Solved: Chemistry Question for NEET

NEET ChemistryMedium

An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is

(1) $\frac{\sqrt{3}}{4} \times 288 \text{ pm}$

(2) $\frac{\sqrt{2}}{4} \times 288 \text{ pm}$

(3) $\frac{4}{\sqrt{3}} \times 288 \text{ pm}$

(4) $\frac{4}{\sqrt{2}} \times 288 \text{ pm}$

Step-by-Step Solution

For a BCC structure, the relationship between edge length 'a' and radius 'r' is $\sqrt{3}a = 4r$ . Therefore, $r = \frac{\sqrt{3}a}{4}$ . Given $a = 288 \text{ pm}$ , $r = \frac{\sqrt{3}}{4} \times 288 \text{ pm}$ .

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