Question

Sodium metal crystallizes in a body-centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is:

• A. 1.86 Å
• B. 3.22 Å
• C. 5.72 Å
• D. 0.93 Å

Hint:

In a body-centred cubic lattice, the relationship between the unit cell edge and atomic radius is \( 4r = \sqrt{3}a \).

Answer 1

The Correct Option is A

In a body-centred cubic (BCC) unit cell, the relationship between the unit cell edge \( a \) and the atomic radius \( r \) is given by: \[ \text{Diagonal} = \sqrt{3}a = 4r \] For sodium, the unit cell edge \( a = 4.29 \, \text{Å} \), so we can solve for \( r \): \[ 4r = \sqrt{3} \times 4.29 \, \text{Å} = 7.44 \, \text{Å} \] \[ r = \frac{7.44}{4} = 1.86 \, \text{Å} \] Thus, the radius of the sodium atom is \( 1.86 \, \text{Å} \). Therefore, the correct answer is option (A).