Sodium has body centred packing. Distance between two nearest atoms is `3.7 Å`. The lattice parameter is

Options

A. `8.6Å`
B. 6.8Å
C. 4.3Å
D. 3.0Å

Explanation

To find the lattice parameter of sodium, which has a body-centered cubic (BCC) structure, we can follow these steps:

### Step 1: Understand the relationship between the nearest neighbor distance and lattice parameter
In a body-centered cubic (BCC) structure, the nearest neighbor distance \( d \) is related to the lattice parameter \( a \) by the formula:
\[ d = \frac{\sqrt{3}}{2} a \]

### Step 2: Substitute the given value of nearest neighbor distance
We are given that the distance between two nearest atoms is \( d = 3.7 \, \text{Å} \). We can substitute this value into the equation:
\[ 3.7 = \frac{\sqrt{3}}{2} a \]

### Step 3: Solve for the lattice parameter \( a \)
To isolate \( a \), we can rearrange the equation:
\[ a = \frac{2 \times 3.7}{\sqrt{3}} \]

### Step 4: Calculate the value of \( a \)
Now, we can calculate \( a \):
\[ a = \frac{7.4}{\sqrt{3}} \approx \frac{7.4}{1.732} \approx 4.27 \, \text{Å} \]

### Step 5: Round the answer
Rounding the answer gives us:
\[ a \approx 4.3 \, \text{Å} \]

### Final Answer
The lattice parameter \( a \) is approximately \( 4.3 \, \text{Å} \).

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