Question

Metal M of radius $50$ nm is crystallized in fcc type and made cubical crystal such that the face of unit cells is aligned with the face of cubical crystal. If the total number of metal atoms of M at all faces of cubical crystal is $6\times {10}^{30}$, then the area of one face of cubical crystal is $A\times {10}^{16}$ m${}^2$. Find the value of A.

Answer 1

The total number of metal atoms (M) at all faces of cubical crystal is $6\times {10}^{30}$.
The total number of metal atoms (M) at one face of cubical crystal is $1\times {10}^{30}$.

$\sqrt{2}a=4r$
The edge length is $a=\frac{r}{0.3535}=\frac{50\times {10}^{-9}}{0.3535}$
The area of one face of one unit cell is $a^2={\left(\frac{r}{0.3535}\right)}^2={\left(\frac{50\times {10}^{-9}}{0.3535}\right)}^2=2\times {10}^{-14}$ m${}^2$
The area of one face of cubic crystal is $1\times {10}^{30}\times 2\times {10}^{-14}$ m${}^2=2\times {10}^{16}=A\times {10}^{16}$
Hence, $A=2$