The atomic radius of strontium ( Sr ) is 215 pm and it crystallizes in FCC . Edge length of the cube is

Options

A. `430` pm
B. `608.2` pm
C. `496.53` pm
D. `304.1` pm

Answer

To find the edge length of the cube (a) for strontium (Sr) crystallizing in a face-centered cubic (FCC) structure, we can follow these steps:

### Step 1: Understand the relationship in FCC
In a face-centered cubic (FCC) structure, the atoms at the corners and the face centers are arranged in such a way that the diagonal of the face of the cube contains 4 atomic radii (4r). This is because there are two atomic radii from the corner atom to the face center atom and two more from the face center atom to the opposite corner atom.

### Step 2: Write the equation for the face diagonal
The relationship between the edge length (a) and the atomic radius (r) in an FCC structure is given by the equation:
\[ \text{Face diagonal} = \sqrt{2} \times a = 4r \]

### Step 3: Rearrange the equation to find edge length (a)
From the equation above, we can rearrange it to solve for the edge length (a):
\[ a = \frac{4r}{\sqrt{2}} \]

### Step 4: Substitute the value of r
Given that the atomic radius (r) of strontium (Sr) is 215 picometers (pm), we can substitute this value into the equation:
\[ a = \frac{4 \times 215 \, \text{pm}}{\sqrt{2}} \]

### Step 5: Calculate the edge length (a)
Now we can perform the calculation:
1. Calculate \( 4 \times 215 = 860 \, \text{pm} \)
2. Calculate \( \sqrt{2} \approx 1.414 \)
3. Now divide:
\[ a = \frac{860 \, \text{pm}}{1.414} \approx 608.2 \, \text{pm} \]

### Final Answer
The edge length of the cube (a) is approximately **608.2 pm**.

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