Question

Edge length of unit cell of BCC structure is 352 pm. What is radius of the atom?

• A. 176.3 pm
• B. 304.8 pm
• C. 152.4 pm
• D. 252.4 pm

Hint:

Memorize the three standard core cubic radius relationships to save valuable time during exams:
Simple Cubic: $r = \frac{a}{2} = 0.5a$
BCC: $r = \frac{\sqrt{3}}{4}a \approx 0.433a$
FCC: $r = \frac{a}{2\sqrt{2}} \approx 0.354a$

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question gives the unit cell edge length of a body-centered cubic (BCC) crystal lattice as 352 pm. We need to determine the corresponding atomic radius of the element.

Step 2: Key Formula or Approach:
In a body-centered cubic (BCC) unit cell, the atoms touch along the principal body diagonal of the cube.
The length of the body diagonal of a cube with edge length $a$ is given by $\sqrt{3}a$.
Since this diagonal spans across one full atom in the center and two half-atoms at the corners, it equals 4 atomic radii ($4r$).
Therefore, the relationship is:
$$\sqrt{3}a = 4r \implies r = \frac{\sqrt{3}}{4}a$$

Step 3: Detailed Explanation:
Given the edge length, $a = 352\text{ pm}$.
Substitute the value of $a$ into our geometric relationship formula:
$$r = \frac{\sqrt{3}}{4} \times 352$$ We know that $\sqrt{3} \approx 1.732$. Let's divide 352 by 4 first to make the mental calculation easier:
$$\frac{352}{4} = 88$$ Now, perform the multiplication:
$$r = 88 \times 1.732$$ $$r \approx 152.416\text{ pm}$$ Rounding to one decimal place gives 152.4 pm, which matches option (C).

Step 4: Final Answer:
The radius of the atom is 152.4 pm, which corresponds to option (C).