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An element has a body centred cubic structure with a cell edge of 360pm. Atomic radius is __pm

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$\sqrt{3} \times 180$

$\frac{360}{\sqrt{2}}$

$\frac{180}{\sqrt{2}}$

$\sqrt{3} \times 90$

answer is C.

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Detailed Solution

for bcc metallic crystal

$4 r = \sqrt{3} a$

$r = \frac{\sqrt{3}}{4} a$

$r = \sqrt{3} \times \frac{360}{4} pm$

$r = \sqrt{3} \times 90 pm$

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