A Natural Nuclear Reactor — Question 9

Section 43-4 · Physics pages 7228–7229

Question 9 Common radioactive aging preserves a normalized isotope deficit Immediately after an ancient event, a sample’s 235U/238U ratio was a fraction b of a matched unfissioned reference ratio. Both subsequently undergo the same radioactive decay for time t and remain closed. Their measured present ratio quotient is b = .920 ± .010. t r reference sample Schematic; labels, not drawing lengths, specify the data. (a) Show that the quotient of isotope ratios remains b at every later time. (b) Infer the depletion fraction and its standard uncertainty, and explain why this comparison does not determine the event age.
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Question 9 - Solution Common radioactive aging preserves a normalized isotope deficit t r reference sample (a) Method and calculation rs(t) = brr(0)e−(λ235−λ238)t, rr(t) = rr(0)e−(λ235−λ238)t. rs(t)/rr(t) = b the common decay factor cancels. (b) Interpretation and further calculation d = 1 −b = .080 ± .010 or (8.0 ± 1.0)%. ∂b/∂t = 0 under matched-history assumptions. Physical check The cancellation fails if samples have different gains, losses or isotope histories. Depletion evidence and age evidence are distinct constraints.

Question 9

Common radioactive aging preserves a normalized isotope deficit

Immediately after an ancient event, a sample’s 235^{235}U/238^{238}U ratio was a fraction b of a matched unfissioned reference ratio. Both subsequently undergo the same radioactive decay for time t and remain closed. Their measured present ratio quotient is b=.920±.010b=.920\pm.010.

Diagram shown in the original SVG worksheet.

  1. Show that the quotient of isotope ratios remains b at every later time.

  2. Infer the depletion fraction and its standard uncertainty, and explain why this comparison does not determine the event age.

Question 9 - Solution

Common radioactive aging preserves a normalized isotope deficit

Diagram shown in the original SVG worksheet.

(a) Method and calculation

rs(t)=brr(0)e−(λ235−λ238)t,rr(t)=rr(0)e−(λ235−λ238)t.r_s(t)=br_r(0)e^{-(\lambda_{235}-\lambda_{238})t},\quad r_r(t)=r_r(0)e^{-(\lambda_{235}-\lambda_{238})t}.rs(t)/rr(t)=bthe common decay factor cancels.\boxed{r_s(t)/r_r(t)=b}\quad\text{the common decay factor cancels}.

(b) Interpretation and further calculation

d=1−b=.080±.010or (8.0±1.0)%.\boxed{d=1-b=.080\pm.010}\quad\text{or }(8.0\pm1.0)\%.∂b/∂t=0under matched-history assumptions.\boxed{\partial b/\partial t=0}\quad\text{under matched-history assumptions}.

Physical check

The cancellation fails if samples have different gains, losses or isotope histories. Depletion evidence and age evidence are distinct constraints.