Chemistry - Solid State Question with Solution | TestHub

$\frac{{\text{d}}_{\gamma }}{{\text{d}}_{\beta }}=\frac{{\text{d}}_{\text{fcc}}}{{\text{d}}_{\text{bcc}}}$

$=\frac{4\times \frac{\text{A}}{{\text{N}}_{\text{a}}}\times \frac{1}{{\text{a}}_{\text{fcc}}^3}}{2\times \frac{\text{A}}{{\text{N}}_{\text{a}}}\times \frac{1}{{\text{a}}_{\text{bcc}}^3}}=2{\left(\frac{{\text{a}}_{\text{bcc}}}{{\text{a}}_{\text{fcc}}}\right)}^3$

$=2\times {\left(\frac{290}{386}\right)}^3=\text{0.848}$