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

Options

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

A

Explanation

Let m is the total mass of the uranium mixture. The masses of the istopes `"" _(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. 281)/(100)m`
If `N_(A)` is the Avogadro number, then number of atoms of three istopes are,`N_(1) = (m_(3) N_(A))/(M_(3))` ?
Activity of radioactive sample `A = lamda N.`
As ` lamda = ( 0.693)/(t _(1//2)) , therefore 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 (2.5 xx 10 ^(5))): ( 0.71)/(235 (7.5 xx 10 ^(8))): (99.284)/(238 (4.5 xx 10 ^(9)))`
`= 51. 41 % :2.13% 46.46%`