Find the radius of an atom in fcc unit cell having edge length $393 \mathrm{pm}$.

Find the radius of an atom in fcc unit cell having edge length $393 \mathrm{pm}$.
  1. $196.51 \mathrm{pm}$
  2. $170.22 \mathrm{pm}$
  3. 78.63 $\mathrm{pm}$
  4. $138.93 \mathrm{pm}$

Solution

For foc crystal structure, $\begin{aligned} 4 \mathrm{r} & =\sqrt{2} \mathrm{a} \\ \therefore \quad \mathrm{r} & =\frac{\sqrt{2} \mathrm{a}}{4}=\frac{1.414 \times 393}{4}=138.93 \mathrm{pm} \end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 2)

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