Question

A sample of uranium is a mixture of three isotopes `._(92)U^(234), ._(92)U^(235)` and `._(92)U^(238)` present in the ratio `0.006%, 0.71%` and `99.284%` respectively. The half lives of then isotopes are `2.5xx10^(5)` years, `7.1xx10^(8)` years and `4.5xx10^(9)` years respectively. The contribution to activity (in `%`) of each isotope in the sample respectively
A. `51.41%, 2.13%,46.46%`
B. `51.41%,46.46%,2.13%`
C. `2.13%, 51.41%,46.46%`
D. `46.46%,2.13%,51.41%`

Answer 1

Correct Answer - A
Let `m` is the total mass of the uranium mixture. The masses of the isotopes `._(92)U^(234), ._(92)U^(235)` and `._(92)U^(238)` in the mixture are `m_(1)=(0.006)/(100)m` ,
`m_(2)=(0.71)/(100)m`, and `m_(3)=(99.284)/(100)m`.
If `N_(A)` is the Avogadro number, then number of atoms of threeisotopes are, `N_(1)=(m_(1)N_(A))/(M_(1)`
`N_(2)=(m_(2)N_(A))/(M_(2))`, and `N_(3)=(m_(3)N_(A))/(M_(3))`
Activity of radioactive sample `A= lambda N`
As `lambda=(0.693)/(t_(1//2)), :. A=(0.693)/(t_(1//2))N`
If `t_(1,t_(2)` and `t_(3)` be the half-lives, then
`A_(1):A_(2):A_(3)=(N_(1))/(t_(1)):(N_(2))/(t_(2)):(N_(3))/(t_(3))`
or `A_(1):A_(2):A_(3)=(m_(1))/(M_(1)t_(1)):(m_(2))/(M_(2)t_(2)), (m_(3))/(M_(3)t_(3))`
`=(0.006)/(234(2.5xx10^(5))):(0.71)/(235(7.5xx10^(8))):(99.284)/(238(4.5xx10^(9)))`
`51.41%:2.13%:46.46%`