Compare the rates of diffusion of ` '^(235)UF_(6)` and ` '^(238)UF_(6)`

Explanation

Let the rate of diffusion of ` '^(235)UF_(6)` be `r_(1)` and that of ` '^(238)UF_(6)` be `r_(2)`.
Now, the molecular mass of ` '^(235)UF_(6)` is
`M_(1)=235+6xx19=349`
The molecular mass of `'^(238)UF_(6)` is
`M_(2)=238+6xx19=352`
According to Graham's law,
`(r_(1))/(r_(2))=sqrt((M_(2))/(M_(1)))=sqrt((35)/(349))=1.004`
Thus, `r('^(235)UF_(6)) : r('^(238)UF_(6))=1.004 : 1`.