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The atomic radius of an ion which crysta...

The atomic radius of an ion which crystallizes in fcc structure is `9/7` Å. Calculate the lattice constant.

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The correct Answer is:
`a=3.6xx10^(-10)` m
Atomic radius, `r = 9/7 Å =9/7 xx 10^(-10)` m.
For fcc structure, `r = a//(2sqrt2)` or lattice constant, = `9/7 xx 10^(-10) m xx 2 xx 1.4142 = 3.6 xx 10^(-10)` m.
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