A Natural Nuclear Reactor — Question 3

Section 43-4 · Physics pages 7216–7217

Question 3 Stable isotope mixing separates a fission signature from background A hypothetical stable element has isotope-X atom fraction .100 in geological background and .250 in a fission-derived endmember. A sample has fraction .160. Let w be the fraction of this element’s atoms contributed by the fission endmember. w fX Schematic; labels, not drawing lengths, specify the data. (a) Infer w by atom-number mixing. (b) If the measured fraction has standard uncertainty .003 and the endmembers are exact, find the uncertainty in w and the admissible measurement interval.
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Question 3 - Solution Stable isotope mixing separates a fission signature from background w fX (a) Method and calculation f = (1 −w)fb + wfF ⇒w = f −fb fF −fb . w = (.160 −.100)/(.250 −.100) = .400 . (b) Interpretation and further calculation σw = σf/|fF −fb| = .003/.150 = .020 . 0 ≤w ≤1 ⇒.100 ≤f ≤.250 . Physical check The inferred w is an atom fraction of this element, not a rock mass fraction. Values outside the endmember interval falsify this two-component model.

Question 3

Stable isotope mixing separates a fission signature from background

A hypothetical stable element has isotope-X atom fraction .100 in geological background and .250 in a fission-derived endmember. A sample has fraction .160. Let w be the fraction of this element’s atoms contributed by the fission endmember.

Diagram shown in the original SVG worksheet.

  1. Infer w by atom-number mixing.

  2. If the measured fraction has standard uncertainty .003 and the endmembers are exact, find the uncertainty in w and the admissible measurement interval.

Question 3 - Solution

Stable isotope mixing separates a fission signature from background

Diagram shown in the original SVG worksheet.

(a) Method and calculation

f=(1−w)fb+wfF⇒w=f−fbfF−fb.f=(1-w)f_b+wf_F\Rightarrow w=\frac{f-f_b}{f_F-f_b}.w=(.160−.100)/(.250−.100)=.400.\boxed{w=(.160-.100)/(.250-.100)=.400}.

(b) Interpretation and further calculation

σw=σf/|fF−fb|=.003/.150=.020.\boxed{\sigma_w=\sigma_f/|f_F-f_b|=.003/.150=.020}.0≤w≤1⇒.100≤f≤.250.0\le w\le1\Rightarrow\boxed{.100\le f\le.250}.

Physical check

The inferred w is an atom fraction of this element, not a rock mass fraction. Values outside the endmember interval falsify this two-component model.