqrt{2}}$.