Question:
The ratio of effective number of atoms in a unit cell of FCC and BCC lattices is:
The ratio of effective number of atoms in a unit cell of FCC and BCC lattices is:
Show Hint
Remember the number of atoms per unit cell:
\[
\text{Simple Cubic (SC)} = 1
\]
\[
\text{Body Centered Cubic (BCC)} = 2
\]
\[
\text{Face Centered Cubic (FCC)} = 4
\]
These values are frequently used in solid-state chemistry problems.
Updated On: Jun 26, 2026
- \(1:2\)
- \(4:1\)
- \(1:4\)
- \(2:1\)
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The Correct Option is D
Solution and Explanation
Step 1: Calculate the effective number of atoms in an FCC unit cell.
In a Face Centered Cubic (FCC) unit cell, atoms are present at: \[ 8 \text{ corners} \] and \[ 6 \text{ face centres} \] Contribution of corner atoms: \[ 8\times \frac{1}{8}=1 \] Contribution of face-centred atoms: \[ 6\times \frac{1}{2}=3 \] Therefore, the effective number of atoms in an FCC unit cell is \[ 1+3=4 \] \[ N_{FCC}=4 \]
Step 2: Calculate the effective number of atoms in a BCC unit cell.
In a Body Centered Cubic (BCC) unit cell, atoms are present at: \[ 8 \text{ corners} \] and \[ 1 \text{ body centre} \] Contribution of corner atoms: \[ 8\times \frac{1}{8}=1 \] Contribution of body-centred atom: \[ 1\times 1=1 \] Therefore, the effective number of atoms in a BCC unit cell is \[ 1+1=2 \] \[ N_{BCC}=2 \]
Step 3: Find the required ratio.
\[ N_{FCC}:N_{BCC} = 4:2 \] Dividing both terms by \(2\), \[ N_{FCC}:N_{BCC} = 2:1 \]
Step 4: Final conclusion.
Therefore, the ratio of effective number of atoms in FCC and BCC unit cells is \[ \boxed{2:1} \] Hence, the correct option is \[ \boxed{(4)} \]
In a Face Centered Cubic (FCC) unit cell, atoms are present at: \[ 8 \text{ corners} \] and \[ 6 \text{ face centres} \] Contribution of corner atoms: \[ 8\times \frac{1}{8}=1 \] Contribution of face-centred atoms: \[ 6\times \frac{1}{2}=3 \] Therefore, the effective number of atoms in an FCC unit cell is \[ 1+3=4 \] \[ N_{FCC}=4 \]
Step 2: Calculate the effective number of atoms in a BCC unit cell.
In a Body Centered Cubic (BCC) unit cell, atoms are present at: \[ 8 \text{ corners} \] and \[ 1 \text{ body centre} \] Contribution of corner atoms: \[ 8\times \frac{1}{8}=1 \] Contribution of body-centred atom: \[ 1\times 1=1 \] Therefore, the effective number of atoms in a BCC unit cell is \[ 1+1=2 \] \[ N_{BCC}=2 \]
Step 3: Find the required ratio.
\[ N_{FCC}:N_{BCC} = 4:2 \] Dividing both terms by \(2\), \[ N_{FCC}:N_{BCC} = 2:1 \]
Step 4: Final conclusion.
Therefore, the ratio of effective number of atoms in FCC and BCC unit cells is \[ \boxed{2:1} \] Hence, the correct option is \[ \boxed{(4)} \]
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