Question

A cubic structure of uniform spheres has the unit cell edge length of 0.8 mm. If the radius of the molecule in a simple cubic lattice is $40\times {10}^{-x}$, then what is the value of x?

Answer 1

As we know, for simple cubic lattice,

$r=\frac{a}{2}=\frac{0.8}{2}=0.4mm$

$r=40\times {10}^{-2}$

By comparing, $r=40\times {10}^{-2}=40\times {10}^{-x}$
we get, $x=2$
Hence, the correct answer is 2.