The edge length of unit cell of metal having molecular weight 75g/mol is 5A⁰ which crystallises in simple cubic lattice. If the density is 2g/cc then the radius of metal atom in pm is x × 10² then ‘x’ is ____ [NA = 6 × 10²³]

$\mathrm{d}=\mathrm{ZM}/{\mathrm{a}}^3\mathrm{N};2=1\times 75/{\mathrm{a}}^3\times 6\times {10}^{23}$
in simple cube 2r = a; r = a/2
original density = 1
$\begin{array}{r}\frac{{\mathrm{d}}_1}{{\mathrm{d}}_2}=\frac{{\left({\mathrm{a}}_2\right)}^3}{5\times {10}^{-8}\times 5\times {10}^{-8}\times 5\times {10}^{-8}}\\ \frac{1}{2}=\frac{{\mathrm{a}}_2^3}{5\times {10}^{-8}\times 5\times {10}^{-8}\times 5\times {10}^{-8}}\end{array}$
So, it is body centered cubic (BCC) structure. For BCC, if a is the edge length of the cube and r is the radius of the atom then,

4r=$\sqrt{3}$​a

⇒r=$\frac{\sqrt{3}}{4}$a

=$\frac{\sqrt{3}}{4}$​​×5 A˚

⇒r=2.165 A˚=2.165×10⁻⁸ cm=216 pm=2.16x10² pm