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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

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

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...