Question

Calculate the volume of bcc unit cell if radius of an atom present in it is \( 1.86 \times 10^{-8} \) cm.

• A. \( 5.391 \times 10^{-23}\text{ cm}^{3} \)
• B. \( 8.995 \times 10^{-23}\text{ cm}^{3} \)
• C. \( 7.951 \times 10^{-23}\text{ cm}^{3} \)
• D. \( 6.453 \times 10^{-23}\text{ cm}^{3} \)

Hint:

For BCC: $a = \frac{4r}{\sqrt{3}}$. For FCC: $a = 2\sqrt{2}r$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
For a Body-Centered Cubic (BCC) unit cell, we need to find the edge length (\( a \)) from the atomic radius (\( r \)) and then calculate the volume (\( V = a^{3} \)).
Step 2: Key Formula or Approach:
Relationship between \( a \) and \( r \) in BCC: \( \sqrt{3} a = 4r \Rightarrow a = \frac{4r}{\sqrt{3}} \).
Volume of unit cell: \( V = a^{3} \).
Step 3: Detailed Explanation:
Given: \( r = 1.86 \times 10^{-8} \text{ cm} \).
Calculate edge length \( a \):
\[ a = \frac{4 \times 1.86 \times 10^{-8}}{1.732} \]
\[ a = \frac{7.44}{1.732} \times 10^{-8} \approx 4.2956 \times 10^{-8} \text{ cm} \]
Calculate volume \( V \):
\[ V = a^{3} = (4.2956 \times 10^{-8})^{3} \]
\[ V \approx 79.25 \times 10^{-24} \text{ cm}^{3} \]
\[ V = 7.925 \times 10^{-23} \text{ cm}^{3} \]
Matching with the closest given option, we get \( 7.951 \times 10^{-23} \text{ cm}^{3} \).
Step 4: Final Answer:
The volume of the bcc unit cell is \( 7.951 \times 10^{-23}\text{ cm}^{3} \).