To determine the density of vanadium (V) in a bcc structure, we can follow these steps:
1) Calculate the volume of the unit cell:
- The edge length of the bcc unit cell is given as 4r/√3, where r is the atomic radius.
- Plugging in the value, we have: edge length = 4(131 pm)/√3 = 4(131 pm)/1.732 = 379.6 pm.
- Convert the edge length to meters: 379.6 pm * (1 m / 10^12 pm) = 3.796 x 10^-10 m.
- The volume of a bcc unit cell is (edge length)^3.
- Volume = (3.796 x 10^-10 m)^3 = 5.744 x 10^-29 m^3.
2) Determine the number of atoms in the bcc unit cell:
- In a bcc structure, there are 2 atoms per unit cell.
- Therefore, the number of atoms in the bcc unit cell is 2.
3) Calculate the volume occupied by one atom:
- Since there are 2 atoms in the unit cell, each atom occupies half of the unit cell volume.
- Volume per atom = Volume of unit cell / Number of atoms = 5.744 x 10^-29 m^3 / 2 = 2.872 x 10^-29 m^3.
4) Calculate the density:
- The density is defined as mass divided by volume.
- Mass of one atom = atomic mass / Avogadro's number (NA).
- Mass per atom = 50.94 g/mol / (6.022 x 10^23 mol^-1) = 8.467 x 10^-23 g.
- Convert the mass to kilograms: 8.467 x 10^-23 g * (1 kg / 1000 g) = 8.467 x 10^-26 kg.
- Density = Mass per atom / Volume per atom = 8.467 x 10^-26 kg / 2.872 x 10^-29 m^3 = 2.947 x 10^3 kg/m^3.
Therefore, the density of vanadium (V) in a bcc structure is approximately 2.947 x 10^3 kg/m^3