Silver crystallises in fcc structure, if edge length of unit cell is $316.5 \mathrm{pm}$. What is the radius…
Silver crystallises in fcc structure, if edge length of unit cell is $316.5 \mathrm{pm}$. What is the radius of silver atom?
- $137.04 \mathrm{pm}$
- $111.91 \mathrm{pm}$
- $121.91 \mathrm{pm}$
- $158.25 \mathrm{pm}$
Solution
$\mathrm{a}=316.5 \mathrm{pm}$
For fcc unit cell, $r=\frac{a}{2 \sqrt{2}}$
$\therefore r=\frac{316.5 \mathrm{pm}}{2 \times 1.414}=111.91 \mathrm{pm}$
Asked in: MHT CET 2020 (16 Oct Shift 2)
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