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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

Expert verified

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

Step by step solution

01

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 \({m_{\rm{F}}} = 18.998403\;{\rm{u}}\).

02

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.

03

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\(1.0086\).

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