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(B) In a $bcc$ (body-centered cubic) lattice,the atoms touch each other along the body diagonal.The length of the body diagonal is $\sqrt{3}a$,where $

Vedclass · Accepted Answer

$a\frac{\sqrt{3}}{2}$

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(B) In a $bcc$ (body-centered cubic) lattice,the atoms touch each other along the body diagonal.The length of the body diagonal is $\sqrt{3}a$,where $a$ is the edge length of the unit cell.Since the body diagonal consists of two atomic radii $(r)$ from the corner atoms and the full diameter $(2r)$ of the central atom,we have $4r = \sqrt{3}a$,which gives the atomic radius $r = \frac{\sqrt{3}a}{4}$.The nearest distance $(d)$ between two atoms is the distance between a corner atom and the central atom,which is $2r$.Therefore,$d = 2 	imes \left( \frac{\sqrt{3}a}{4} ight) = \frac{\sqrt{3}a}{2}$.

The nearest distance between two atoms in case of a $bcc$ lattice is equal to

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