Question

Find the radius of a BCC molecule having an edge length of \( 2.0 \times 10^{-11} \) m.

• A. \( 1.0 \times 10^{-11} \) m
• B. \( 1.5 \times 10^{-11} \) m
• C. \( 2.0 \times 10^{-11} \) m
• D. \( 3.0 \times 10^{-11} \) m

Hint:

In a BCC unit cell, the body diagonal connects two corner atoms and the center atom. The relationship between the edge length and the radius of the atoms is given by \( r = \frac{\sqrt{3}}{4} \, a \).

Answer 1

The Correct Option is A

In a Body-Centered Cubic (BCC) unit cell, the atoms are arranged such that there is one atom at each corner of the cube and one atom at the center of the cube. The distance between the centers of two atoms along the body diagonal of the cube is equal to four times the radius of an atom.
Step 1: Relationship between edge length and radius Let the edge length of the BCC unit cell be \( a \). In a BCC unit cell, the body diagonal is related to the edge length \( a \) by the Pythagorean theorem: \[ \text{Body diagonal} = \sqrt{3} \, a \] Since the body diagonal of a BCC unit cell passes through two atoms (one at the center and one at the corner), the body diagonal is also equal to \( 4r \), where \( r \) is the radius of the atoms. Thus, we have the equation: \[ \sqrt{3} \, a = 4r \]
Step 2: Solve for the radius Rearranging the equation to solve for \( r \): \[ r = \frac{\sqrt{3}}{4} \, a \]
Step 3: Substitute the given edge length We are given that the edge length \( a = 2.0 \times 10^{-11} \, \text{m} \). Substituting this into the equation: \[ r = \frac{\sqrt{3}}{4} \times 2.0 \times 10^{-11} = 1.0 \times 10^{-11} \, \text{m} \] Thus, the radius of the BCC molecule is \( \boxed{1.0 \times 10^{-11}} \, \text{m} \).