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Calculate the radius of molybdenum atom if the element crystallizes as body-centred cubic crystals.Given : Density of $Mo=10.3gcm^{-3}$ and molar mass of $Mo=95.94gmol^{-1}$ .

$\begin{array}{1 1}135.5pm\\125.5pm\\313pm\\145.5pm\end{array} $

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Answer: 135.5pm
In a body-centred cubic unit cell,there are two atoms per unit cell.Hence,from the expression
$r=\large\frac{N}{a^3}\big(\frac{M}{N_A}\big)$
We get $a^3=\large\frac{N}{\rho}\big(\frac{M}{N_A}\big)=\big(\frac{2}{10.3gcm^{-3}}\big)\big(\frac{95.94gmol^{-1}}{6.023\times 10^{23}mol^{-1}}\big)$$=3.09\times 10^{-23}cm^3$
$a=3.13\times 10^{-8}cm=313$pm
Now since in the body-centred cubic unit cell,atoms touch other along the cross diagonal of the cube we have
$4r=\sqrt 3a$ or $\;\;r=(\sqrt 3)(313pm)/4=135.5$pm
answered Jul 16, 2014 by sreemathi.v
 

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