Question

If metallic atoms of mass 197 and radius 166 pm are arranged in ABCABC fashion then what is the surface area of each unit cell?

• A. $1.32\times {10}^{6}p{m}^2$
• B. $1.32\times {10}^{-18}p{m}^2$
• C. $2.20\times {10}^{5}p{m}^2$
• D. $2.20\times {10}^{-19}p{m}^2$

Answer 1

The correct option is A $1.32\times {10}^{6}p{m}^2$

ABCABC arrangement is found in CCP.
In closed cubic packing, relation between edge length of unit cell, a, and radius of particle, r, is given as a=$2\sqrt{2}r$.
Surface area (S.A.) = $6a^2$
From the relationship,
$a^2=8r^2$
$S.A.=6a^2=48r^2$
When r = 166 pm,
$S.A.=48(166pm{)}^2=1.32\times {10}^{6}p{m}^2$.