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InterviewSolution
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Niobium is found to crystallise with bcc structureand found to have density of 8.55g cm-3.Determine the atomic radius of niobium if itsatomic mass is 93 (14.29 nm) |
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Answer» Given density (d) = 8.55 g/cm³Atomic mass (M) = 93u = 93 g/molWe know, Avogadro number, N= 6.022 × 10²³given lattice is body - centered cubic,so, number of atoms per unit cell (z) = 2we know,\bf{d=\frac{zM}{a^3N}}d=a3NzM=> 8.55 = (2 × 93)/(a³ × 6.023 × 10²³)=> 8.55 × a³ × 6.023 × 10²³ = 186=> a³ = 186/(8.55 × 6.023 × 10²³)=> a³ = 36.124 × 10^-24=> a = 3.3057 × 10^-8 cm for BCC unit cell, radius = √3a/4= (√3 × 3.3057 × 10^-8)/4= (1.732 × 3.3057)/4 × 10^-8= 1.431 × 10^-8 cm = 1.431 × 10^-10 m= 1.431 A° hence, atomic radius of Niobium is 1.431A |
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