Problem 28 - Precalculus : Building Concepts and Connections

Question

The half-life of plutonium-238 is 88 years. (a) Given an initial amount of \(A_{0}\) grams of plutonium238 at time \(t=0,\) find an exponential decay model, \(A(t)=A_{0} e^{k t},\) that gives the amount of plutonium238 at time \(t, t \geq 0\). (b) Calculate the time required for \(A_{0}\) grams of plutonium- 238 to decay to \(\frac{1}{3} A_{0}\).

Step-by-Step Solution

Answer

The exponential decay model for plutonium-238 is \(A(t) = A_{0} e^{\frac{ln(\frac{1}{2})}{88} t}\). Meanwhile, the time required to decay to \(\frac{1}{3} A_{0}\) can be found using the formula \(t = \frac{ln(\frac{1}{3})}{\frac{ln(\frac{1}{2})}{88}}\), which should be calculated to get a numerical time value.

Step 1: Determine the Rate of Decay

Firstly, we know the half-life of plutonium-238 is 88 years. Half-life means the time it takes for half of the initial substance to decay. So we can say that \(A_{0}\) will go down to \(\frac{1}{2} A_{0}\) in 88 years. We use the decay model \(A(t) = A_{0} e^{k t}\) where \(k\) is the rate of decay. Half-life indicates that \(A(88) = \frac{1}{2} A_{0}\), so we substitute these values into our decay model, getting \(A_{0} e^{k * 88} = \frac{1}{2} A_{0}\). We solve this equation for \(k\).

Step 2: Solve for Decay Rate

Now we solve our equation \(A_{0} e^{k * 88} = \frac{1}{2} A_{0}\) for \(k\). We start by cancelling out \(A_{0}\) from both sides of the equation, which gives us \(e^{k * 88} = \frac{1}{2}\). Taking the natural logarithm (ln) on both sides to get rid of \(e\), we have \(k * 88 = ln(\frac{1}{2})\). Finally, we find \(k\) by dividing both sides by 88, so \(k = \frac{ln(\frac{1}{2})}{88}\).

Step 3: Subsitute Decay Rate into Model

Now we have the decay constant \(k\), so we can fill it into our decay model \(A(t) = A_{0} e^{k t}\). Therefore, our exponential decay model becomes \(A(t) = A_{0} e^{\frac{ln(\frac{1}{2})}{88} t}\).

Step 4: Calculate Time for Decay to One Third

We're asked to find out the time needed for the plutonium-238 to decay to \(\frac{1}{3} A_{0}\). That means we set \(A(t)\) to \(\frac{1}{3} A_{0}\) in our decay model and solve for \(t\). Our equation now is \(\frac{1}{3} A_{0} = A_{0} e^{\frac{ln(\frac{1}{2})}{88} t}\). We cancel \(A_{0}\) out from both sides and solve for \(t\).

Step 5: Solve for Time

We proceed to solve the equation \(\frac{1}{3} = e^{\frac{ln(\frac{1}{2})}{88} t}\) for \(t\). Again, we take the natural logarithm of both sides to get \(\frac{ln(\frac{1}{3})}{\frac{ln(\frac{1}{2})}{88}} = t\). Thus, after simplifying, we find the time \(t\).

Key Concepts

Half-life

Understanding the concept of half-life is crucial when studying radioactive substances such as plutonium-238. Half-life is a period required for a quantity to reduce to half its initial value due to decay. This concept is particularly useful in the context of exponential decay models because it serves as a key time point for predicting how long it takes for a substance to undergo significant decay.

For instance, if the half-life of plutonium-238 is 88 years, it means that every 88 years, the mass of a sample of plutonium-238 will be half of what it was at the beginning of that period. Hence, if you started with 100 grams of plutonium-238, you'd expect to have 50 grams remaining after 88 years. Half-life is a fixed characteristic of radioactive isotopes and is used to establish the rate of decay in an exponential decay model.

Rate of Decay

The rate of decay, denoted by the constant 'k' in the exponential decay model, acts as a multiplier in the formula to express the rate at which the substance decays over time. It's derived from the natural logarithm of the half-life and characterizes how quickly the process happens.

The decay model is represented mathematically as A(t) = A0ekt, where A(t) is the amount at time t, A0 is the initial amount, and e is the base of the natural logarithm. By substituting the half-life into the model and rearranging the equation, you obtain the precise rate at which the substance is decaying. This rate is consistent over time, leading to an exponential decrease in the remaining quantity of the substance.

Natural Logarithm

The natural logarithm, often abbreviated as 'ln', is a mathematical operation that's the inverse of exponentiation with base e, roughly equal to 2.71828. It plays a vital part in calculating the decay rate and analyzing decay processes. In equations dealing with exponential growth or decay, taking the natural logarithm of both sides allows us to solve for the exponent, which often represents time or the rate of decay.

For example, when calculating the rate of decay for plutonium-238, we take the natural logarithm of the half-life condition to isolate the decay rate constant 'k'. In practice, this means that if we know the amount of substance present after a time period and the half-life, we can work out the length of the time period by essentially 'reversing' the exponential function with the help of natural logarithms.