The unit cell of Na is bcc and its density is 0.97 g/c`m^(3)`. What is the radius of a sodium atom if the molar mass of Na is 23 g/mol?

Answer

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

### Step 1: Understand the relationship between density, molar mass, and edge length.
The density (\(d\)) of a crystal can be expressed using the formula:
\[ d = \frac{Z \cdot M}{N_A \cdot A^3} \]
where:
- \(Z\) = number of atoms per unit cell (for BCC, \(Z = 2\))
- \(M\) = molar mass (for Na, \(M = 23 \, \text{g/mol}\))
- \(N_A\) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))
- \(A\) = edge length of the unit cell (in cm)

### Step 2: Rearrange the formula to solve for edge length \(A\).
Rearranging the density formula gives:
\[ A^3 = \frac{Z \cdot M}{d \cdot N_A} \]
Substituting the known values:
- \(Z = 2\)
- \(M = 23 \, \text{g/mol}\)
- \(d = 0.97 \, \text{g/cm}^3\)
- \(N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}\)

### Step 3: Plug in the values to calculate \(A^3\).
\[ A^3 = \frac{2 \cdot 23}{0.97 \cdot 6.022 \times 10^{23}} \]
Calculating this gives:
\[ A^3 = \frac{46}{5.83234 \times 10^{23}} \approx 7.89 \times 10^{-23} \, \text{cm}^3 \]

### Step 4: Calculate the edge length \(A\).
Taking the cube root:
\[ A = (7.89 \times 10^{-23})^{1/3} \approx 4.286 \times 10^{-8} \, \text{cm} \]

### Step 5: Use the relationship between radius \(r\) and edge length \(A\) for BCC.
In a BCC structure, the relationship between the radius \(r\) and the edge length \(A\) is given by:
\[ r = \frac{\sqrt{3}}{4} A \]

### Step 6: Substitute \(A\) into the formula to find \(r\).
\[ r = \frac{\sqrt{3}}{4} \cdot 4.286 \times 10^{-8} \, \text{cm} \]
Calculating this gives:
\[ r \approx 1.86 \times 10^{-8} \, \text{cm} \]

### Final Answer:
The radius of a sodium atom is approximately \(1.86 \times 10^{-8} \, \text{cm}\).

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