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The edge length of a cubic unit cell of metallic calcium is 556pm.If the density of calcium is $1.54gcm^{-3}$ and its molar mass is $40.08gmol^{-1}$,calculate the radius of calcium atom

$\begin{array}{1 1}186.6pm\\196.6pm\\176.6pm\\206.6pm\end{array} $

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Answer : 196.6pm
From the expression
$\rho=\large\frac{N}{a^3}\big(\frac{M}{N_A}\big)$
We get $N=\large\frac{\rho a^3N_A}{M}=\frac{(1.54gcm^{-3})(556\times 10^{-12}m)^3(6.023\times 10^{23} mol^{-1})}{(40.08gmol^{-1})}$$=3.98\simeq 4$
For N=4,the unit cell is face-centred cube.Since atoms touch each other along the face diagonal in a face-centred cubic unit cell,we have
$4r=\sqrt 2a$ or $\;\;r=\large\frac{a}{2\sqrt 2}=\frac{556pm}{2\sqrt 2}$$=196.6$pm
answered Jul 16, 2014 by sreemathi.v
 

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