Sodium metal crystllizes in a body centred cubic lattice with a unit cell edge of `4.56Å`. The radius of sodium atom is approximately

Options

A. `5.72 Å`
B. `1.98Å`
C. `1.86 Å`
D. `3.22 Å`

Explanation

To find the radius of a sodium atom in a body-centered cubic (BCC) lattice, we can follow these steps:

### Step 1: Understand the BCC Structure
In a body-centered cubic lattice, atoms are located at each corner of the cube and one atom is present at the center of the cube.

### Step 2: Identify the Unit Cell Edge Length
The edge length (a) of the unit cell is given as \(4.56 \, \text{Å}\).

### Step 3: Relate the Body Diagonal to Atomic Radius
In a BCC lattice, the body diagonal of the cube can be expressed in terms of the atomic radius (R). The body diagonal spans across three atomic radii:
\[ \text{Body diagonal} = 4R \]
The length of the body diagonal can also be expressed using the edge length (a) as:
\[ \text{Body diagonal} = a\sqrt{3} \]

### Step 4: Set Up the Equation
Equating the two expressions for the body diagonal gives:
\[ 4R = a\sqrt{3} \]

### Step 5: Solve for the Radius (R)
Substituting the value of a into the equation:
\[ R = \frac{a\sqrt{3}}{4} \]
Now substituting \(a = 4.56 \, \text{Å}\):
\[ R = \frac{4.56 \times \sqrt{3}}{4} \]

### Step 6: Calculate the Value
Calculating \(\sqrt{3} \approx 1.732\):
\[ R = \frac{4.56 \times 1.732}{4} \]
Calculating the numerator:
\[ R \approx \frac{7.90632}{4} \approx 1.97658 \, \text{Å} \]
Rounding this value gives:
\[ R \approx 1.98 \, \text{Å} \]

### Step 7: Conclusion
Thus, the radius of the sodium atom is approximately \(1.98 \, \text{Å}\).

### Final Answer
The radius of the sodium atom is approximately \(1.98 \, \text{Å}\).

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