Question

The fraction of the total volume occupied by atoms in a Simple Cubic (SC) structure is:

• A. \( \frac{\pi}{6} \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{3\sqrt{2}} \)
• D. \( \frac{\pi}{3\sqrt{3}} \)

Hint:

Standard packing efficiencies to remember:

• SC = \( \frac{\pi}{6} \approx 0.52 \)

• BCC = \( \frac{\pi\sqrt{3}}{8} \approx 0.68 \)

• FCC = \( \frac{\pi}{3\sqrt{2}} \approx 0.74 \)

Answer 1

The Correct Option is A

Step 1: Understand packing fraction.
The packing fraction tells us what part of the box is actually filled by the atoms. It is the volume taken by the atoms divided by the total volume of the unit cell.

Step 2: Count atoms in a simple cubic cell.
In a simple cubic structure there is one atom at each of the eight corners. Each corner atom is shared by eight cells, so each contributes one eighth. The total is $8 \times \frac{1}{8} = 1$ atom per cell.

Step 3: Relate atom radius to cell edge.
In a simple cubic cell, atoms touch along the edge. So the edge length equals two radii: \[ a = 2r \]

Step 4: Write the two volumes.
The volume of one atom is $\frac{4}{3}\pi r^3$. The volume of the cell is: \[ a^3 = (2r)^3 = 8 r^3 \]

Step 5: Form the packing fraction.
Divide atom volume by cell volume: \[ \text{fraction} = \frac{\frac{4}{3}\pi r^3}{8 r^3} \] The $r^3$ cancels.

Step 6: Simplify.
This becomes: \[ \frac{4\pi}{3 \times 8} = \frac{4\pi}{24} = \frac{\pi}{6} \] So about $52\%$ of the space is filled, and the fraction is $\frac{\pi}{6}$. \[ \boxed{\dfrac{\pi}{6}} \]