Question

If the distance between $C{s}^{+}$ and $C{l}^{-}$ in $CsCl$ cubic lattice is $2\sqrt{3}\mathring{A}$, then the edge length (in $\mathring{A}$) of the cube is :

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (A)

$4$

$CsCl$ crystallizes in bcc.
Therefore, $\sqrt{3}a=2(r_{+}+r_{-})$
$a=4\mathring{A}$