Question

use the exponential decay model ( a = a_0 e^{kt} ) to solve the following. the half - life of a certain substance is 29 years. how long will it take for a sample of this substance to decay to 74% of its original amount? it will take approximately (\boldsymbol{\text{blank}}) for the sample of the substance to decay to 74% of its original amount. round all intermediate values to four decimal places as needed. round the final answer to one decimal place as needed.

Explanation:

Step1: Recall the exponential decay formula

The exponential decay model is \( A = A_0 e^{kt} \), and for half - life, when \( t = T_{1/2}\) (half - life), \( A=\frac{A_0}{2}\). Given \( T_{1/2}=29\) years, we substitute into the formula: \(\frac{A_0}{2}=A_0e^{k\times29}\). Divide both sides by \( A_0\) (since \( A_0
eq0\)) to get \(\frac{1}{2}=e^{29k}\). Take the natural logarithm of both sides: \(\ln(\frac{1}{2}) = 29k\), so \( k=\frac{\ln(0.5)}{29}\approx\frac{- 0.6931}{29}\approx - 0.0239\).

Step2: Find the time to decay to 74% of original

We want to find \( t\) when \( A = 0.74A_0\). Substitute into the decay formula \( A = A_0e^{kt}\): \( 0.74A_0=A_0e^{kt}\). Divide both sides by \( A_0\) to get \( 0.74 = e^{kt}\). Take the natural logarithm of both sides: \(\ln(0.74)=kt\). We know \( k=\frac{\ln(0.5)}{29}\), so \( t=\frac{\ln(0.74)}{k}=\frac{\ln(0.74)}{\frac{\ln(0.5)}{29}}=29\times\frac{\ln(0.74)}{\ln(0.5)}\). Calculate \(\ln(0.74)\approx - 0.3011\) and \(\ln(0.5)\approx - 0.6931\). Then \( t = 29\times\frac{- 0.3011}{- 0.6931}\approx29\times0.4344\approx12.6\) (rounded to one decimal place).

Answer:

\(12.6\) years