Question

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

• A. $\frac{\sqrt{3}}{4}\times 288 \,pm$
• B. $\frac{\sqrt{2}}{4}\times 288 \,pm$
• C. $\frac{4}{\sqrt{3}}\times 288 \,pm$
• D. $\frac{4}{\sqrt{2}}\times 288 \,pm$

Answer 1

The Correct Option is A

To find the atomic radius of an element with a body-centered cubic (bcc) structure, we need to understand the relation between the atomic radius and the cell edge length.

In a body-centered cubic (bcc) structure, the unit cell consists of atoms at each of the eight corners of a cube and one atom at the center of the cube. The atoms are touching along the body diagonal of the cube.

The body diagonal of a cube can be related to the cell edge length (\(a\)) and the atomic radius (\(r\)). The relation is given by:

\[\sqrt{3} \times a = 4 \times r\]

From this equation, we can solve for the atomic radius \(r\):

\[r = \frac{\sqrt{3}}{4} \times a\]

Given that the cell edge length (\(a\)) is 288 pm, we substitute this value into the equation:

\[r = \frac{\sqrt{3}}{4} \times 288 \, \text{pm}\]

Therefore, the atomic radius is \(\frac{\sqrt{3}}{4} \times 288 \, \text{pm}\).

This matches the correct answer given, which is \(\frac{\sqrt{3}}{4} \times 288 \, \text{pm}\).