A Natural Nuclear Reactor — Question 5

Section 43-4 · Physics pages 7220–7221

Question 5 Extinct radioactive products leave stable descendants as evidence A hypothetical fission product P decays entirely to stable D with half-life 30.0 yr. It was produced in a brief episode 2.00 × 109 yr ago. Initially there was no D; assume full retention and no other production. P stable D emission Schematic; labels, not drawing lengths, specify the data. (a) Express the surviving P fraction using a base-ten logarithm. (b) Determine present D relative to initial P and explain why absence of P does not disprove ancient fission.
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Question 5 - Solution Extinct radioactive products leave stable descendants as evidence P stable D emission (a) Method and calculation NP /N0 = 2−t/T ⇒log10(NP /N0) = −(t/T) log10 2. log10(NP /N0) = −2.00 × 109 30.0 (.30103) = −2.00687 × 107 . (b) Interpretation and further calculation ND = N0 −NP ≃N0 . NP + ND = N0 the stable daughter preserves the atom count. Physical check The surviving-parent fraction is effectively zero. Stable isotope anomalies can preserve evidence long after short-lived parents disappear.

Question 5

Extinct radioactive products leave stable descendants as evidence

A hypothetical fission product P decays entirely to stable D with half-life 30.0 yr. It was produced in a brief episode 2.00×1092.00\times10^9 yr ago. Initially there was no D; assume full retention and no other production.

Diagram shown in the original SVG worksheet.

  1. Express the surviving P fraction using a base-ten logarithm.

  2. Determine present D relative to initial P and explain why absence of P does not disprove ancient fission.

Question 5 - Solution

Extinct radioactive products leave stable descendants as evidence

Diagram shown in the original SVG worksheet.

(a) Method and calculation

NP/N0=2−t/T⇒log⁡10(NP/N0)=−(t/T)log⁡102.N_P/N_0=2^{-t/T}\Rightarrow\log_{10}(N_P/N_0)=-(t/T)\log_{10}2.log⁡10(NP/N0)=−2.00×10930.0(.30103)=−2.00687×107.\boxed{\log_{10}(N_P/N_0)=-\frac{2.00\times10^9}{30.0}(.30103)=-2.00687\times10^7}.

(b) Interpretation and further calculation

ND=N0−NP≃N0.\boxed{N_D=N_0-N_P\simeq N_0}.NP+ND=N0the stable daughter preserves the atom count.\boxed{N_P+N_D=N_0}\quad\text{the stable daughter preserves the atom count}.

Physical check

The surviving-parent fraction is effectively zero. Stable isotope anomalies can preserve evidence long after short-lived parents disappear.