Question

An element with simple cubic close structure has edge length of unit cell $3.86\ \text{\AA}$. What is the radius of atom?

• A. $5.79 \times 10^{-8}\ \text{cm}$
• B. $1.93 \times 10^{-8}\ \text{cm}$
• C. $3.86 \times 10^{-8}\ \text{cm}$
• D. $2.43 \times 10^{-8}\ \text{cm}$

Hint:

Keep the radius-edge length relationships for the three cubic cells handy:
Simple Cubic (SC): $r = \frac{a}{2}$
Face-Centered Cubic (FCC): $r = \frac{a}{2\sqrt{2}}$
Body-Centered Cubic (BCC): $r = \frac{\sqrt{3}a}{4}$

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the edge length ($a$) of a simple cubic unit cell and must calculate the atomic radius ($r$) in centimeters.

Step 2: Key Formula or Approach:
In a Simple Cubic (SC) lattice, the constituent atoms touch each other directly along the edges of the cubic unit cell.
Therefore, the edge length ($a$) is exactly equal to twice the atomic radius ($r$).
$$a = 2r \implies r = \frac{a}{2}$$
Also, to convert Angstroms () to centimeters (cm), use the conversion factor: $1\ \text{\AA} = 10^{-8}\ \text{cm}$.

Step 3: Detailed Explanation:
First, convert the given edge length to centimeters:
$$a = 3.86\ \text{\AA} = 3.86 \times 10^{-8}\ \text{cm}$$
Now, substitute this value into the radius formula:
$$r = \frac{3.86 \times 10^{-8}\ \text{cm}}{2}$$
$$r = 1.93 \times 10^{-8}\ \text{cm}$$

Step 4: Final Answer:
The radius of the atom is $1.93 \times 10^{-8}\ \text{cm}$, which corresponds to option (B).