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

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

1 Answer

From the expression $\rho=\large\frac{n}{a^3}\big(\large\frac{M}{N_A}\big)$
We get $n=\large\frac{\rho a^3N_A}{M}=\frac{1.54\times (556\times 10^{-10})^3\times 6.023\times 10^{23}}{40.08}$
$\Rightarrow 3.98\simeq 4$
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
$4r=\sqrt 2a$
$r=\large\frac{a}{\sqrt 2}$
$\;\;=\large\frac{556}{2\sqrt 2}$pm
$\;\;=196.6$pm
answered Aug 1, 2014 by sreemathi.v
 

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