K Concentration on Earth Over Time

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SUMMARY

The discussion centers on the calculation of potassium (K) concentration on Earth over time, specifically using the half-life of 1.25 billion years. Given the Earth's age of 5 billion years, the concentration of K has decreased by a factor of 8, as confirmed through the calculation of successive half-lives. The method involves determining the remaining concentration after three half-lives, leading to the conclusion that the concentration is now X/8 of its original value. Participants sought simpler methods for this calculation but confirmed the accuracy of the exponential decay approach.

PREREQUISITES
  • Understanding of radioactive decay and half-life concepts
  • Basic knowledge of exponential functions
  • Familiarity with the concept of geological time scales
  • Ability to perform calculations involving fractions and powers
NEXT STEPS
  • Study radioactive decay equations and their applications in geology
  • Learn about the implications of half-life in radiometric dating techniques
  • Explore the concept of isotopes and their relevance in Earth sciences
  • Investigate other elements with known half-lives and their concentration changes over geological time
USEFUL FOR

Students in Earth sciences, geologists, and anyone interested in understanding the principles of radioactive decay and its impact on elemental concentrations over geological time scales.

ilona
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Homework Statement


Hei!Half life of the reaction is about 1,25 billion years. Assuming that the age of Earth is 5 billion years,the concentration of K on Earth since its formation decreased:
a)2times b)4times c)8 times d)16times


The Attempt at a Solution


So I just do like that T[itex]\underline{}1/2[/itex]=1,25 bilion years
age=5 bilion years
so in the beginning we have X and then it is X/2 so we 1,25 bilion year
then X/4 it is 2,5 bilion year and then X/8 so it is 5 bilion years
so the answer i think is it decreased 8 times
but my question is: is there easier way to do it?

thanks in advance!
 
Physics news on Phys.org
After half life you are left with [itex]\frac 1 2[/itex] of the initial amount, after two half lives you are left with half of that or [itex]\frac 1 2 \times \frac 1 2 = (\frac 1 2)^2[/itex] of the initial amount, after three half lives you are left with half of that or [itex](\frac 1 2)^3[/itex] of the initial amount and so on.
 

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