Question

Lithium crystallises into a body centered cubic structure. What is the radius of lithium if the edge length of its unit cell is \(351~\text{pm}\)?

• A. \(151.98~\text{pm}\)
• B. \(300.50~\text{pm}\)
• C. \(75.50~\text{pm}\)
• D. \(240.80~\text{pm}\)

Hint:

In BCC structures, atoms touch along the body diagonal, not the edge.

Answer 1

The Correct Option is A

Step 1: Relation between edge length and atomic radius in BCC.
For a body centered cubic (BCC) lattice, the atoms touch along the body diagonal. The relation is: \[ \sqrt{3}\,a = 4r \] Step 2: Substitute the given value.
\[ a = 351~\text{pm} \] \[ r = \frac{\sqrt{3}\times 351}{4} \] Step 3: Calculate the radius.
\[ r = \frac{1.732 \times 351}{4} = 151.98~\text{pm} \] Step 4: Conclusion.
Hence, the radius of lithium atom is \(151.98~\text{pm}\).