Question

A sample of $Ne$ is originally 20% by moles $N{e}^{20}$ isotope and the rest is $N{e}^{24}$ isotopes. In how many steps of effusion can a 33.33% ratio of $N{e}^{20}$ be achieved?

Answer 1

${\left(\frac{n_A}{n_B}\right)}_2={\left(\frac{n_A}{n_B}\right)}_1\times {\left(\frac{M_B}{M_A}\right)}^{\frac{1}{2}}{\left(\frac{n_A}{n_B}\right)}_n={\left(\frac{n_A}{n_B}\right)}_1\times {\left(\frac{M_B}{M_A}\right)}^{\frac{n-1}{2}}\Rightarrow \frac{1}{2}=\frac{1}{4}\times {\left(\frac{24}{20}\right)}^{\frac{n-1}{2}}\Rightarrow {\left(\frac{24}{20}\right)}^{\frac{n-1}{2}}=2({1.2}^4\approx 2)\Rightarrow \frac{n-1}{2}=4\Rightarrow n=9$