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# The edge length of cubic unit cell of metallic calcium is $556\;pm$ If the density of calcium is $1.54 \;g\;cm^{-3}$ and its molar mass is $40.08];g\;mol^{-1}$ calculate the radius of calcium atom.

$\begin{array}{1 1} 135.5\;pm \\ 415\;pm \\ 678\;pm \\ 196.6\;pm \end{array}$

Answer :$196.6\;pm$
From the expression $\rho = \large\frac{N}{a^3} \bigg( \large\frac{M}{N_A}\bigg)$
We get,
$N= \large\frac{\rho a^3 N_A}{M}$
$\qquad= \large\frac{(1.54 \;g\;cm^{-3})(556 \times 10^{-12}m)^3(6.023 \times 10^{23} \;mol^{-1} )}{(40.08\;g\;mol^{-1} )}$$=3.98 \approx 4$
For $N=4$ the unit cell is face- centered cube.
Since atoms touch each other along the face diagonal in a face - centred cubic unit cell, we have
$4r= \sqrt 2 a$
or $r = \large\frac{a}{2\sqrt 2} =\frac{556 \;pm}{2 \sqrt 2}$
$\qquad = 196.6\;pm$