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NUMERICAL) Niobium is found to crystallise with bee structurecm"and found to have density of 8.55g mDetermine the atomic radius of niobium if itsatomic mass is 93 (14 29 m)

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) = 2

we know, d = z×M/a³×N=> 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 = 1.429 × 10^-8 cm = 14.29 × 10^-9 m = 14.29 nm


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