Problem 78 - Fundamentals of Physics

Question

Radioactive element \(A A\) can decay to either element \(B B\) or element \(C C\). The decay depends on chance, but the ratio of the resulting number of \(B B\) atoms to the resulting number of \(C C\) atoms is always \(2 / 1 .\) The decay has a half-life of 8.00 days. We start with a sample of pure \(A A .\) How long must we wait until the number of \(C C\) atoms is 1.50 times the number of \(A A\) atoms?

Step-by-Step Solution

Answer

The wait time is approximately 8.758 days.

Step 1: Understand the problem

The problem involves a radioactive decay where element \(AA\) decays to \(BB\) and \(CC\) at a fixed ratio. We need to find out when \(CC\) is 1.5 times \(AA\). The half-life of \(AA\) is 8 days.

Step 2: Establish the decay ratio

Given the ratio of \(BB\) to \(CC\) is 2:1, for every 3 atoms of \(AA\) that decay, 2 become \(BB\) and 1 becomes \(CC\). Therefore, if \(x\) atoms have decayed to \(BB\), \(x/2\) atoms have decayed to \(CC\).

Step 3: Relate remaining AA to formed CC

Let \(N_0\) be the initial number of \(AA\) atoms, and \(t\) be the time in days. At any time \(t\), the number of \(AA\) atoms is \(N_0 e^{-kt}\), where \(k = \frac{\ln(2)}{8} \). The number of \(CC\) atoms is \(\frac{1}{3}(N_0 - N_0 e^{-kt})\) since a third of all decayed \(AA\) becomes \(CC\).

Step 4: Set up the equation using given condition

We are asked to find \(t\) such that \(\text{Number of CC} = 1.5 \times \text{Number of AA}\). This means \(\frac{1}{3}(N_0 - N_0 e^{-kt}) = 1.5 \cdot N_0 e^{-kt}\).

Step 5: Solve for t

Simplify the equation: \(N_0 - N_0 e^{-kt} = 4.5 N_0 e^{-kt}\). This leads to \(1 - e^{-kt} = 4.5 e^{-kt}\). Rearranging gives \(1 = 5.5 e^{-kt}\), so \(e^{kt} = \frac{5.5}{4.5} = \frac{11}{9}\). Taking natural logs, \(kt = \ln(\frac{11}{9})\), thus \(t = \frac{\ln(\frac{11}{9})}{\ln(2)/8}\), which simplifies to \(t \approx 8.758\) days.

Key Concepts

Half-life

In radioactive decay, the concept of half-life is fundamental. It describes the time required for half of the radioactive nuclei in a sample to decay into other elements.
This property is characteristic of each radioactive isotope. If we have a substance with a half-life of 8 days, this means that if you start with a certain amount of this substance, half of it will have decayed to its daughter products in that time span.

• It is important to note that after two half-lives, you won't have the original substance entirely used up but only reduced to a quarter of the initial amount.
• The remaining quantity of the substance follows an exponential decay pattern.

Understanding half-life helps us model decay processes precisely over time. This property remains constant regardless of the quantity of the substance present.

Decay ratio

The decay ratio is a way to express the likelihood of a radioactive element transforming into different products. For instance, the problem states that element AA decays to either BB or CC, and the decay ratio of BB to CC is 2:1.
This implies that for every three decaying atoms of AA, two will become BB, and one will become CC.

• This means that knowing the initial amount of AA, you can calculate the amount of BB and CC formed over time.
• It's crucial to maintain this fixed ratio in order to predict how many atoms of each product will be formed as time elapses.

Understanding these ratios is key to harnessing further calculations and predicting phenomena in processes like radioactive decay in nuclear chemistry.

Natural logarithm

The natural logarithm, denoted as \ln(x)\, is a useful mathematical function in equations involving exponential decay, such as those found in radioactive processes.
In the original exercise, the natural logarithm helps convert exponential equations into linear ones, which are easier to solve.

• Essentially, \ln\ is the inverse function of exponentially raised Euler’s number \(e\).
• Using natural logs can simplify the calculation of time required for decay, as shown through the equation transformation in the solution.

By utilizing \ln\, we can effectively "undo" the exponential function, allowing us to isolate variables like time in decay equations and solve them efficiently.

Exponential decay

Exponential decay describes how quantities decrease at a rate proportional to their current value. In the realm of radioactive decay, this means that the amount of a substance decreases exponentially over time.

• The formula for exponential decay is given by \(N(t) = N_0 e^{-kt}\).Where \(N(t)\) is the quantity of the substance at time \(t\), \(N_0\) is the initial amount, and \(k\) is the decay constant related to the half-life.
• The decay constant \(k\) can be computed using the half-life: \(k = \frac{\ln(2)}{\text{half-life}}\).
• An important characteristic of exponential decay is that it does not stop or slow down but always decreases by a constant percentage in equal increments of time.

This dynamic is what makes modeling decay in radioactive materials so precise and reliable. By understanding exponential decay, one can predict how much of a substance will remain after any given period.