Percentage of free space in a body-centred cu | vedclass

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(D) In a $bcc$ unit cell,the number of atoms $Z = 2$.The volume of atoms in the unit cell $(v) = 2 	imes \frac{4}{3} \pi r^3$.For a $bcc$ structure,t

Vedclass · Accepted Answer

$32$

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(D) In a $bcc$ unit cell,the number of atoms $Z = 2$.The volume of atoms in the unit cell $(v) = 2 	imes \frac{4}{3} \pi r^3$.For a $bcc$ structure,the relationship between radius $(r)$ and edge length $(a)$ is $r = \frac{\sqrt{3}}{4} a$.Substituting $r$ in the volume formula: $v = 2 	imes \frac{4}{3} \pi (\frac{\sqrt{3}}{4} a)^3 = \frac{\sqrt{3}}{8} \pi a^3 \approx 0.68 a^3$.The volume of the unit cell $(V) = a^3$.The packing efficiency (percentage of volume occupied) $= \frac{v}{V} 	imes 100 = 68 \%$.Therefore,the percentage of free space $= 100 \% - 68 \% = 32 \%$.

Percentage of free space in a body-centred cubic unit cell is ................ $\%$

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