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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Elit · 8 yıl öğretmeniUzman doğrulaması
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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.<br /><br />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:<br /><br />Mass of unit cell = 2 × Atomic weight of molybdenum / Avogadro's number<br /><br />Mass of unit cell = 2 × 95.94 g/mol / 6.02 × 10^23 atoms/mol<br /><br />Mass of unit cell = 1.59 × 10^-22 g<br /><br />Next, 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:<br /><br />Body diagonal length = 4 × Atomic radius<br /><br />Body diagonal length = 4 × 0.1363 nm<br /><br />Body diagonal length = 0.5452 nm<br /><br />The volume of the unit cell can be calculated using the formula for the volume of a cube:<br /><br />Volume of unit cell = (Body diagonal length / √3)^3<br /><br />Volume of unit cell = (0.5452 nm / √3)^3<br /><br />Volume of unit cell = 0.0824 nm^3<br /><br />Finally, we can calculate the theoretical density of molybdenum:<br /><br />Theoretical density = Mass of unit cell / Volume of unit cell<br /><br />Theoretical density = 1.59 × 10^-22 g / 0.0824 nm^3<br /><br />Theoretical density = 1.93 × 10^23 g/m^3<br /><br />Therefore, the theoretical density of molybdenum is 1.93 × 10^23 g/m^3.
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