`Na` and `Mg` crystallize in bcc- and fcc-type crystals, the ratio of number of atoms present in the unit cell of their respective crystal is

Options

A. 4 and 2
B. 9 and 14
C. 14 and 9
D. 2 and 4

Explanation

To solve the problem, we need to determine the number of atoms present in the unit cells of sodium (Na) and magnesium (Mg) in their respective crystal structures, which are body-centered cubic (BCC) and face-centered cubic (FCC).

### Step-by-Step Solution:

1. **Identify the Crystal Structure of Sodium (Na)**:
- Sodium crystallizes in a BCC structure.
- In a BCC unit cell, there are atoms located at the eight corners and one atom at the body center.

2. **Calculate the Contribution of Atoms in BCC**:
- Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell.
- There are 8 corners, so the contribution from the corners is:
\[ \text{Contribution from corners} = 8 \times \frac{1}{8} = 1 \text{ atom} \]
- The body-centered atom contributes fully (1 atom).
- Therefore, the total number of sodium atoms in the BCC unit cell is:
\[ \text{Total Na atoms} = 1 + 1 = 2 \text{ atoms} \]

3. **Identify the Crystal Structure of Magnesium (Mg)**:
- Magnesium crystallizes in an FCC structure.
- In an FCC unit cell, there are atoms located at the eight corners and six face centers.

4. **Calculate the Contribution of Atoms in FCC**:
- Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell.
- There are 8 corners, so the contribution from the corners is:
\[ \text{Contribution from corners} = 8 \times \frac{1}{8} = 1 \text{ atom} \]
- Each face-centered atom contributes \( \frac{1}{2} \) of an atom to the unit cell.
- There are 6 face centers, so the contribution from the face centers is:
\[ \text{Contribution from face centers} = 6 \times \frac{1}{2} = 3 \text{ atoms} \]
- Therefore, the total number of magnesium atoms in the FCC unit cell is:
\[ \text{Total Mg atoms} = 1 + 3 = 4 \text{ atoms} \]

5. **Calculate the Ratio of Atoms in the Unit Cells**:
- We have found that there are 2 sodium atoms and 4 magnesium atoms in their respective unit cells.
- The ratio of the number of atoms present in the unit cell of sodium to magnesium is:
\[ \text{Ratio} = \frac{\text{Number of Na atoms}}{\text{Number of Mg atoms}} = \frac{2}{4} = \frac{1}{2} \]

### Final Answer:
The ratio of the number of atoms present in the unit cell of sodium to magnesium is \( 1:2 \).