Chemistry - Solid State Question with Solution | TestHub

From FCC density: $a_1={\left(\frac{Z\times M}{\rho \times N_A}\right)}^{1/3}={\left(\frac{4\times 80}{4\times 6\times 10^{23}}\right)}^{1/3}={\left(\frac{400}{3}\right)}^{1/3}\times 10^{-8}=3\sqrt{3}\times 10^{-8}$ cm.

For FCC, $4r=a_1\sqrt{2}$, so $r=\frac{a_1\sqrt{2}}{4}$.

For BCC, distance between next nearest neighbors is $d=a_2$.

For BCC, $4r=a_2\sqrt{3}$, so $a_2=\frac{4r}{\sqrt{3}}=\frac{a_1\sqrt{2}}{\sqrt{3}}=\frac{3\sqrt{3}\times 10^{-8}\times \sqrt{2}}{\sqrt{3}}=3\sqrt{2}\times 10^{-8}$ cm.

Thus, $3\sqrt{2}\times 10^{7}d=3\sqrt{2}\times 10^7\times (3\sqrt{2}\times 10^{-8})=18\times 10^{-1}=1.8$.