Question

A metal element ‘M’ of radius 50 nm is crystallized in FCC format cubical crystal such that the face of unit cells align with the face of cubical crystal. If the total number of metal atoms of ‘M’ at the face of the cubical crystal is $6\times {10}^{30}$, then the area of one face of the cubical crystal is $A\times {10}^{16}{m}^2$, the value of ‘A’ is:___

Answer 1

No. of atoms at one face of crystal $=\frac{1}{6}\times 6\times {10}^{30}={10}^{30}$ atoms.
So, no. of unit cells of edge of crystal $=\sqrt{{10}^{30}}={10}^{15}$
Now edge length of one unit cell $=\frac{4}{\sqrt{2}}\times 50\times {10}^{15}nm$
So, area of face of crystal $={(\frac{4}{\sqrt{2}}\times 50\times {10}^{15})}^{2}n{m}^2$
$=\frac{16}{2}\times 25\times {10}^2\times {10}^{30}=2\times {10}^{34}n{m}^2$
$=2\times {10}^{34}\times {10}^{-18}{m}^2=2\times {10}^{16}{m}^2$