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What is the edge length of the unit cell of a body centred cubic crystal of an element whose atomic radius is \(75 \mathrm{pm}\) ?
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The correct answer is:
\(173.2 \mathrm{pm}\)
Hint : In BCC, \(4 r=\sqrt{3} a\)
\(\therefore \mathrm{a}=\frac{4 \mathrm{r}}{\sqrt{3}}=\frac{4 \times 75}{\sqrt{3}}=\frac{300}{\sqrt{3}}=\sqrt{3} \times 100 \mathrm{pm}=173.2 \mathrm{pm}\)
\(\therefore \mathrm{a}=\frac{4 \mathrm{r}}{\sqrt{3}}=\frac{4 \times 75}{\sqrt{3}}=\frac{300}{\sqrt{3}}=\sqrt{3} \times 100 \mathrm{pm}=173.2 \mathrm{pm}\)
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