Question

Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm −3 , calculate atomic radius of niobium using its atomic mass 93 u.

Answer 1

Given density (d) = 8.55 g/cm³
Atomic mass (M) = 93u = 93 g/mol
We know, Avogadro number, N = 6.022 × 10²³
given lattice is body - centered cubic,
so, number of atoms per unit cell (z) = 2
we know, $\bf{d=\frac{zM}{a^3N}}$
=> 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°.

Answer 2

Answer:

Explanation:

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