Question

For a molecule of benzene, $n=2.76\times {10}^{19}{\text{cm}}^{-3}$ and mean free path $\lambda =2.2\times {10}^{-10}\text{m}$. Calculate diameter of benzene molecule.

• A. $6.06\times {10}^{-9}m$
• B. $3.03\times {10}^{-9}m$
• C. $3.03\times {10}^{-7}m$
• D. $3.03\times {10}^{-11}m$

Answer 1

The correct option is A $6.06\times {10}^{-9}m$

Here, $n=2.76\times {10}^{19}{\text{cm}}^{-3}=2.76\times {10}^{25}{\text{m}}^{-3}$
$\lambda =2.2\times {10}^{-10}\text{m},\sigma =?$
As
$\lambda =\frac{1}{\sqrt{2}\pi {\sigma }^{2}n}$
$\therefore {\sigma }^2=\frac{1}{\sqrt{2}\pi \lambda n}$
$=\frac{1}{1.414\times 3.14\times 2.2\times {10}^{-10}\times 2.76\times {10}^{25}}$
${\sigma }^2=\frac{1000\times {10}^{-18}}{1.414\times 3.14\times 2.2\times 2.76}$
$=\sqrt{36.76\times {10}^{-18}}$
$\sigma =\sqrt{36.76\times {10}^{-18}}=6.06\times {10}^{-9}\text{m}$
Hence, diameter of benzene molecule is $6.06\times {10}^{-9}m$