A Natural Nuclear Reactor — Question 8

Section 43-4 · Physics pages 7226–7227

Question 8 Spatial confinement places a bound on long-term diffusion Assume a stable tracer’s one-dimensional rms displacement follows ℓ= √ 2Dt. A fossil pattern remains confined to an rms scale at most 1.00 m over 2.00 × 109 yr. Use 1 yr = 3.156 × 107 s and constant D. x c spread Schematic; labels, not drawing lengths, specify the data. (a) Find the upper bound on D. (b) If D were four times this bound, find the predicted rms displacement and explain one geological limitation of the inference.
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Question 8 - Solution Spatial confinement places a bound on long-term diffusion x c spread (a) Method and calculation t = (2.00 × 109)(3.156 × 107) = 6.312 × 1016 s. D ≤ℓ2/(2t) = 7.92142 × 10−18 m2/s . (b) Interpretation and further calculation ℓ′ = √ 2(4Dmax)t = 2.00 m . ℓ2 = 2 ∫t 0 D(t′)dt′ ⇒ the constant-D bound is time averaged. Physical check Advection, chemical trapping, changing temperature and measurement geometry can invalidate a simple free-diffusion interpretation.

Question 8

Spatial confinement places a bound on long-term diffusion

Assume a stable tracer’s one-dimensional rms displacement follows ℓ=2Dt\ell=\sqrt{2Dt}. A fossil pattern remains confined to an rms scale at most 1.00 m over 2.00×1092.00\times10^9 yr. Use 1yr=3.156×1071\,\mathrm{yr}=3.156\times10^7 s and constant D.

Diagram shown in the original SVG worksheet.

  1. Find the upper bound on D.

  2. If D were four times this bound, find the predicted rms displacement and explain one geological limitation of the inference.

Question 8 - Solution

Spatial confinement places a bound on long-term diffusion

Diagram shown in the original SVG worksheet.

(a) Method and calculation

t=(2.00×109)(3.156×107)=6.312×1016s.t=(2.00\times10^9)(3.156\times10^7)=6.312\times10^{16}\,\mathrm s.D≤ℓ2/(2t)=7.92142×10−18m2/s.\boxed{D\le\ell^2/(2t)=7.92142\times10^{-18}\,\mathrm{m^2/s}}.

(b) Interpretation and further calculation

ℓ′=2(4Dmax)t=2.00m.\boxed{\ell\prime=\sqrt{2(4D_{\max})t}=2.00\,\mathrm m}.ℓ2=2∫0tD(t′)dt′⇒the constant-D bound is time averaged.\ell^2=2\int_0^tD(t\prime)dt\prime\quad\Rightarrow\quad\text{the constant-D bound is time averaged}.

Physical check

Advection, chemical trapping, changing temperature and measurement geometry can invalidate a simple free-diffusion interpretation.