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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.

$\begin{array}{1 1}123pm\\143pm\\163pm\\173pm\end{array} $

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In a body centred cubic structure ,the number of atoms per unit cell = 2
Density =$\large\frac{z\times M}{a^3\times N_A\times 10^{-30}}$$gcm^{-3}$
$a^3=\large\frac{M\times z}{d\times N_A\times 10^{-30}}$$gcm^{-3}$
$\;\;\;\;=\large\frac{2\times 93}{6.02\times 10^{23}\times 10^{-30}\times 8.55}$
$\;\;\;\;=3.61\times 10^7$
$a=3.305\times 10^2pm=330.5pm$
But in body-centered cubic unit cell body diagonal is equal to 4 times the radius of atom.
$4r=\sqrt 3a=\sqrt 3\times 330.5$
$\therefore$ radius of niobium atom =143pm
answered Jul 30, 2014 by sreemathi.v

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