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

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

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