Question

Question: (II)Two isotopes of uranium,\({}^{{\bf{235}}}{\bf{U}}\)and\({}^{{\bf{238}}}{\bf{U}}\)(the superscripts refer to their atomic masses), can be separated by a gas diffusion process by combining them with fluorine to make the gaseous compound\({\bf{U}}{{\bf{F}}_{\bf{6}}}\). Calculate the ratio of the rms speeds of these molecules for the two isotopes, at constant T. Use Appendix B for masses.

Short answer

The ratio of the rms speeds of two molecules for the two isotopes is \(1.0086\).

Step-by-step solution

Given data

The mass of uranium\({}^{{\rm{235}}}{\rm{U}}\)is\({m_{235}} = 235.043930\;{\rm{u}}\).

The mass of uranium\({}^{{\rm{238}}}{\rm{U}}\)is\({m_{238}} = 238.050788\;{\rm{u}}\).

The mass of uranium \({\rm{F}}\) is \({m_{\rm{F}}} = 18.998403\;{\rm{u}}\).

Understanding isotopes

The root-mean-square speed of gas molecules depends on the mass of the molecule and the absolute temperature of the gas.

The root-mean-square speed is given as follows:

\({v_{{\rm{rms}}}} = \sqrt {\frac{{3kT}}{m}} \) … (i)

Here, k is the Boltzmann constant, T is the temperature, and m is the mass of the molecule.

Determination of the ratio of root-mean-square speeds of the molecules

The two compounds formed in the process are\(^{{\rm{235}}}{\rm{U}}{{\rm{F}}_{\rm{6}}}\)and\(^{{\rm{238}}}{\rm{U}}{{\rm{F}}_{\rm{6}}}\).

As the temperature is constant, from equation (i),

\(\begin{aligned}{c}\frac{{{v_{{\rm{rms}}}}\left( {^{{\rm{235}}}{\rm{U}}{{\rm{F}}_{\rm{6}}}} \right)}}{{{v_{{\rm{rms}}}}\left( {^{{\rm{238}}}{\rm{U}}{{\rm{F}}_{\rm{6}}}} \right)}} &= \frac{{\sqrt {\frac{{3kT}}{{{m_{^{{\rm{235}}}{\rm{U}}{{\rm{F}}_{\rm{6}}}}}}}} }}{{\sqrt {\frac{{3kT}}{{{m_{^{{\rm{238}}}{\rm{U}}{{\rm{F}}_{\rm{6}}}}}}}} }}\\ &= \sqrt {\frac{{{m_{^{{\rm{238}}}{\rm{U}}{{\rm{F}}_{\rm{6}}}}}}}{{{m_{^{{\rm{235}}}{\rm{U}}{{\rm{F}}_{\rm{6}}}}}}}} \\ &= \sqrt {\frac{{{m_{^{{\rm{235}}}{\rm{U}}}} + 6{m_{\rm{F}}}}}{{{m_{^{{\rm{238}}}{\rm{U}}}} + 6{m_{\rm{F}}}}}} \end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}\frac{{{v_{{\rm{rms}}}}\left( {^{{\rm{235}}}{\rm{U}}{{\rm{F}}_{\rm{6}}}} \right)}}{{{v_{{\rm{rms}}}}\left( {^{{\rm{238}}}{\rm{U}}{{\rm{F}}_{\rm{6}}}} \right)}} &= \left( {\sqrt {\frac{{238.050788\;{\rm{u}} + 6\left( {18.998403\;{\rm{u}}} \right)}}{{235.043930\;{\rm{u}} + 6\left( {18.998403\;{\rm{u}}} \right)}}} } \right)\\ &= \sqrt {\frac{{352.041206}}{{349.034348}}} \\ &= 1.0086\end{aligned}\)

Thus, the ratio of the rms speeds of two molecules for the two isotopes is\(1.0086\).