Density Calculation for Molybdenum BCC Structure

To find the density ($\rho$) of Molybdenum (Mo) in a body-centered cubic (bcc) structure, we use the formula:

$ \rho = \frac{Z \times M}{V \times N_A} $

Where:

• $Z$ is the number of atoms per unit cell. For a bcc structure, $Z = 2$.
• $M$ is the molar mass of the element. Given as 96 g/mol.
• $V$ is the volume of the unit cell. For a cubic unit cell, $V = a^3$, where $a$ is the edge length.
• $N_A$ is Avogadro's number ($6.022 \times 10^{23}$ mol⁻¹).

Step-by-Step Calculation

1. Convert Units:
   • Unit cell edge length, $a = 0.314$ nm $= 0.314 \times 10^{-9}$ m.
   • Molar mass, $M = 96$ g/mol $= 96 \times 10^{-3}$ kg/mol.
2. Calculate Unit Cell Volume (V):

   $ V = a^3 = (0.314 \times 10^{-9} \text{ m})^3 $

   $ V \approx 0.030959 \times 10^{-27} \text{ m}^3 $

3. Calculate Density ($\rho$):

   Using the density formula with $Z=2$ for bcc:

   $ \rho = \frac{2 \times (96 \times 10^{-3} \text{ kg/mol})}{(0.030959 \times 10^{-27} \text{ m}^3) \times (6.022 \times 10^{23} \text{ mol}^{-1})} $

   $ \rho = \frac{0.192}{1.864 \times 10^{-4}} \text{ kg/m}^3 $

   $ \rho \approx 10300 \text{ kg/m}^3 $

The calculated density is approximately $10300$ kg/m³, which lies between 10000 and 10500 kg/m³.