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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
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The correct answer is:
$3 \cdot 07 \times 10^{-8} \mathrm{~cm}$
For bcc crystal, $\quad \mathrm{r}=\frac{\sqrt{3}}{4} \mathrm{a}$
$\therefore \quad a=\frac{4 \mathrm{r}}{\sqrt{3}}=\frac{4 \times 1.33 \times 10^{-8} \mathrm{~cm}}{1.732}$
$\therefore \quad a=3.07 \times 10^{-8} \mathrm{~cm} .$
$\therefore \quad a=\frac{4 \mathrm{r}}{\sqrt{3}}=\frac{4 \times 1.33 \times 10^{-8} \mathrm{~cm}}{1.732}$
$\therefore \quad a=3.07 \times 10^{-8} \mathrm{~cm} .$
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