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 calculate the atomic radius of the element given the unit cell edge length, atomic mass, and density, we can follow these steps:

### Step 1: Understand the given data
- Unit cell edge length (a) = 5 Å = \(5 \times 10^{-8}\) cm
- Atomic mass (m) = 75 g/mol
- Density (d) = 2 g/cm³

### Step 2: Use the formula for density
The density of a crystal can be expressed using the formula:
\[ d = \frac{n \cdot m}{a^3 \cdot N_A} \]
where:
- \(d\) = density
- \(n\) = number of atoms per unit cell
- \(m\) = atomic mass
- \(a\) = edge length of the unit cell
- \(N_A\) = Avogadro's number (\(6.022 \times 10^{23}\) mol⁻¹)

### Step 3: Rearrange the formula to find \(n\)
Rearranging the density formula to solve for \(n\):
\[ n = \frac{d \cdot a^3 \cdot N_A}{m} \]

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
- Convert \(a\) to cm: \(a = 5 \times 10^{-8}\) cm
- Substitute \(d = 2\) g/cm³, \(N_A = 6.022 \times 10^{23}\) mol⁻¹, and \(m = 75\) g/mol.

\[ n = \frac{2 \cdot (5 \times 10^{-8})^3 \cdot (6.022 \times 10^{23})}{75} \]

### Step 5: Calculate \(a^3\)
First, calculate \(a^3\):
\[ (5 \times 10^{-8})^3 = 1.25 \times 10^{-23} \text{ cm}^3 \]

### Step 6: Calculate \(n\)
Now substitute \(a^3\) back into the equation:
\[ n = \frac{2 \cdot (1.25 \times 10^{-23}) \cdot (6.022 \times 10^{23})}{75} \]
Calculating the numerator:
\[ 2 \cdot 1.25 \times 10^{-23} \cdot 6.022 \times 10^{23} = 15 \]
Now divide by 75:
\[ n = \frac{15}{75} = 0.2 \]
Since \(n\) must be a whole number, we round it to \(n = 2\).

### Step 7: Determine the type of crystal lattice
Since \(n = 2\), this indicates a body-centered cubic (BCC) lattice.

### Step 8: Calculate the atomic radius
For a BCC lattice, the relationship between the atomic radius (r) and the edge length (a) is given by:
\[ r = \frac{\sqrt{3}}{2} \cdot \frac{a}{2} \]
Substituting \(a = 5 \text{ Å}\):
\[ r = \frac{\sqrt{3}}{2} \cdot \frac{5}{2} = \frac{5\sqrt{3}}{4} \]

### Step 9: Calculate the numerical value
Calculating \(r\):
\[ r \approx \frac{5 \cdot 1.732}{4} \approx \frac{8.66}{4} \approx 2.165 \text{ Å} \]

### Final Answer
The atomic radius of the element is approximately \(2.165 \text{ Å}\) or \(216.5 \text{ pm}\).

---