Find the radius of an atom in fcc unit cell having edge length $405 \mathrm{pm}$.

Find the radius of an atom in fcc unit cell having edge length $405 \mathrm{pm}$.
  1. $202.5 \mathrm{pm}$
  2. $175.3 \mathrm{pm}$
  3. $143.2 \mathrm{pm}$
  4. $181.0 \mathrm{pm}$

Solution

For fcc crystal structure, $r=\frac{\sqrt{2}}{4} a$ $\therefore \quad \mathrm{r}=\frac{1.414 \times 405}{4}=143.2 \mathrm{pm}$

Asked in: MHT CET 2023 (11 May Shift 2)

Practice more Solid State questions on Aicharya