Question

\(^{214} \mathrm{Pb}\) has a half-life of \(26.8 \mathrm{~min}\). How many minutes must elapse for \(90.0 \%\) of a given sample of \({ }^{214} \mathrm{~Pb}\) atoms to decay?

Short answer

Answer: It takes approximately 89.16 minutes for 90% of a given sample of 214Pb atoms to decay.

Step-by-step solution

Understand the given information and the half-life formula

We are given the half-life of \(^{214}\mathrm{Pb}\) atoms as 26.8 minutes, and we need to find the time it takes for 90% of the sample to decay. We can use the half-life formula: \(N_t = N_0 \times \left( \frac{1}{2} \right)^\frac{t}{t_{1/2}}\) where: - \(N_t\) is the remaining amount of substance at time \(t\) - \(N_0\) is the initial amount of substance - \(t\) is the time elapsed - \(t_{1/2}\) is the half-life of the substance Since 90% of the substance has decayed, it means that 10% is remaining. So, we can write \(N_t\) as \(0.1N_0\). Now, we need to find the time \(t\).

Setup the equation and isolate time

We can rewrite the half-life formula with the given values for the initial and remaining amounts of \(^{214}\mathrm{Pb}\) atoms: \(0.1N_0 = N_0 \times \left( \frac{1}{2} \right)^\frac{t}{26.8}\) Since we are looking for the time, we can eliminate the initial amount \(N_0\) from both sides: \(0.1 = \left( \frac{1}{2} \right)^\frac{t}{26.8}\) Now, we need to solve for \(t\).

Solve for time

To solve for \(t\), we can take the natural logarithm of both sides of the equation: \(\ln(0.1) = \ln \left(\frac{1}{2}\right)^{\frac{t}{26.8}}\) Using the properties of logarithms, we can bring the exponent to the front: \(\ln(0.1) = \frac{t}{26.8} \times \ln \left( \frac{1}{2} \right)\) Now, we can solve for \(t\) by isolating it: \(t = \frac{26.8 \times \ln(0.1)}{\ln\left(\frac{1}{2}\right)}\) Calculating the value for \(t\): \(t \approx 89.16 \mathrm{~minutes}\)

Conclusion

It takes approximately 89.16 minutes for 90% of a given sample of \(^{214}\mathrm{Pb}\) atoms to decay.

Key concepts

Half-life

In the context of radioactive decay, half-life is a fundamental concept. It describes the time required for half of a radioactive isotope within a sample to decay. This means that after one half-life, 50% of the original substance will remain. For isotopes like \(^{214}\mathrm{Pb}\), which has a half-life of 26.8 minutes, this means that every 26.8 minutes, half of any given sample of \(^{214}\mathrm{Pb}\) will have decayed into another element or isotope.
Knowing the half-life allows us to predict how long it will take for a certain percentage of the isotope to decay. The decay follows an exponential pattern, which is why half-life is a consistent value regardless of the amount of the substance present. This makes half-life a powerful tool in radioactive aging and dating techniques, as well as in understanding the behavior of radioactive substances.

Natural Logarithm

The natural logarithm is a crucial mathematical function when dealing with exponential decay processes, like radioactive decay. In the half-life formula, we often use the natural logarithmic transformation to solve for variables like time.
The natural logarithm, denoted as \(\ln\), is the base \(e\) (approximately 2.718) logarithm. It comes into play in problems involving exponential change because it effectively linearizes the problem. For example, in the equation \(\ln(0.1) = \frac{t}{26.8} \times \ln \left( \frac{1}{2} \right)\), the logarithms allow us to isolate \(t\) by taking advantage of the property that \(\ln(a^b) = b \cdot \ln(a)\). This conversion makes it much easier to solve exponential equations and find unknown variables.

Decay Percentage

Decay percentage tells us how much of the original sample of a radioactive isotope has transformed into another substance over a given period of time. It is expressed as a percentage and is crucial for understanding both the scale and implications of radioactive decay.
In this exercise, we focus on a 90% decay percentage. This means that 90% of the \(^{214}\mathrm{Pb}\) sample has decayed, leaving only 10% remaining. Understanding decay percentage is important because it describes the decay process's progress and helps in determining variables such as the time required to reach a certain level of decay.

• This is often determined by collecting decay data and applying decay equations that include the logarithm and half-life concepts.
• In practical terms, knowing the percentage of decay helps in fields from archaeology to nuclear medicine, supporting dating methods and safety protocols.