Question

Question: (II)Two isotopes of uranium,${}^{\mathbf{235}}\mathbf{U}$and${}^{\mathbf{238}}\mathbf{U}$(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$\mathbf{U}{\mathbf{F}}_{\mathbf{6}}$. 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${}^{235}\mathrm{U}$is$m_{235}=235.043930\mathrm{u}$.

The mass of uranium${}^{238}\mathrm{U}$is$m_{238}=238.050788\mathrm{u}$.

The mass of uranium $\mathrm{F}$ is $m_{\mathrm{F}}=18.998403\mathrm{u}$.

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:

$v_{\mathrm{r}\mathrm{m}\mathrm{s}}=\sqrt{\frac{3kT}{m}}$ … (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${}^{235}\mathrm{U}{\mathrm{F}}_6$and${}^{238}\mathrm{U}{\mathrm{F}}_6$.

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

$\begin{array}{rl}c\frac{v_{\mathrm{r}\mathrm{m}\mathrm{s}}\left({}^{235}\mathrm{U}{\mathrm{F}}_6\right)}{v_{\mathrm{r}\mathrm{m}\mathrm{s}}\left({}^{238}\mathrm{U}{\mathrm{F}}_6\right)}& =\frac{\sqrt{\frac{3kT}{m_{{}^{235}\mathrm{U}{\mathrm{F}}_6}}}}{\sqrt{\frac{3kT}{m_{{}^{238}\mathrm{U}{\mathrm{F}}_6}}}}\\ & =\sqrt{\frac{m_{{}^{238}\mathrm{U}{\mathrm{F}}_6}}{m_{{}^{235}\mathrm{U}{\mathrm{F}}_6}}}\\ & =\sqrt{\frac{m_{{}^{235}\mathrm{U}}+6m_{\mathrm{F}}}{m_{{}^{238}\mathrm{U}}+6m_{\mathrm{F}}}}\end{array}$

Substitute the values in the above equation.

$\begin{array}{rl}c\frac{v_{\mathrm{r}\mathrm{m}\mathrm{s}}\left({}^{235}\mathrm{U}{\mathrm{F}}_6\right)}{v_{\mathrm{r}\mathrm{m}\mathrm{s}}\left({}^{238}\mathrm{U}{\mathrm{F}}_6\right)}& =\left(\sqrt{\frac{238.050788\mathrm{u}+6\left(18.998403\mathrm{u}\right)}{235.043930\mathrm{u}+6\left(18.998403\mathrm{u}\right)}}\right)\\ & =\sqrt{\frac{352.041206}{349.034348}}\\ & =1.0086\end{array}$

Thus, the ratio of the rms speeds of two molecules for the two isotopes is$1.0086$.