Q111 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Chemistry
What is the packing efficiency of a metal crystal if it crystallizes in a face centered cubic lattice ?
- A$52.4\%$
- B$68.0\%$
- Correct answerC$74.0\%$
- D$60.0\%$
Solution
For a face-centered cubic (fcc) lattice, the atoms touch each other along the face diagonal of the unit cell.
Let the edge length of the unit cell be $a$ and the radius of each atom be $r$.
The length of the face diagonal is $\sqrt{2}a$.
Since the atoms touch along the face diagonal, we have $\sqrt{2}a = 4r$, which gives $a = 2\sqrt{2}r$.
The number of atoms per unit cell in an fcc lattice is $Z = 4$.
The volume occupied by the atoms is $4 \times \dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi r^3$.
The total volume of the unit cell is $a^3 = (2\sqrt{2}r)^3 = 16\sqrt{2}r^3$.
Packing efficiency = $\dfrac{\text{Volume occupied by atoms}}{\text{Volume of unit cell}} \times 100$
Packing efficiency = $\dfrac{\dfrac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100$
Packing efficiency = $\dfrac{\pi}{3\sqrt{2}} \times 100$
Substituting the values of $\pi$ and $\sqrt{2}$, we get $\dfrac{3.14}{3 \times 1.414} \times 100 = 74.0\%$
Answer: $74.0\%$
Let the edge length of the unit cell be $a$ and the radius of each atom be $r$.
The length of the face diagonal is $\sqrt{2}a$.
Since the atoms touch along the face diagonal, we have $\sqrt{2}a = 4r$, which gives $a = 2\sqrt{2}r$.
The number of atoms per unit cell in an fcc lattice is $Z = 4$.
The volume occupied by the atoms is $4 \times \dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi r^3$.
The total volume of the unit cell is $a^3 = (2\sqrt{2}r)^3 = 16\sqrt{2}r^3$.
Packing efficiency = $\dfrac{\text{Volume occupied by atoms}}{\text{Volume of unit cell}} \times 100$
Packing efficiency = $\dfrac{\dfrac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100$
Packing efficiency = $\dfrac{\pi}{3\sqrt{2}} \times 100$
Substituting the values of $\pi$ and $\sqrt{2}$, we get $\dfrac{3.14}{3 \times 1.414} \times 100 = 74.0\%$
Answer: $74.0\%$
Hence, the correct answer is option: 'C'.