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30/12/2021 6:13 pm
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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.
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30/12/2021 6:22 pm
It is given that the density of niobium, d = 8.55 g cm-3
Atomic mass, M = 93 gmol-1
As the lattice is bcc type, the number of atoms per unit cell, z = 2
We also know that, NA = 6.022 × 1023 mol-1
Applying the relation:
d = \(\frac{zM}{a^3N_A}\)
⇒ a3 = \(\frac{zM}{d N_A}\)
= \(\frac{2 \times 93 gmol^{-1}}{8.55gcm^{-3} \times 6.022 \times 10^{23}mol^{-1}}\)
= 3.612 × 10-23 cm3
So, a = 3.306 × 10-8 cm
For body-centred cubic unit cell:
r = \(\frac{\sqrt 3}{4}a\)
= \(\frac{\sqrt 3}{4}\)x 3.306 x 10-8 cm
= 1.432 × 10-8 cm
= 14.32 × 10-9 cm
= 14.32 nm