MCQ

$CsCl$ has bcc arrangement, its unit cell edge length is 400 pm, it's inter atomic distance is

  • A
    $400 pm$
  • B
    $800 pm$
  • C
     $\sqrt{3} \times 100 pm$
  • $\left(\frac{\sqrt{3}}{2}\right) \times 400 pm$

Answer

Correct option: D.
$\left(\frac{\sqrt{3}}{2}\right) \times 400 pm$
$\left(\frac{\sqrt{3}}{2}\right) \times 400 pm$

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