A Natural Nuclear Reactor — Question 1

Section 43-4 · Physics pages 7212–7213

Question 1 Different uranium half-lives increase the ancient fissile fraction Ancient natural fission at Oklo, Gabon, motivates this reference calculation. For an unfissioned closed reference sample, use present 235U atom fraction f0 = .00720 among 235U and 238U only. Their half-lives are .704 and 4.47 billion years. Ignore other isotopes. Let t be positive lookback time. tlookback f Schematic; labels, not drawing lengths, specify the data. (a) Derive the ancient isotope ratio r(t) = N235/N238 and atom fraction f(t). (b) Evaluate f at t=2.00 billion years and compare it with the present fraction.
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Question 1 - Solution Different uranium half-lives increase the ancient fissile fraction tlookback f (a) Method and calculation r0 = f0 1 −f0 , r(t) = r0e(λ235−λ238)t . f(t) = r(t) 1 + r(t), λi = ln 2/Ti . (b) Interpretation and further calculation r(2) = .00720 .99280 exp [ 2 ln 2 ( 1 .704 − 1 4.47 )] = .0381049. f(2) = .0367062 = 3.67062%, f(2)/f0 = 5.0981 . Physical check The shorter-lived isotope was relatively more abundant in the past. Abundance alone does not establish whether a natural chain reaction occurred.

Question 1

Different uranium half-lives increase the ancient fissile fraction

Ancient natural fission at Oklo, Gabon, motivates this reference calculation. For an unfissioned closed reference sample, use present 235^{235}U atom fraction f0=.00720f_0=.00720 among 235^{235}U and 238^{238}U only. Their half-lives are .704 and 4.47 billion years. Ignore other isotopes. Let t be positive lookback time.

Diagram shown in the original SVG worksheet.

  1. Derive the ancient isotope ratio r(t)=N235/N238r(t)=N_{235}/N_{238} and atom fraction f(t).

  2. Evaluate f at t=2.00 billion years and compare it with the present fraction.

Question 1 - Solution

Different uranium half-lives increase the ancient fissile fraction

Diagram shown in the original SVG worksheet.

(a) Method and calculation

r0=f01−f0,r(t)=r0e(λ235−λ238)t.r_0=\frac{f_0}{1-f_0},\quad\boxed{r(t)=r_0e^{(\lambda_{235}-\lambda_{238})t}}.f(t)=r(t)1+r(t),λi=ln⁡2/Ti.\boxed{f(t)=\frac{r(t)}{1+r(t)},\quad\lambda_i=\ln2/T_i}.

(b) Interpretation and further calculation

r(2)=.00720.99280exp⁡[2ln2(1.704−14.47)]=.0381049.r(2)=\frac{.00720}{.99280}\exp\!\left[2\ln2\left(\frac1{.704}-\frac1{4.47}\right)\right]=.0381049.f(2)=.0367062=3.67062%,f(2)/f0=5.0981.\boxed{f(2)=.0367062=3.67062\%,\quad f(2)/f_0=5.0981}.

Physical check

The shorter-lived isotope was relatively more abundant in the past. Abundance alone does not establish whether a natural chain reaction occurred.