Soru
(c) Molybdenum has a BCC (Body Centred Cubic)crystal structure, has an atomic radius of 0.1363 nmand atomic weight of 95.94g/mol Avogadro's number is 6.02times 10^23 atoms/mole and 1 nm is 10^-9m Compute the theoretical density of molybdenum.
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To compute the theoretical density of molybdenum, we need to find the mass of the unit cell and divide it by the volume of the unit cell.The BCC unit cell has two atoms per unit cell. The mass of the unit cell can be calculated using the atomic weight of molybdenum:Mass of unit cell = 2 × Atomic weight of molybdenum / Avogadro's numberMass of unit cell = 2 × 95.94 g/mol / 6.02 × 10^23 atoms/molMass of unit cell = 1.59 × 10^-22 gNext, we need to find the volume of the unit cell. In a BCC structure, the length of the body diagonal is equal to four times the atomic radius. Therefore, the length of the body diagonal is:Body diagonal length = 4 × Atomic radiusBody diagonal length = 4 × 0.1363 nmBody diagonal length = 0.5452 nmThe volume of the unit cell can be calculated using the formula for the volume of a cube:Volume of unit cell = (Body diagonal length / √3)^3Volume of unit cell = (0.5452 nm / √3)^3Volume of unit cell = 0.0824 nm^3Finally, we can calculate the theoretical density of molybdenum:Theoretical density = Mass of unit cell / Volume of unit cellTheoretical density = 1.59 × 10^-22 g / 0.0824 nm^3Theoretical density = 1.93 × 10^23 g/m^3Therefore, the theoretical density of molybdenum is 1.93 × 10^23 g/m^3.