Question

An element occurs in two crystalline forms $\alpha$ and $\beta$. The $\alpha$-form a fcc arrangement with $a=3.68$ $\mathring{A}$ and the $\beta$-form has a bcc arrangement with $a=2.92\mathring{A}$. Calculate the ratio of their densities.

• A. 2 : 1
• B. 1 : 2
• C. 1 : 3
• D. 1 : 1

Answer 1

The correct option is D 1 : 1

$UnitcelllengthinFcc=3.68A^0\left(a_1\right)UnitcelllengthinBcc=2.92A^0\left(a_2\right)Density\alpha \left(Fcc\right)=\frac{M\times Z_1}{N_0\times a_1^3}Density\beta \left(Bcc\right)=\frac{M\times Z_2}{N_0\times a_2^3}\frac{d\alpha }{d\beta }=\frac{Z_1\times ({a_2)}^3}{Z_2\times (a_1{)}^3}=\frac{4\times ({2.92)}^3}{2\times {\left(3.68\right)}^3}=0.999\sim 1Henceratioofdensity=1:1$