Half-life of radioactive substance

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SUMMARY

The half-life of a radioactive substance, where 20% disappears in 70 days, is calculated to be 217.44 days. The decay model used is represented by the equation y = C*e^(k*t), with k determined to be -3.188 * 10^-3. By substituting this k value back into the equation to find the time at which 50% of the substance remains, the result confirms the half-life calculation. This method effectively demonstrates the application of exponential decay in radioactive substances.

PREREQUISITES
  • Understanding of exponential decay models
  • Familiarity with the equation y = C*e^(k*t)
  • Basic algebra for solving equations
  • Knowledge of radioactive decay concepts
NEXT STEPS
  • Study the derivation of the exponential decay formula y = C*e^(k*t)
  • Learn about different methods for calculating half-life in various contexts
  • Explore applications of radioactive decay in real-world scenarios
  • Investigate the impact of varying decay constants on half-life calculations
USEFUL FOR

Students studying physics or chemistry, educators teaching radioactive decay, and anyone interested in mathematical modeling of decay processes.

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Homework Statement



If 20% of a radioactive substance disappears in 70 days, what is its half-life?

Homework Equations



y = C*e^(k*t)
where t is time in days
k is the constant of proportionality?
y is the current amount of substance


The Attempt at a Solution



20% disappears, so 100% - 20% = 80% remaining = 0.80 after 70 days
.8*C = C*e^(70*k)
Solve for k
k = -3.188 * 10^-3

Plug k value back into equation to find t at the substance's half-life (.5)
.5*C = C*e^((-3.188 * 10^-3)*t)
Solve for t
t = 217.44

217.44 days is the substance's half-life.

Is that correct?
Thank you.
 
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