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

$\sqrt{2}\mathrm{a}$

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B

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

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C

$2a$

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D

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Solution

## The correct option is B $\frac{a}{\sqrt{2}}$Explanation for the correct option:(B)) $\frac{\mathbf{a}}{\sqrt{\mathbf{2}}}$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.It is represented as: $\frac{1}{2}×\mathrm{length}\mathrm{of}\mathrm{face}\mathrm{diagonal}$ ……………$\left(1\right)$Length of face diagonal for FCC is given by: $\sqrt{2}\mathrm{a}$ and the edge length of unit cell is $\text{'}\mathrm{a}\text{'}$.Putting the values for length of face diagonal and edge length in equation $\left(1\right)$ we get: $\frac{1}{2}×\sqrt{2}\mathrm{a}=\frac{\sqrt{2}\mathrm{a}}{2}=\frac{\mathrm{a}}{\sqrt{2}}$Explanation for incorrect options:(A) $\sqrt{\mathbf{2}}\mathbf{a}$For FCC the closest approach between two atoms in metallic crystal is half of the length of the face diagonal. $\sqrt{2}\mathrm{a}$ does not represent the half of the length of the face diagonal. (C) $\mathbf{2}\mathbf{a}$For FCC the closest approach between two atoms in metallic crystal is half of the length of the face diagonal. $2\mathrm{a}$ does not represent the half of the length of the face diagonal. (D) $\mathbf{2}\sqrt{\mathbf{2}}\mathbit{a}$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. Hence option B is correct, for FCC the closest approach between two atoms in metallic crystal will be $\frac{\mathrm{a}}{\sqrt{2}}$.

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