Moldydenum `("At. mass=96g//mol"^(-1))` crystallizes as bcc crystal. If density of crystal is `10.3g//cm^(3)`, then radius of Mo atoms `(use N_(A)= 6 xx 10^(23))`:

Options

A. 111PM
B. 314PM
C. 135.96PM
D. none of these

Answer

To find the radius of molybdenum (Mo) atoms in a body-centered cubic (BCC) crystal structure, we can follow these steps:

### Step 1: Understand the relationship between density, atomic mass, and unit cell volume
The formula for density (ρ) is given by:
\[ \rho = \frac{Z \cdot m}{N_A \cdot a^3} \]
where:
- \( \rho \) = density of the crystal (10.3 g/cm³)
- \( Z \) = number of atoms per unit cell (for BCC, \( Z = 2 \))
- \( m \) = molar mass of the element (96 g/mol for Mo)
- \( N_A \) = Avogadro's number (\( 6 \times 10^{23} \) mol⁻¹)
- \( a \) = edge length of the unit cell in cm

### Step 2: Rearrange the formula to find \( a^3 \)
Rearranging the density formula gives us:
\[ a^3 = \frac{Z \cdot m}{\rho \cdot N_A} \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ a^3 = \frac{2 \cdot 96 \, \text{g/mol}}{10.3 \, \text{g/cm}^3 \cdot 6 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 4: Calculate \( a^3 \)
Calculating the numerator:
\[ 2 \cdot 96 = 192 \, \text{g/mol} \]
Calculating the denominator:
\[ 10.3 \cdot 6 \times 10^{23} = 61.8 \times 10^{23} \, \text{g/cm}^3 \cdot \text{mol}^{-1} \]
Now substituting these values:
\[ a^3 = \frac{192}{61.8 \times 10^{23}} \approx 3.1 \times 10^{-22} \, \text{cm}^3 \]

### Step 5: Calculate \( a \)
Taking the cube root:
\[ a \approx (3.1 \times 10^{-22})^{1/3} \approx 3.14 \times 10^{-8} \, \text{cm} = 3.14 \, \text{Å} = 314 \, \text{pm} \]

### Step 6: Relate edge length to atomic radius in BCC
For a BCC structure, the relationship between the edge length \( a \) and the atomic radius \( r \) is given by:
\[ a = \frac{4r}{\sqrt{3}} \]
Rearranging gives:
\[ r = \frac{\sqrt{3}}{4} a \]

### Step 7: Substitute \( a \) to find \( r \)
Substituting \( a \):
\[ r = \frac{\sqrt{3}}{4} \cdot 314 \, \text{pm} \approx 135.96 \, \text{pm} \]

### Final Answer
The radius of molybdenum atoms is approximately:
\[ \boxed{135.96 \, \text{pm}} \]