Natural uranium is a mixture of three isotopes `._92^234U`, `._92^235U` and `._92^238U` with mass percentage `0.01%, 0.71% and 99.28%` respectively. The half-life of three isotopes are `2.5xx10^5yr`, `7.1xx10^8yr` and `4.5xx10^9yr` respectively.
Determine the share of radioactivity of each isotope into the total activity of the natural uranium.

Answer

`86.4%`

Explanation

Let `A_(1),A_(2)` and `A_(3)` be activites of `._(92)U^(234)` , `._(92)U^(235)` and `._(92)U^(238)` respectively. Total activity,
`A=A_(1) +A_(2) +A_(3)`
Share of `._(92)U^(234)` is
`(A_(1))/(A)=(lambda_(1) N_(1))/(lambda_(1)N_(1) +lambda_(2)N_(2) +lambda_(3)N_(3))`
Let m be the total mass of natural uranium.Then,
`m_(1) (0.006)/(100) m, m_(2) =(0.71)/(100) m,.m_(3) =(99.28)/(100) m`
Now,
` N_(1) =(m_(1))/(m_(1)), N_(2), =(m_(2))/(M_(2)),N_(3)=(m_(3))/(M_(3))`
where `M_(1) M_(2)` and `M_(3)` are atomic weights.
`because (A_(1))/(A_(2)) =(((m_(1))/(M_(1)))(1)/(T_(1)))/((m_(1))/(M_(1)T_(1))+(m_(2))/(M_(2)T_(2))+(m_(3))/(M_(2)T_(3))) =0.99~~9.0%`
`(A_(2))/(A)=0.039 =3.9%`
`(A_(3))/(A)=0.864=86.4%`.