Question

A metal forms fcc structure. Calculate the volume of fcc unit cell in \(\text{cm}^3\) if void volume is \(1.66\times 10^{-23} \text{cm}^3\).

• A. \(4.912\times 10^{-23} \text{cm}^3\)
• B. \(8.151\times 10^{-23} \text{cm}^3\)
• C. \(9.346\times 10^{-23} \text{cm}^3\)
• D. \(6.385\times 10^{-23} \text{cm}^3\)

Hint:

fcc packing efficiency is 74 percent, so voids occupy 26 percent of the cell volume.

Answer 1

The Correct Option is D

Step 1: Void fraction:
Packing efficiency of fcc $= 74\%$, so the void fraction $= 100 - 74 = 26\% = 0.26$.

Step 2: Solve for the cell volume:
$0.26 \times V = 1.66\times10^{-23}$.
$V = 1.66\times10^{-23}/0.26 = 6.385\times10^{-23}$ cm$^3$.

Step 3: Quick check:
$0.74V = 4.725\times10^{-23}$ cm$^3$ is the volume of the 4 atoms, so $V_{atoms} + V_{void} = 6.385\times10^{-23}$ cm$^3$. This matches option (D) only.

Step 4: Meaning of the void fraction:
The void fraction counts the empty space left between the four atoms of the fcc cell. A packing efficiency of 74 percent is a fixed property of close packing, so the void fraction is always 26 percent, whatever the metal.

Final Answer:
$V = 6.385\times10^{-23}$ cm$^3$, option (D). \[ \boxed{6.385\times10^{-23}\text{ cm}^3 \text{ (D)}} \]