Question

A closed packed structure of uniform spheres has the unit cell edge length of 0.8 nm. If the radius of the molecule in fcc lattice, is $28\times {10}^{-x}$, then what is the value of x?

Answer 1

As we know, for fcc lattice,

$r=\frac{a}{2\sqrt{2}}=\frac{0.8}{2\sqrt{2}}=0.2828nm$

$r=28\times {10}^{-2}nm$

By comparing, $r=28\times {10}^{-x}=28\times {10}^{-2}nm$

We get, $x=2$

Hence, the correct answer is 2.