Radioactive Dating — Question 4

Section 42-6 · Physics pages 7098–7099

Question 4 A branching decay clock uses the total decay constant A parent has total half-life T and branch probability b=.100 for producing stable daughter D. Other branches produce different nuclides. Initially D=0, the system stays closed, and the present number ratio is D/P=.150. parent b : D other branches Schematic; labels, not drawing lengths, specify the data. (a) Derive the age relation accounting for branching. (b) Find age in units of T and compare with the unbranched formula.
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Question 4 - Solution A branching decay clock uses the total decay constant parent b : D other branches (a) Method and calculation P = P0e−λt, D = b(P0 −P) ⇒D/P = b(eλt −1). t = λ−1 ln[1 + D/(bP)] . (b) Interpretation and further calculation t/T = ln(1 + .150/.100)/ ln 2 = 1.32193 . tnaive/T = ln(1 + .150)/ ln 2 = .20163 too young. Physical check The branching probability controls daughter production, while all branches deplete the parent. Replacing the total half-life with a partial one would miscount parent survival.

Question 4

A branching decay clock uses the total decay constant

A parent has total half-life T and branch probability b=.100 for producing stable daughter D. Other branches produce different nuclides. Initially D=0, the system stays closed, and the present number ratio is D/P=.150.

Diagram shown in the original SVG worksheet.

  1. Derive the age relation accounting for branching.

  2. Find age in units of T and compare with the unbranched formula.

Question 4 - Solution

A branching decay clock uses the total decay constant

Diagram shown in the original SVG worksheet.

(a) Method and calculation

P=P0e−λt,D=b(P0−P)⇒D/P=b(eλt−1).P=P_0e^{-\lambda t},\quad D=b(P_0-P)\Rightarrow D/P=b(e^{\lambda t}-1).t=λ−1ln⁡[1+D/(bP)].\boxed{t=\lambda^{-1}\ln[1+D/(bP)]}.

(b) Interpretation and further calculation

t/T=ln⁡(1+.150/.100)/ln⁡2=1.32193.\boxed{t/T=\ln(1+.150/.100)/\ln2=1.32193}.tnaive/T=ln⁡(1+.150)/ln⁡2=.20163too young.\boxed{t_{\mathrm{naive}}/T=\ln(1+.150)/\ln2=.20163}\quad\text{too young}.

Physical check

The branching probability controls daughter production, while all branches deplete the parent. Replacing the total half-life with a partial one would miscount parent survival.