Question

A sample of $Ne$ is originally $10\% $ by mole in $N{e^{20}}$ isotopes and remaining are in $N{e^{22}}$ isotope. In $n$ diffusion steps $25\% $, $N{e^{20}}$ can be achieved. Find ($\dfrac{n}{6}$​) :
(Given\[\;{\mathbf{log3}} = {\mathbf{0}}.{\mathbf{48}},{\mathbf{log1}}.{\mathbf{1}} = {\mathbf{0}}.{\mathbf{04}}\])

A) 0
B) 1
C) 2
D) 3

Answer 1

Hint:Isotopes are the elements which have same Atomic number but different Mass number which implies that they will have similar chemical properties but different physical properties
As in the given question the isotopes will have different rate of diffusion which is a physical property depending upon molecular mass or mass number of a given element
Formula Used:
 \[final{\text{ }}purity{\text{ }} = {\text{ }}\left( {initial{\text{ }}purity} \right){\left( {separation} \right)^n}\]
Here $n$is the steps required for diffusion
initial purity is the fraction of a particular isotope present initially in the sample
final purity is the fraction of a particular isotope present finally in the sample
While, separation is the ratio of diffusion rate of the isotope whose purity is asked $\left( {{R_1}} \right)$ to the diffusion rate of the other isotope $\left( {{R_2}} \right)$
$separation = \dfrac{{{R_1}}}{{{R_2}}}$

Complete step by step answer:
As we all are familiar with Graham’s law of diffusion which states that “Under similar conditions of temperature and pressure the rates of diffusion is inversely proportional to the square root of their molecular weight”
Therefore in the given question the ratio of diffusion rates of both the isotope will be constant and is given by
$\dfrac{{{R_1}}}{{{R_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $

Where ${M_1}$ is molecular weight of isotope with diffusion rate ${R_1}$
And ${M_2}$ is molecular weight of isotope with diffusion rate ${R_2}$
This ratio is also known as separation and hence the value of separation can be calculated as follow
By using ${M_1} = 20\,and\,{M_2} = 22$
\[separation = \sqrt {\dfrac{{22}}{{20}}} \]
\[separation = \sqrt {1.1} \]
Also initial purity of $N{e^{20}}\,\operatorname{in} \,fraction\,is\,0.1$
While final purity of $N{e^{20}}\,\operatorname{in} \,fraction\,is\,0.25$
Let $n$ be the diffusion steps them using the below formula
\[final{\text{ }}purity{\text{ }} = {\text{ }}\left( {initial{\text{ }}purity} \right){\left( {separation} \right)^n}\]
$0.25 = \left( {0.1} \right){\left( {\sqrt {1.1} } \right)^n}$
Taking $\log $ both sides we get
$\log 0.25 = \log 0.1 + \dfrac{n}{2}\log 1.1$
$1 - \log 4 = \dfrac{n}{2}\log 1.1$
On solving further we get $n = 19.23$
And finally $\dfrac{n}{6} \approx 3$
Hence the correct option for the given question is ‘D’.

Note:
Rate of diffusion can also be expressed in terms of vapour density
Using $vapour\,density = \dfrac{{Molecular\,weight}}{2}$

Hence separation can also be defined as the ratio of square root of vapour density $\left( {{D_2}} \right)$ of second isotope to the square root of vapour density $\left( {{D_1}} \right)$ of the first isotope
\[separation = \left( {\sqrt {\dfrac{{{D_2}}}{{{D_1}}}} } \right)\]