A Natural Nuclear Reactor — Question 2

Section 43-4 · Physics pages 7214–7215

Question 2 An isotope deficit must be normalized to the unchanged reference isotope In a hypothetical fossil sample, present 235U fraction is .00660, compared with .00720 in a matched unfissioned reference. Suppose 238U is unchanged by the event, only 235U was removed, and both samples subsequently remained closed. Neglect other isotopes. reference: rr sample: rs rs/rr Schematic; labels, not drawing lengths, specify the data. (a) Find the fraction of 235U removed using ratios to 238U. (b) Compare it with the shortcut 1 −fs/fr and explain the discrepancy.
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Question 2 - Solution An isotope deficit must be normalized to the unchanged reference isotope reference: rr sample: rs rs/rr (a) Method and calculation rs = .00660 .99340, rr = .00720 .99280, rs = (1 −d)rr. d = 1 −rs rr = .0838870 = 8.38870% . (b) Interpretation and further calculation dshortcut = 1 −.00660/.00720 = .0833333 = 8.33333% . d −dshortcut = .00055365 or .055365 percentage point. Physical check Removing atoms changes both the numerator and total in an atom fraction. Ratios to an unchanged isotope provide the appropriate normalization.

Question 2

An isotope deficit must be normalized to the unchanged reference isotope

In a hypothetical fossil sample, present 235^{235}U fraction is .00660, compared with .00720 in a matched unfissioned reference. Suppose 238^{238}U is unchanged by the event, only 235^{235}U was removed, and both samples subsequently remained closed. Neglect other isotopes.

Diagram shown in the original SVG worksheet.

  1. Find the fraction of 235^{235}U removed using ratios to 238^{238}U.

  2. Compare it with the shortcut 1−fs/fr1-f_s/f_r and explain the discrepancy.

Question 2 - Solution

An isotope deficit must be normalized to the unchanged reference isotope

Diagram shown in the original SVG worksheet.

(a) Method and calculation

rs=.00660.99340,rr=.00720.99280,rs=(1−d)rr.r_s=\frac{.00660}{.99340},\quad r_r=\frac{.00720}{.99280},\quad r_s=(1-d)r_r.d=1−rsrr=.0838870=8.38870%.\boxed{d=1-\frac{r_s}{r_r}=.0838870=8.38870\%}.

(b) Interpretation and further calculation

dshortcut=1−.00660/.00720=.0833333=8.33333%.\boxed{d_{\mathrm{shortcut}}=1-.00660/.00720=.0833333=8.33333\%}.d−dshortcut=.00055365or .055365 percentage point.\boxed{d-d_{\mathrm{shortcut}}=.00055365}\quad\text{or .055365 percentage point}.

Physical check

Removing atoms changes both the numerator and total in an atom fraction. Ratios to an unchanged isotope provide the appropriate normalization.