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CsCl crystall...

Question

$C s C l$ crystallizes in a cubic lattice that has a $C l^{-}$ at each corner and $C s^{+}$ at the centre of the unit cell. If $(r_{C s^{+}}) = 1.69 o A$ and $(r_{C l^{-}}) = 1.81 o A$ , what is the value of edge length a of the cube?

4.04

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3.50

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3.03

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1.95

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Solution

The correct option is B $4.04$
In CsCl type of crystal atoms touch along the body diagonal

$\sqrt{3} a = 2 (r^{+} + r^{-})$

$a = \frac{2 \times (1.69 + 1.81)}{\sqrt{3}} = 4.04 \times 10^{- 10} m$

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