An element crystallises in bcc type having atomic radius $1 \cdot 33 \times 10^{-8} \mathrm{~cm}$, the edge…
An element crystallises in bcc type having atomic radius $1 \cdot 33 \times 10^{-8} \mathrm{~cm}$, the edge length of unit cell will be
- $2 \cdot 17 \times 10^{-8} \mathrm{~cm}$
- $2 \cdot 66 \times 10^{-8} \mathrm{~cm}$
- $4.08 \times 10^{-8} \mathrm{~cm}$
- $3 \cdot 07 \times 10^{-8} \mathrm{~cm}$
Solution
For bcc crystal, $\quad \mathrm{r}=\frac{\sqrt{3}}{4} \mathrm{a}$
$\therefore \quad a=\frac{4 \mathrm{r}}{\sqrt{3}}=\frac{4 \times 1.33 \times 10^{-8} \mathrm{~cm}}{1.732}$
$\therefore \quad a=3.07 \times 10^{-8} \mathrm{~cm} .$
Asked in: MHT CET 2020 (13 Oct Shift 2)
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