Uranium has two isotopes of masses 235 and 23 | vedclass

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(C) Given:Mass of heavier uranium isotope $(M_1)$ $= 238$ unitsMass of lighter uranium isotope $(M_2)$ $= 235$ unitsWe know that the root mean square

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$0.64$

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(C) Given:Mass of heavier uranium isotope $(M_1)$ $= 238$ unitsMass of lighter uranium isotope $(M_2)$ $= 235$ unitsWe know that the root mean square velocity $(v_{	ext{rms}})$ is given by:$v_{	ext{rms}} = \sqrt{\frac{3RT}{M}}$Thus, $v_{	ext{rms}} \propto \frac{1}{\sqrt{M}}$.The percentage ratio of the difference in rms velocities to the rms velocity of the heavier isotope is:$	ext{Ratio} = \left( \frac{v_{	ext{rms, lighter}} - v_{	ext{rms, heavier}}}{v_{	ext{rms, heavier}}} ight) 	imes 100$Substituting the proportionality $v_{	ext{rms}} \propto \frac{1}{\sqrt{M}}$:$	ext{Ratio} = \left( \frac{\frac{1}{\sqrt{235}} - \frac{1}{\sqrt{238}}}{\frac{1}{\sqrt{238}}} ight) 	imes 100 = \left( \frac{\sqrt{238}}{\sqrt{235}} - 1 ight) 	imes 100$$	ext{Ratio} = \left( \sqrt{\frac{238}{235}} - 1 ight) 	imes 100 \approx (\sqrt{1.01276} - 1) 	imes 100$$	ext{Ratio} \approx (1.00636 - 1) 	imes 100 = 0.00636 	imes 100 = 0.636 \% \approx 0.64 \%$

Uranium has two isotopes of masses $235$ and $238$ units. If both of them are present in uranium hexafluoride gas, find the percentage ratio of the difference in rms velocities of the two isotopes to the rms velocity of the heavier isotope.

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