AP EAMCET · Chemistry · Solid State
If the radius of an atom of an element which forms a body centered cubic unit cell is 173.2 \(\mathrm{pm}\), the volume of unit cell in \(\mathrm{cm}^3\) is
- A \(3.12 \times 10^{-23}\)
- B \(6.4 \times 10^{-23}\)
- C \(3.2 \times 10^{-24}\)
- D \(2.13 \times 10^{-23}\)
Answer & Solution
Correct Answer
(B) \(6.4 \times 10^{-23}\)
Step-by-step Solution
Detailed explanation
Given, Radius of an atom in body centered cubic (bcc) unit cell \(=173.2 \mathrm{pm}\). \(\because\) For bcc structure \[ \sqrt{3} \cdot a=4 r \] where, \(a=\) edge-length \(r=\) radius of atom and \(a^3=V\) (volume of cubic unit cell).…
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