Question

Natural uranium is a mixture of three isotopes ${}_{92}{U}^{234}{,}_{92}{U}^{235}$, and ${}_{92}{U}^{238}$ with the shares of negligible $0.0006$%, $0.71$%, and $99.28$%, respectively. The half-life of the three isotopes are $2.5\times {10}^{5}years,7.1\times {10}^{8}years$, and $4.5\times {10}^{9}years$, respectively. Determine the share of radioactivity(in %) of each isotope into the total activity of the natural uranium

Answer 1

$Activity\left(A\right)=\frac{ln2}{halflife}\times Numberofparticles\left(N\right)-\left(1\right)$

Given that for ${}_{92}{U}^{234}$, ${}_{92}{U}^{235}$ and ${}_{92}{U}^{238}$:

The number of particles are in ratio, $0.0006:0.71:99.28$ and half-life are in ratio, $2.5:7100:45000$

From the equation $\left(1\right)$, the activities are in ratio: $2.4:1:22$

So, percentage share in total activity of natural uranium are $9.45\%$, $3.94\%$ and $86.61\%$ for the ${}_{92}{U}^{234}$, ${}_{92}{U}^{235}$ and ${}_{92}{U}^{238}$ respectively.