Age of Sample: U-235 & U-238 Abundances

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SUMMARY

The discussion centers on calculating the age of a sample based on the isotopic abundances of Uranium-235 (U-235) and Uranium-238 (U-238), which are 0.72% and 99.27%, respectively. The relevant decay constant formula, λ = ln(2) / t1/2, is utilized to derive the relationship between the isotopes. By assuming equal initial abundances, the ratio of current isotopes is set to 99.27/0.72, allowing for the calculation of time since formation (t). Participants confirm that this approach is correct and provide guidance on solving for t.

PREREQUISITES
  • Understanding of radioactive decay and half-life concepts
  • Familiarity with the decay constant formula λ = ln(2) / t1/2
  • Knowledge of isotopic abundance and its implications in radiometric dating
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Research the half-lives of U-235 and U-238 for accurate decay constant calculations
  • Learn about the application of radiometric dating techniques in geology
  • Explore the significance of isotopic ratios in determining geological time scales
  • Study advanced topics in nuclear chemistry related to isotope behavior
USEFUL FOR

Students and professionals in geology, nuclear physics, and radiometric dating, as well as anyone interested in understanding the principles of isotopic abundances and their applications in determining the age of geological samples.

Purple Baron
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Homework Statement


The isostopc abundances of a sample is U-235 and U-238 are 0.72 and 99.27 respectively; what is the age of the sample? (assume isotope abundance was equal when sample was formed)

Homework Equations


\lambda=\frac{ln2}{ t_{\frac{1}{2}}}

The Attempt at a Solution


for U-238 N_{238}(T)=N_{238}(t)e^{\lambda _{238}t}
U-235 N_{235}(T)=N_{235}(t)e^{\lambda _{235}t}
T is time at present and t is time of sample formation.

diving the two equations gives
\frac{N_{238}(T)}{N_{235}(T)}=\frac{N_{238}(t)}{N_{235}(t)}e^{(\lambda _{238}- \lambda _{235})t}

From the assumption, one can say \frac{N_{238}(t)}{N_{235}(t)}=1
It's here where I'm not sure; do i just say that \frac{N_{238}(T)}{N_{235}(T)}=\frac{99.27}{0.72} and solve for t or am I missing something? Thanks

EDIT: Sorry, I can't get my latex to work, I can't seem to fix it
 
Last edited:
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I fixed the two broken equations, but the first one looks odd.
Yes, just do that and solve for t.
 
Thanks for that, I fixed the first equation.
 

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