Question

Sodium metal crystallises in a body centred cubic lattice with edge length of x Å. If the radius of sodium atom is 1.86 Å, the value of x is

• A. 4.29
• B. 3.29
• C. 2.39
• D. 3.93

Hint:

Memorize the relationship between edge length (a) and atomic radius (r) for the main cubic lattices: - Simple Cubic (SC): \(a = 2r\) (atoms touch along the edge) - Body-Centered Cubic (BCC): \(a\sqrt{3} = 4r\) (atoms touch along the body diagonal) - Face-Centered Cubic (FCC): \(a\sqrt{2} = 4r\) (atoms touch along the face diagonal)

Answer 1

The Correct Option is A

To find the edge length of a sodium atom crystallised in a body-centred cubic (BCC) lattice, we can use the relationship between the atomic radius and the edge length in a BCC lattice. In a BCC lattice, the body diagonal of the cube is occupied by atoms in the following arrangement:

• An atom at each corner of the cube contributes 1/8th of its volume to the unit cell.
• One atom is located at the centre of the cube.

The relationship between the atomic radius \( r \) and the edge length \( a \) in a BCC structure is given by:

\(4r = \sqrt{3} \cdot a\)

Given, the radius of the sodium atom \( r = 1.86 \, \text{Å} \). Substituting the value of \( r \) into the equation:

\(4 \times 1.86 = \sqrt{3} \cdot a\)

\(a = \frac{4 \times 1.86}{\sqrt{3}}\)

Calculating the above equation:

\(a = \frac{7.44}{1.732} \approx 4.29 \, \text{Å}\)

This shows that the edge length \( x \) is approximately 4.29 Å, which matches the correct answer option.