`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. `1`
B. `0.5`
C. `3`
D. `4`

Explanation

To solve the problem of finding the ratio of the number of atoms present in the unit cell of sodium (Na) and magnesium (Mg), we will analyze the crystal structures of both elements.

### Step-by-Step Solution:

1. **Identify the Crystal Structures**:
- Sodium (Na) crystallizes in a body-centered cubic (BCC) structure.
- Magnesium (Mg) crystallizes in a face-centered cubic (FCC) structure.

2. **Calculate the Number of Atoms in the BCC Unit Cell (Na)**:
- In a BCC unit cell, there are:
- 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- 1 atom at the body center contributing 1 atom.
- Total contribution from corner atoms:
\[ \text{Contribution from corners} = 8 \times \frac{1}{8} = 1 \]
- Total contribution from the body center:
\[ \text{Contribution from body center} = 1 \]
- Therefore, the total number of atoms in the BCC unit cell for Na is:
\[ Z_{Na} = 1 + 1 = 2 \]

3. **Calculate the Number of Atoms in the FCC Unit Cell (Mg)**:
- In an FCC unit cell, there are:
- 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom.
- 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom.
- Total contribution from corner atoms:
\[ \text{Contribution from corners} = 8 \times \frac{1}{8} = 1 \]
- Total contribution from face-centered atoms:
\[ \text{Contribution from face centers} = 6 \times \frac{1}{2} = 3 \]
- Therefore, the total number of atoms in the FCC unit cell for Mg is:
\[ Z_{Mg} = 1 + 3 = 4 \]

4. **Calculate the Ratio of Atoms**:
- Now, we need to find the ratio of the number of atoms in the unit cell of Na to that in Mg:
\[ \text{Ratio} = \frac{Z_{Na}}{Z_{Mg}} = \frac{2}{4} = \frac{1}{2} = 0.5 \]

### Final Answer:
The ratio of the number of atoms present in the unit cell of sodium (Na) to magnesium (Mg) is **0.5**.