Percentage of free space in a body centred cubic unit cell is
A. 30%
B. 32%
C. 34%
D. 28%
Answer
544.6k+ views
Hint: Unit cells are basically the smallest repeating atoms in a crystal. There are different types of unit cell like body centred unit cell represented as BCC, face centred unit cell (FCC). BCC has atoms at each corner of a cube that is total 8 and one atom in the centre of the cube. The percentage of free space can be calculated with the formula: Percentage of Free space = 100 - packing efficiency.
Complete step by step answer:
- Firstly we will see the arrangement of atoms in bcc. There are atoms present at each corner and one at the centre of the cube.
- If we will see the percentage of free space = 100- packing efficiency.
- So, for this we should know what is the packing efficiency of a body centred unit cell.
- Body centred unit cells contain two atoms per unit cell. That is Z = 2;
- And there is a relation in between r value and edge length (a) that is:
\[r=\frac{\sqrt{3}}{4}a\]
- So, we know that packing efficiency =
\[packing\text{ }efficiency=\frac{Z\times volume\text{ }of\text{ }sphere}{volume\text{ }of\text{ }cube}\]
Number of atoms Z = 2,
Volume of sphere = $\frac{4}{3}\pi {{r}^{3}}$
Volume of cube = ${{a}^{3}}$
\[=\frac{2\times \frac{4}{3}\pi {{r}^{3}}}{{{a}^{3}}}\]
- We can convert r into a, and do further calculation:
\[\begin{align}
& ={{\frac{2\times \frac{4}{3}\times \pi \times \left[ \frac{\sqrt{3}}{4}a \right]}{{{a}^{3}}}}^{3}} \\
& =2\times \frac{4}{3}\times \frac{22}{7}\times \frac{\sqrt{3}\times \sqrt{3}\times \sqrt{3}}{4\times 4\times 4} \\
& = 0.68 \\
\end{align}\]
Now,
\[\begin{align} & 0.68\times 100 \\ & =68 \text{%} \\ \end{align}\]
- Now the percentage of free space will be = 100 - packing efficiency
= 100 - 68
= 32%
Hence, we can conclude that the correct option is (B) that is the percentage of free space in a body centred cubic unit cell is 32%.
Note: We must not get confused in terms of FCC and BCC. FCC is face centred cubic unit cell, it contains four atoms per unit cell. Whereas, BCC is a body centred unit cell, it contains 2 atoms per unit cell.
Complete step by step answer:
- Firstly we will see the arrangement of atoms in bcc. There are atoms present at each corner and one at the centre of the cube.
- If we will see the percentage of free space = 100- packing efficiency.
- So, for this we should know what is the packing efficiency of a body centred unit cell.
- Body centred unit cells contain two atoms per unit cell. That is Z = 2;
- And there is a relation in between r value and edge length (a) that is:
\[r=\frac{\sqrt{3}}{4}a\]
- So, we know that packing efficiency =
\[packing\text{ }efficiency=\frac{Z\times volume\text{ }of\text{ }sphere}{volume\text{ }of\text{ }cube}\]
Number of atoms Z = 2,
Volume of sphere = $\frac{4}{3}\pi {{r}^{3}}$
Volume of cube = ${{a}^{3}}$
\[=\frac{2\times \frac{4}{3}\pi {{r}^{3}}}{{{a}^{3}}}\]
- We can convert r into a, and do further calculation:
\[\begin{align}
& ={{\frac{2\times \frac{4}{3}\times \pi \times \left[ \frac{\sqrt{3}}{4}a \right]}{{{a}^{3}}}}^{3}} \\
& =2\times \frac{4}{3}\times \frac{22}{7}\times \frac{\sqrt{3}\times \sqrt{3}\times \sqrt{3}}{4\times 4\times 4} \\
& = 0.68 \\
\end{align}\]
Now,
\[\begin{align} & 0.68\times 100 \\ & =68 \text{%} \\ \end{align}\]
- Now the percentage of free space will be = 100 - packing efficiency
= 100 - 68
= 32%
Hence, we can conclude that the correct option is (B) that is the percentage of free space in a body centred cubic unit cell is 32%.
Note: We must not get confused in terms of FCC and BCC. FCC is face centred cubic unit cell, it contains four atoms per unit cell. Whereas, BCC is a body centred unit cell, it contains 2 atoms per unit cell.
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