A Natural Nuclear Reactor — Question 4

Question 4

Unequal retention alters a fossil fission-product ratio

Two stable products X and Y formed in ratio $N_X^0/N_Y^0=2.00$. A mineral receives no new input but retains fractions $r_X$ and $r_Y$ of the original products. Today $N_X/N_Y=1.20$ and an independent tracer gives $r_Y=.800$.

1. Infer $r_X$ and the fraction of X lost.

2. Without the independent value of $r_Y$, what remains inferable? Give the allowed range of $r_X$ for $0<r_Y\le1$.

Question 4 - Solution

Unequal retention alters a fossil fission-product ratio

(a) Method and calculation

$$\frac{N_X}{N_Y}=\frac{r_X}{r_Y}\frac{N_X^0}{N_Y^0}\Rightarrow\boxed{r_X/r_Y=1.20/2.00=.600}.$$$$\boxed{r_X=.600(.800)=.480,\quad1-r_X=.520}.$$

(b) Interpretation and further calculation

$$\boxed{r_X=.600r_Y}\quad\text{only relative retention is determined}.$$$$0<r_Y\le1\Rightarrow\boxed{0<r_X\le.600}.$$

Physical check

An anomalous ratio can reflect selective migration as well as production. A second constraint is needed to identify absolute retention.