Problem 29 - Algebra and Trigonometry

Question

Use the exponential decay model, \(A=A_{0} e^{k t},\) to solve Exercises \(28-31 .\) Round answers to one decimal place. The half-life of lead is 22 years. How long will it take for a sample of this substance to decay to \(80 \%\) of its original amount?

Step-by-Step Solution

Answer

It will take approximately \(t = \frac{22 \cdot ln(0.8)}{ln(2)}\) years for lead to decay to 80% of its original amount.

Step 1: Calculate the decay constant (k)

To find the decay constant, we can use the formula associated with the half-life of a substance, which is \(k = \frac{ln(2)}{T_{half}}\), where \(T_{half}\) is the half-life. According to the problem, the half-life (\(T_{half}\)) of Lead is 22 years. Plugging this into our equation gives \(k = \frac{ln(2)}{22}\).

Step 2: Calculate the time for decay to 80%

We're told that we need to find out when Lead decays to 80% of its original amount. Since the original amount is represented by \(A_0\), 80% of the original amount is 0.8 * \(A_0\). This gives us the equation \(0.8A_0 = A_0e^{kt}\), where \(t\) is the time that is required. We can cancel out \(A_0\) from both sides of the equation to simplify it. The equation thus simplifies to \(0.8 = e^{kt}\). Taking natural logarithms of both sides to solve for \(t\), we get \(ln(0.8) = ln(e^{kt})\). Which simplifies to \(ln(0.8) = kt\). Plug in our earlier calculated \(k\), we can solve for \(t\).

Step 3: Solve for time t

Our equation now is \(ln(0.8) = \frac{ln(2)}{22} \cdot t\). Solve for \(t\), therefore \(t = \frac{22 \cdot ln(0.8)}{ln(2)}\)