Step 1: Understanding the Concept:
In a simple cubic (SC) unit cell, particles (atoms) are present only at the corners.
Since each corner atom is shared by 8 adjacent unit cells, the total number of particles per simple cubic unit cell (\( Z \)) is \( 8 \times \frac{1}{8} = 1 \).
The packing efficiency relates the volume occupied by the particles to the total volume of the unit cell.
Step 2: Key Formula or Approach:
For a simple cubic lattice:
The relation between edge length (\( a \)) and particle radius (\( r \)) is \( a = 2r \).
Total volume of the unit cell (\( V_{\text{cell}} \)) = \( a^3 = (2r)^3 = 8r^3 \).
Volume of one particle (a sphere) (\( V_{\text{particle}} \)) = \( \frac{4}{3}\pi r^3 \).
We are given \( V_{\text{particle}} \) and need to find \( V_{\text{cell}} \).
We can create a ratio:
\[ \frac{V_{\text{particle}}}{V_{\text{cell}}} = \frac{\frac{4}{3}\pi r^3}{8r^3} = \frac{\pi}{6} \]
This ratio is the packing fraction, which is approximately 0.524.
Rearranging to solve for \( V_{\text{cell}} \):
\[ V_{\text{cell}} = V_{\text{particle}} \times \frac{6}{\pi} \]
Step 3: Detailed Explanation:
We are given:
\( V_{\text{particle}} = 2.1 \times 10^{-23} \text{ cm}^3 \)
Using the formula derived above:
\[ V_{\text{cell}} = (2.1 \times 10^{-23} \text{ cm}^3) \times \frac{6}{\pi} \]
Substitute the approximate value of \( \pi \approx 3.1416 \):
\[ V_{\text{cell}} = 2.1 \times 10^{-23} \times \frac{6}{3.1416} \]
\[ V_{\text{cell}} \approx 2.1 \times 10^{-23} \times 1.9098 \]
\[ V_{\text{cell}} \approx 4.01 \times 10^{-23} \text{ cm}^3 \]
The problem states the volume of the unit cell is \( x \times 10^{-23} \text{ cm}^3 \).
Comparing our result to this expression, we find:
\[ x \approx 4.01 \]
Looking at the options, 4.0 is the closest match.
Step 4: Final Answer:
The value of \( x \) is 4.0.