The atomic radius of strontium `(Sr)` is `215 p m` and it crystallizes with a cubic close packing . Edge length of the cube is `:`

Options

A. 4.30 pm
B. 608.02 pm
C. 496.53 pm
D. none of these

Answer

To find the edge length of the cube for strontium (Sr) which crystallizes in a cubic close packing (FCC), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**:
- Strontium crystallizes in a face-centered cubic (FCC) arrangement. In this arrangement, atoms are located at the corners and the centers of each face of the cube.

2. **Determine the Number of Atoms in FCC**:
- In an FCC unit cell, there are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell, and 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom.
- Total contribution from corner atoms: \( 8 \times \frac{1}{8} = 1 \)
- Total contribution from face-centered atoms: \( 6 \times \frac{1}{2} = 3 \)
- Therefore, the total number of atoms (Z) in an FCC unit cell is \( 1 + 3 = 4 \).

3. **Relate Atomic Radius to Edge Length**:
- The relationship between the atomic radius (R) and the edge length (a) of the cube in FCC is given by the formula:
\[ a = 2\sqrt{2}R \]

4. **Substitute the Given Atomic Radius**:
- The atomic radius of strontium (R) is given as \( 215 \, \text{pm} \).
- Substitute R into the formula:
\[ a = 2\sqrt{2} \times 215 \, \text{pm} \]

5. **Calculate the Edge Length**:
- Calculate \( \sqrt{2} \):
\[ \sqrt{2} \approx 1.414 \]
- Now calculate \( a \):
\[ a = 2 \times 1.414 \times 215 \, \text{pm} \]
\[ a \approx 2 \times 1.414 \times 215 \approx 608.02 \, \text{pm} \]

6. **Final Result**:
- The edge length of the cube is approximately \( 608.02 \, \text{pm} \).

### Final Answer:
The edge length of the cube is \( 608.02 \, \text{pm} \).

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