Question

Natural uranium is a mixture of three isotopes `_92^234U`, `_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 1

Correct Answer - A::B::C::D
Let `R_1`, `R_2` and `R_3` be the activities of `U_234`, `U^235` and `U^238` respectively.
Total activity, `R=R_1+R_2+R_3`
Share of `U^234`, `R_1/R=(lambda_1N_1)/(lambda_1N_1+lambda_2N_2+lambda_3N_3`
`Let m be the mass of natural uranium.
Then, `m_1=0.01/100m`, `m_2=0.71/100m` and `m_3=99.28/100m`
Now, `N_1=m_1/M_1`, `N_2=m_2/M_2` and `N_3=m_3/M_3`
where `M_1`, `M_2` and `M_3` are atomic weights.
:. `R_1/R=((m_1/M_1)1/T_1)/(m_1/M_1 1/T_1+m_2/M_2*1/T_2+m_3/M_3*1/T_3)`
`=(((0.01//100))/(234)xx(1)/(2.5xx10^5years))/(((0.01//100)/(234))((1)/(25.xx10^5))+((0.71//100)/(235))((1)/(7.1xx10^8))+((99.28//100)/(238))((1)/(4.5xx10^9)))`
`=0.648=64.8%`
Similarly, share of `U^235=0.016%`
and of `U^238=35.184%`