Question

use the exponential decay model, $a = a_0 e^{kt}$, to solve the following. the half-life of a certain substance is 27 years. how long will it take for a sample of this substance to decay to 68% of its original amount? it will take approximately \\(\square\\) \\(\boldsymbol{\bigtriangledown}\\) for the sample of the substance to decay to 68% of its original amount. (round the final answer to on as needed.) round all intermediate values to four decimal places percent percent per year years

Explanation:

Step1: Find the value of \(k\)

When \(t = 27\) (half - life), \(A=\frac{A_0}{2}\). Substitute into \(A = A_0e^{kt}\):
\(\frac{A_0}{2}=A_0e^{27k}\)
Divide both sides by \(A_0\): \(\frac{1}{2}=e^{27k}\)
Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{27k})\)
Since \(\ln(e^{x}) = x\), we have \(-\ln(2)=27k\)
So, \(k=\frac{-\ln(2)}{27}\approx\frac{- 0.6931}{27}\approx - 0.0257\)

Step2: Find the time \(t\) when \(A = 0.68A_0\)

Substitute \(A = 0.68A_0\) and \(k\approx - 0.0257\) into \(A = A_0e^{kt}\):
\(0.68A_0=A_0e^{-0.0257t}\)
Divide both sides by \(A_0\): \(0.68 = e^{-0.0257t}\)
Take the natural logarithm of both sides: \(\ln(0.68)=\ln(e^{-0.0257t})\)
Since \(\ln(e^{x}) = x\), we have \(\ln(0.68)=- 0.0257t\)
Then \(t=\frac{\ln(0.68)}{-0.0257}\)
\(\ln(0.68)\approx - 0.3857\)
\(t=\frac{-0.3857}{-0.0257}\approx15.0\)

Answer:

\(15.0\) years