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

1. √2/4 × 288 pm
2. 4/√3 × 288 pm
3. 4/√2 × 288 pm
4. √3/4 × 288 pm

Correct answer: √3/4 × 288 pm

Solution

In a body-centered cubic (bcc) structure, the relationship between the atomic radius (r) and the edge length (a) is r = √3/4 × a. Substituting a = 288 pm, the atomic radius is √3/4 × 288 pm.

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