Question

An element has bcc structure with atomic radius 200 pm. Find the volume of unit cell.

• A. \(5.625\times 10^{-23}\ \text{cm}^{3}\)
• B. \(7.513\times 10^{-23}\ \text{cm}^{3}\)
• C. \(8.41\times 10^{-23}\ \text{cm}^{3}\)
• D. \(9.848\times 10^{-23}\ \text{cm}^{3}\)

Hint:

In a bcc cell the atoms touch along the body diagonal, so the edge is 4r divided by root 3.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
In a bcc unit cell, atoms touch along the body diagonal. The body diagonal is \(\sqrt{3}\,a\) and equals \(4r\).

Step 2: Find the edge length
\(a = \dfrac{4r}{\sqrt{3}} = \dfrac{4\times 200}{1.732}\approx 461.9\ \text{pm} = 4.619\times 10^{-8}\ \text{cm}\).

Step 3: Calculation
\(V = a^{3} = (4.619\times 10^{-8})^{3}\approx 9.848\times 10^{-23}\ \text{cm}^{3}\).

Final Answer:
The volume of the unit cell is about 9.848 times 10 to the power minus 23 cubic centimetre. \[ \boxed{\text{(D) }9.848\times 10^{-23}\ \text{cm}^{3}} \]