Proving the half-life of Potassium-40

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SUMMARY

The discussion centers on calculating the half-life of Potassium-40 (K-40), which is established as 1.3 billion years based on the decay processes of electron capture and β-emission. A sample containing 4.0x1018 nuclei emits 68 β-particles and photons per second, leading to the determination of the decay constant (λ) using the equation A=λN. The user initially miscalculated the half-life in seconds but was guided to convert the result to years, confirming the established half-life value.

PREREQUISITES
  • Understanding of radioactive decay processes, specifically electron capture and β-emission.
  • Familiarity with the decay constant (λ) and its relationship to activity (A) and number of nuclei (N).
  • Knowledge of logarithmic functions, particularly natural logarithms (ln).
  • Ability to convert time units from seconds to years.
NEXT STEPS
  • Review the derivation of the half-life formula T1/2 = ln2/λ.
  • Practice calculations involving radioactive decay using different isotopes.
  • Explore the implications of half-life in radiometric dating techniques.
  • Learn about the applications of K-40 in geology and archaeology.
USEFUL FOR

Students studying nuclear chemistry, physicists interested in radioactive decay, and professionals in fields such as geology and archaeology who utilize isotopic dating methods.

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Homework Statement



"K-40... decays by two radioactive processes. It can decay by electron capture or β- emission.

It is found that a sample containing 4.0x10^18 nuclei of K-40 emits a total of 68 β- particles and photons each second. This shows the half life is 1.3x10^9 years."

Use the data in the passage to show that the half life is 1.3x10^9 years.

Homework Equations



I'm assuming...

T1/2 = ln2/λ; A=λN; A=A0e^(-λ)(t)

The Attempt at a Solution



Using A=λN --> A/N=λ --> (68)/(4*10^18) = 1.7^10-17
then ln2/(1.7^10-17) = 4.1*10^16 ->> waayy too big.

I then tried various other combinations of the above, but to no success.

Please help! I know I'm missing something VERY obvious.
 
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You calculated (correctly) the half-life in seconds. Now convert to years.
 
phyzguy said:
You calculated (correctly) the half-life in seconds. Now convert to years.

D'oh!

You legend, thank you :D
 

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