An element (atomic mass=250 u) crystallizes in a simple cubic. If the density of the unit cell is `7.2 g cm^(-3)`,what is the radius of the element ?

Options

A. `1.93times10^(-3)cm`
B. `1.93times10^(-8) cm`
C. `1.93times10^(-8)Å`
D. `1.93times10^(-8)m`

Explanation

To find the radius of the element that crystallizes in a simple cubic structure, we can follow these steps:

### Step 1: Understand the relationship between density, mass, and edge length
The formula for density (D) of a unit cell is given by:
\[ D = \frac{Z \cdot M}{N_A \cdot A^3} \]
where:
- \( D \) = density of the unit cell (in g/cm³)
- \( Z \) = number of atoms per unit cell
- \( M \) = molar mass of the element (in g/mol)
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \) mol⁻¹)
- \( A \) = edge length of the unit cell (in cm)

### Step 2: Identify the values
For a simple cubic structure:
- \( Z = 1 \) (since there is one atom per unit cell)
- \( M = 250 \, \text{u} = 250 \, \text{g/mol} \)
- \( D = 7.2 \, \text{g/cm}^3 \)

### Step 3: Rearrange the density formula to find the edge length (A)
Rearranging the formula gives:
\[ A^3 = \frac{Z \cdot M}{N_A \cdot D} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ A^3 = \frac{1 \cdot 250 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 7.2 \, \text{g/cm}^3} \]

### Step 5: Calculate \( A^3 \)
Calculating the right-hand side:
\[ A^3 = \frac{250}{6.022 \times 10^{23} \cdot 7.2} \]

Calculating the denominator:
\[ 6.022 \times 10^{23} \cdot 7.2 \approx 4.334 \times 10^{24} \]

Thus:
\[ A^3 \approx \frac{250}{4.334 \times 10^{24}} \approx 5.77 \times 10^{-23} \, \text{cm}^3 \]

### Step 6: Find the edge length (A)
Taking the cube root:
\[ A \approx (5.77 \times 10^{-23})^{1/3} \approx 3.86 \times 10^{-8} \, \text{cm} \]

### Step 7: Calculate the radius (r)
For a simple cubic structure, the relationship between the edge length and the radius is:
\[ r = \frac{A}{2} \]
Substituting the value of \( A \):
\[ r = \frac{3.86 \times 10^{-8}}{2} \approx 1.93 \times 10^{-8} \, \text{cm} \]

### Final Answer
The radius of the element is approximately:
\[ r \approx 1.93 \times 10^{-8} \, \text{cm} \]

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