MHT CET · Chemistry · Solid State
Find the radius of an atom in fcc unit cell having edge length \(393 \mathrm{pm}\).
- A \(196.51 \mathrm{pm}\)
- B \(170.22 \mathrm{pm}\)
- C 78.63 \(\mathrm{pm}\)
- D \(138.93 \mathrm{pm}\)
Answer & Solution
Correct Answer
(D) \(138.93 \mathrm{pm}\)
Step-by-step Solution
Detailed explanation
For foc crystal structure,
\(\begin{aligned}
4 \mathrm{r} & =\sqrt{2} \mathrm{a} \\
\therefore \quad \mathrm{r} & =\frac{\sqrt{2} \mathrm{a}}{4}=\frac{1.414 \times 393}{4}=138.93 \mathrm{pm}
\end{aligned}\)
\(\begin{aligned}
4 \mathrm{r} & =\sqrt{2} \mathrm{a} \\
\therefore \quad \mathrm{r} & =\frac{\sqrt{2} \mathrm{a}}{4}=\frac{1.414 \times 393}{4}=138.93 \mathrm{pm}
\end{aligned}\)
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