Question

In any ionic crystal A has formed cubic close packing and B atoms are present at every tetrahedral voids. If any sample of crystal contain `N' number of B atoms then number of A atoms in that sample is

Answer 1

Answer:

Number of A atoms in that sample is$\frac{N}{2}$.

Explanation:

Number of of atoms of A in CCP ,n= 1

Number of tetrahedral voids in CCP = 2n

n = total numbers of spheres in a crystal.

Number of voids= 2 × 1 = 2

Number of B atoms in single crystal = 2

if number N numbers of B atoms are present in the sample

Then numbers of A atoms present will be:

$2n=N$

$n=\frac{N}{2}$