Chemistry - Solid State Question with Solution | TestHub

Given, $\left(\text{a} – 2\text{r}\right)=53 \text{pm}$

The spheres at the corners are not touching
each other. For bcc structure,

$4\mathrm{r}=\sqrt{3}\mathrm{a}$

$2\mathrm{r}=\frac{\sqrt{3}}{2}\mathrm{a}$

$\mathrm{a}-\frac{\sqrt{3}}{2}\mathrm{a}=53$

$\therefore \text{a}=395.6 \text{pm}$

$395.6\times {10}^{-12} \mathrm{m}$

$=395.6\times {10}^{-10} \text{cm}$

$=3.956\times {10}^{-8} \text{cm}$

$\text{Density}\left(\mathrm{d}\right)=\frac{\mathrm{z} \mathrm{M}}{{\mathrm{N}}_{\mathrm{A}} {\mathrm{a}}^3}$

$\text{d}=\frac{2\times 23}{6.023\times {10}^{23}\times {(3.956)}^3\times {10}^{-24}}=1.23\text{g/cc}$