Uranium has two naturally occurring isotopes. ${}^{238}\mathrm{U}$ has a natural abundance of $99.3\%$ and${}^{235}\mathrm{U}$ has an abundance of $0.7\%$. It is the rarer ${}^{235}\mathrm{U}$that is needed for nuclear reactors. The isotopes are separated by forming uranium hexafluoride, role="math" ${\mathrm{UF}}_6$, which is a gas, then allowing it to diffuse through a series of porous membranes. ${}^{235}\mathrm{UF}_6$ has a slightly larger rms speed than ${}^{238}\mathrm{UF}_6$ and diffuses slightly faster. Many repetitions of this procedure gradually separate the two isotopes. What is the ratio of the rms speed of ${}^{235}\mathrm{UF}_6$ to that of${}^{238}\mathrm{UF}_6$$?$

The average translational kinetic energy is

${\mathrm{ϵ}}_{\mathrm{avg}}=\frac{3}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}..........1$

The molecule with mass$\mathrm{m}$and velocity $\mathrm{v}$has an average translational kinetic energy,

${\mathrm{ϵ}}_{\mathrm{avg}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{rms}}^2..........2$

Equating equation$1$and$2$have the same left side,

Root mean square velocity ${\mathrm{v}}_{\text{rms}}$is,

$\frac{1}{2}{\mathrm{mv}}_{\mathrm{rms}}^2=\frac{3}{2}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}$

${\mathrm{v}}_{\mathrm{rms}}^2=\frac{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{m}}$

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{m}}}...................3$

The root square of the molar mass is inversely proportional to root mean square velocity.

localid="1648639940028" ${\mathrm{v}}_{\mathrm{rms}}\propto \frac{1}{\sqrt{\mathrm{m}}}$

The molar mass of fluorinelocalid="1648639957709" $\mathrm{F}$islocalid="1648639970840" $19\mathrm{u}$.

So, the molar mass of localid="1648639989462" ${\mathrm{UF}}_{238}$islocalid="1648639997841" $238\mathrm{u}+6(19\mathrm{u})$,localid="1648640488650" ${\mathrm{UF}}_{235}$islocalid="1648640493790" $235\mathrm{u}+6(19\mathrm{u})$.

localid="1648640499643" $\frac{{\mathrm{v}}_{235}}{{\mathrm{v}}_{238}}=\sqrt{\left(\frac{{\mathrm{m}}_{238}}{{\mathrm{m}}_{235}}\right)}$

localid="1648640507384" $=\sqrt{\left(\frac{238\mathrm{u}+6\left(19\mathrm{u}\right)}{235\mathrm{u}+6\left(19\mathrm{u}\right)}\right)}$

localid="1648640513569" $\frac{{\mathrm{v}}_{235}}{{\mathrm{v}}_{238}}=1.0043$