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प्रश्न
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.
उत्तर
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 = (zM)/(a^3N_A)`
`=>a^3= (zM)/(dN_A)`
= `(2xx93 gmol^(-1))/(8.55 "gcm"^(-3)xx6.022xx10^(23) mol^(-1))`
= 3.612 × 10−23 cm3
So, a = 3.306 × 10−8 cm
For body-centred cubic unit cell:
`r = sqrt3/4a`
=`sqrt3/4xx3.306xx10^(-8) cm`
= 1.432 × 10−8 cm
= 14.32 × 10−9 cm
= 14.32 nm..
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