Question

Calculate the edge length of bcc unit cell if radius of metal atom is 227 pm.

• A. 4.54 x 10^-8 cm
• B. 5.24 x 10^-8 cm
• C. 6.42 x 10^-8 cm
• D. 1.135 x 10^-8 cm

Hint:

For BCC unit cells, the atoms touch along the body diagonal. The relationship is $4r = a\sqrt{3}$. Always pay attention to unit conversions (pm to cm, nm to m, etc.) in final answers.

Answer 1

The Correct Option is C

Step 1: Recall the relationship between atomic radius and edge length for a BCC unit cell.\ In a Body-Centered Cubic (BCC) unit cell, the atoms touch along the body diagonal. The length of the body diagonal is $4r$, where $r$ is the atomic radius.\ The body diagonal of a cube with edge length $a$ is given by $a\sqrt{3}$.\ Therefore, for a BCC unit cell:\ \[4r = a\sqrt{3}\]\
Step 2: Rearrange the formula to solve for edge length ($a$).\ From the relationship, we can solve for $a$:\ \[a = \frac{4r}{\sqrt{3\]\
Step 3: Substitute the given values and calculate the edge length in picometers.\ Given atomic radius $r = 227 \text{ pm}$.\ Given $\sqrt{3} \approx 1.732$.\ \[a = \frac{4 \times 227 \text{ pm{1.732}\]\ \[a = \frac{908 \text{ pm{1.732}\]\ \[a \approx 524.249 \text{ pm}\]\
Step 4: Convert the edge length from picometers to centimeters.\ We know that $1 \text{ pm} = 10^{-12} \text{ m}$ and $1 \text{ m} = 10^{2} \text{ cm}$.\ So, $1 \text{ pm} = 10^{-12} \times 10^{2} \text{ cm} = 10^{-10} \text{ cm}$.\ \[a = 524.249 \times 10^{-10} \text{ cm}\]\ \[a = 5.24249 \times 10^{2} \times 10^{-10} \text{ cm}\]\ \[a = 5.24249 \times 10^{-8} \text{ cm}\]\
Step 5: Compare with the given options.\ The calculated edge length is approximately $5.24 \times 10^{-8} \text{ cm}$, which matches Option B.