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(D) In a $BCC$ unit cell,the atoms touch along the body diagonal,so $\sqrt{3}a = 4R$,where $R$ is the radius of the atom at the corners and center.Thu

Vedclass · Accepted Answer

$0.067$

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(D) In a $BCC$ unit cell,the atoms touch along the body diagonal,so $\sqrt{3}a = 4R$,where $R$ is the radius of the atom at the corners and center.Thus,$R = \frac{\sqrt{3}a}{4}$.For a sphere of radius $r$ to fit at the center of the edge,it must touch the two corner atoms on that edge.The distance between two corner atoms along the edge is $a$,so $2(R + r) = a$.Substituting $R = \frac{\sqrt{3}a}{4}$ into the equation: $2(\frac{\sqrt{3}a}{4} + r) = a$.$\frac{\sqrt{3}a}{2} + 2r = a$.$2r = a - \frac{\sqrt{3}a}{2} = a(1 - \frac{1.732}{2}) = a(1 - 0.866) = 0.134a$.$r = \frac{0.134a}{2} = 0.067a$.

The radius of the largest sphere which fits properly at the centre of the edge of a body-centered cubic $(BCC)$ unit cell is (Edge length is represented by $a$) (in $a$)

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