Question

Calculate the edge length of bcc unit cell if radius of a particle present in it is 186 pm.

• A. $4.296\times10^{-8}$ cm
• B. $7.301\times10^{-8}$ cm
• C. $3.715\times10^{-8}$ cm
• D. $5.419\times10^{-8}$ cm

Hint:

Radius-Edge Length Relationships to memorize:
Simple Cubic (sc): $a = 2r$
Face-Centered Cubic (fcc): $a = 2\sqrt{2}r$
Body-Centered Cubic (bcc): $a = \frac{4r}{\sqrt{3}}$

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the radius of an atom ($r = 186$ pm) arranged in a body-centered cubic (bcc) lattice. We must calculate the edge length ($a$) of the unit cell and convert it to centimeters.

Step 2: Detailed Explanation:
In a body-centered cubic (bcc) unit cell, the atoms touch each other along the body diagonal. The length of the body diagonal is $a\sqrt{3}$, and it is equal to 4 times the atomic radius ($4r$).
The fundamental relationship is:
$a\sqrt{3} = 4r \implies a = \frac{4r}{\sqrt{3}}$
Given $r = 186$ pm and using $\sqrt{3} \approx 1.732$:
$a = \frac{4 \times 186}{1.732}$
$a = \frac{744}{1.732}$
$a \approx 429.56$ pm
Now, convert picometers (pm) to centimeters (cm). We know that $1 \text{ pm} = 10^{-12} \text{ m} = 10^{-10} \text{ cm}$.
$a \approx 429.56 \times 10^{-10} \text{ cm}$
Adjust the decimal to standard scientific notation format:
$a \approx 4.2956 \times 10^{-8} \text{ cm}$
Rounding to three decimal places gives $4.296 \times 10^{-8}$ cm.

Step 3: Final Answer:
The edge length is $4.296\times10^{-8}$ cm, matching option (a).