Q112 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Chemistry
What is the packing efficiency in BCC unit cell ?
- A$24\ \%$
- B$32\ \%$
- Correct answerC$68\ \%$
- D$74\ \%$
Solution
In a body-centered cubic (BCC) unit cell, the atoms touch each other along the body diagonal.
Let the edge length of the unit cell be $a$ and the radius of the atom be $r$.
The length of the body diagonal is $\sqrt{3}a$.
Since the atoms touch along the body diagonal, we have:
$\sqrt{3}a = 4r \Rightarrow a = \dfrac{4r}{\sqrt{3}}$
The number of atoms per unit cell in a BCC structure is $Z = 2$.
Volume of the unit cell is $V = a^3 = \left(\dfrac{4r}{\sqrt{3}}\right)^3 = \dfrac{64r^3}{3\sqrt{3}}$
Volume occupied by the atoms is $v = 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3$
Packing efficiency is given by:
$\text{Packing efficiency} = \dfrac{v}{V} \times 100$
$\text{Packing efficiency} = \dfrac{\dfrac{8}{3}\pi r^3}{\dfrac{64r^3}{3\sqrt{3}}} \times 100 = \dfrac{\sqrt{3}\pi}{8} \times 100$
Substituting the values of $\pi$ and $\sqrt{3}$:
$\text{Packing efficiency} \approx 68\ \%$
Answer: $68\ \%$
Let the edge length of the unit cell be $a$ and the radius of the atom be $r$.
The length of the body diagonal is $\sqrt{3}a$.
Since the atoms touch along the body diagonal, we have:
$\sqrt{3}a = 4r \Rightarrow a = \dfrac{4r}{\sqrt{3}}$
The number of atoms per unit cell in a BCC structure is $Z = 2$.
Volume of the unit cell is $V = a^3 = \left(\dfrac{4r}{\sqrt{3}}\right)^3 = \dfrac{64r^3}{3\sqrt{3}}$
Volume occupied by the atoms is $v = 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3$
Packing efficiency is given by:
$\text{Packing efficiency} = \dfrac{v}{V} \times 100$
$\text{Packing efficiency} = \dfrac{\dfrac{8}{3}\pi r^3}{\dfrac{64r^3}{3\sqrt{3}}} \times 100 = \dfrac{\sqrt{3}\pi}{8} \times 100$
Substituting the values of $\pi$ and $\sqrt{3}$:
$\text{Packing efficiency} \approx 68\ \%$
Answer: $68\ \%$
Hence, the correct answer is option: 'C'.