Question

question 12 • of 24 step 1 of 3 01:58:44 the half - life of gold - 194 is approximately 1.6 days. step 1 of 3: determine a so that ( a(t)=a_0a^t ) describes the amount of gold - 194 left after ( t ) days, where ( a_0 ) is the amount at time ( t = 0 ). round to six decimal places. answer: (to enter your answer (opens in new window) 2 points keypad keyboard shortcuts)

Explanation:

Step1: Recall half - life formula

For a radioactive decay model \(A(t)=A_0a^t\), at the half - life \(t = T_{1/2}\) (where \(T_{1/2}\) is the half - life), the amount of the substance \(A(T_{1/2})=\frac{1}{2}A_0\). We know that the half - life of Gold - 194, \(T_{1/2}=1.6\) days. So we substitute \(t = 1.6\) and \(A(1.6)=\frac{1}{2}A_0\) into the formula \(A(t)=A_0a^t\).

We get \(\frac{1}{2}A_0=A_0a^{1.6}\). Since \(A_0
eq0\) (because \(A_0\) is the initial amount of the substance), we can divide both sides of the equation by \(A_0\). This gives us \(\frac{1}{2}=a^{1.6}\).

Step2: Solve for \(a\)

To solve for \(a\), we take the 1.6 - th root of both sides. We can rewrite the equation \(\frac{1}{2}=a^{1.6}\) as \(a = (\frac{1}{2})^{\frac{1}{1.6}}\).

First, calculate \(\frac{1}{1.6}=\frac{10}{16}=\frac{5}{8} = 0.625\). So \(a = (\frac{1}{2})^{0.625}\).

Using a calculator, \((\frac{1}{2})^{0.625}=2^{- 0.625}\). We know that \(2^{-0.625}=\frac{1}{2^{0.625}}\). Calculating \(2^{0.625}\): \(2^{0.625}=e^{0.625\ln2}\). Since \(\ln2\approx0.693147\), then \(0.625\times\ln2 = 0.625\times0.693147\approx0.433217\). Then \(e^{0.433217}\approx1.542214\), so \(2^{-0.625}=\frac{1}{1.542214}\approx0.648419\).

Answer:

\(0.648419\)