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

Vedclass · Accepted Answer

$0.64$

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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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