For a unit cell edge length `= 5 Å`, the element is of atomic mass 75, has denstiy of 2gm/cc. Calculate atomic radius of the element

Answer

To solve the problem, we need to calculate the atomic radius of the element given its unit cell edge length, atomic mass, and density. Here’s a step-by-step breakdown of the solution:

### Step 1: Write down the known values
- Edge length (a) = 5 Å = \(5 \times 10^{-8}\) cm (since 1 Å = \(10^{-8}\) cm)
- Atomic mass (M) = 75 g/mol
- Density (d) = 2 g/cm³
- Avogadro's number (Nₐ) = \(6.022 \times 10^{23}\) mol⁻¹

### Step 2: Use the density formula
The density of a substance can be expressed using the formula:
\[ d = \frac{Z \cdot M}{N_a \cdot a^3} \]
where:
- \(Z\) = number of atoms per unit cell
- \(M\) = molar mass
- \(N_a\) = Avogadro's number
- \(a\) = edge length of the unit cell

### Step 3: Rearranging the formula to find Z
Rearranging the formula to solve for \(Z\):
\[ Z = \frac{d \cdot N_a \cdot a^3}{M} \]

### Step 4: Substitute the known values into the equation
Substituting the known values:
\[ Z = \frac{2 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} \cdot (5 \times 10^{-8} \, \text{cm})^3}{75 \, \text{g/mol}} \]

### Step 5: Calculate \(a^3\)
Calculate \(a^3\):
\[ (5 \times 10^{-8} \, \text{cm})^3 = 1.25 \times 10^{-22} \, \text{cm}^3 \]

### Step 6: Substitute \(a^3\) back into the equation
Now, substituting this back into the equation for \(Z\):
\[ Z = \frac{2 \cdot 6.022 \times 10^{23} \cdot 1.25 \times 10^{-22}}{75} \]

### Step 7: Calculate \(Z\)
Calculating \(Z\):
\[ Z = \frac{2 \cdot 6.022 \cdot 1.25}{75} \approx 2 \]

### Step 8: Determine the type of unit cell
Since \(Z = 2\), this indicates that the unit cell is a Body-Centered Cubic (BCC) structure.

### Step 9: Use the relationship between atomic radius and edge length for BCC
For a BCC unit cell, the relationship between atomic radius (r) and edge length (a) is given by:
\[ r = \frac{\sqrt{3}}{4} a \]

### Step 10: Substitute the edge length to find the atomic radius
Substituting the edge length into the equation:
\[ r = \frac{\sqrt{3}}{4} \cdot 5 \, \text{Å} = \frac{\sqrt{3} \cdot 5}{4} \approx 2.165 \, \text{Å} \]

### Final Answer
The atomic radius of the element is approximately **2.165 Å**.

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