AP Physics 2: Half-Life Calculations

Understanding radioactive decay is crucial for fields ranging from archaeology to nuclear medicine. At the heart of this understanding is the concept of half-life, the predictable time it takes for half of a radioactive sample to decay. Mastering half-life calculations empowers you to date ancient artifacts, determine safe dosages for medical treatments, and grasp the fundamental behavior of unstable atomic nuclei.

What Half-Life Means

The half-life ($t_{½}$) of a radioactive isotope is defined as the time required for half of the atoms in a given sample to undergo decay. This is a statistical process; you cannot predict which specific atom will decay next, but for a large collection of atoms, the decay rate is remarkably consistent. A key feature of half-life is that it is constant for a given isotope and independent of external conditions like temperature, pressure, or chemical state. Whether you have a gram or a kilogram of Carbon-14, its half-life remains approximately 5,730 years. This constancy allows us to model decay mathematically using an exponential function, leading to reliable predictions about the remaining quantity of a substance over time.

The Core Mathematical Model: Exponential Decay

The exponential nature of radioactive decay is captured by a straightforward formula. If you start with an initial quantity $N_0$ (which could be number of atoms, mass, or activity), the amount $N$ remaining after a time $t$ has elapsed is given by:

$$N=N_0{\left(\frac{1}{2}\right)}^{(t/t_{½})}$$

This is the fundamental half-life formula. The exponent $(t/t_{½})$ represents the number of half-lives that have passed. For example, if three half-lives pass, the exponent is 3, and the remaining fraction is $(\frac{1}{2}{)}^3=\frac{1}{8}$ of the original. This equation directly models the stepwise, halving process characteristic of radioactive decay. It’s important to note that $t$ and $t_{½}$ must be expressed in the same units (e.g., both in years or both in seconds) for the calculation to be valid.

Rearranging the Equation for Different Goals

While the standard formula solves for the remaining quantity $N$, you will often need to solve for other variables: elapsed time $t$, the number of half-lives, or even the half-life $t_{½}$ itself. This requires using logarithms, which allow you to "bring down" the exponent for solving.

• Solving for Elapsed Time ($t$): To find out how much time has passed given a starting amount $N_0$ and a current amount $N$, rearrange and use the natural logarithm (ln) or base-10 log (log).

First, divide both sides by $N_0$: $\frac{N}{N_0}=(\frac{1}{2}{)}^{(t/t_{½})}$. Then take the logarithm of both sides: $\ln \left(\frac{N}{N_0}\right)=\frac{t}{t_{½}}\cdot \ln (\frac{1}{2})$. Finally, solve for $t$: $t=\frac{\ln (N/N_0)}{\ln (1/2)}\cdot t_{½}$. Since $\ln (1/2)=-\ln (2)$, this is often written as $t=\frac{\ln (N_0/N)}{\ln (2)}\cdot t_{½}$.

• Solving for the Half-Life ($t_{½}$): If you know the elapsed time and the fractional amount remaining, you can isolate $t_{½}$. From $\frac{N}{N_0}=(\frac{1}{2}{)}^{(t/t_{½})}$, taking logs gives $t_{½}=\frac{t\cdot \ln (1/2)}{\ln (N/N_0)}$ or, more commonly, $t_{½}=\frac{t\cdot \ln (2)}{\ln (N_0/N)}$.

Application: Carbon Dating

Carbon dating is a classic application of half-life calculations. Living organisms constantly exchange carbon with the atmosphere, maintaining a steady ratio of radioactive Carbon-14 (${}^{14}\text{C}$) to stable Carbon-12. Upon death, this exchange stops, and the ${}^{14}\text{C}$ begins to decay with its known half-life of 5,730 years.

Example Problem: A wooden tool is found to have a ${}^{14}\text{C}$ activity that is 25% of the activity found in a living tree sample. How old is the tool?

1. Identify the fraction remaining: $N/N_0=0.25$.
2. Determine the number of half-lives: $0.25=(\frac{1}{2}{)}^n$. We see that $(\frac{1}{2}{)}^2=0.25$, so $n=2$ half-lives.
3. Calculate the age: $t=n\times t_{½}=2\times 5,730\text{ years}=11,460\text{ years}$.

You could also use the logarithmic method: $t=\frac{\ln (0.25)}{\ln (0.5)}\times 5,730\text{ years}$, which yields the same result. This demonstrates how measuring the remaining ${}^{14}\text{C}$ allows archaeologists to determine the age of organic materials.

Application: Medical Isotopes

In medicine, radioactive isotopes are used for both diagnosis (e.g., PET scans) and treatment (e.g., radiation therapy). Half-life calculations are essential for determining safe and effective dosing. A medical isotope must have a half-life long enough to perform its function but short enough to minimize long-term radiation exposure to the patient.

Example Problem: Technetium-99m, a common diagnostic tracer, has a half-life of 6.0 hours. A sample is prepared with an initial activity of 80 mCi. What activity remains after 18 hours, ready to be disposed of?

1. Calculate the number of half-lives: $n=t/t_{½}=18\text{ hr}/6.0\text{ hr}=3.0$ half-lives.
2. Apply the decay formula: $N=N_0\times (\frac{1}{2}{)}^n=80\text{ mCi}\times (\frac{1}{2}{)}^3=80\text{ mCi}\times \frac{1}{8}=10\text{ mCi}$.

This calculation is vital for hospital radiopharmacies. They must produce enough activity for the procedure while ensuring leftover material has decayed to a safe level for storage and disposal, governed by strict regulations.

Common Pitfalls

1. Misapplying the Exponent: The most frequent error is placing $t$ and $t_{½}$ in the wrong order in the exponent. Remember, the exponent is not $(t_{½}/t)$ but $(t/t_{½})$. The exponent represents "how many half-lives fit into the elapsed time."

• Correction: Always ask: "How many half-life periods have passed?" That number is $t/t_{½}$.

1. Incorrect Units: Using time units that don't match will give a nonsense answer. If the half-life is 12 years and the elapsed time is 24 months, you must convert 24 months to 2 years before dividing.

• Correction: Before plugging numbers into any equation, ensure $t$ and $t_{½}$ are expressed in the same unit (seconds, hours, years, etc.).

1. Confusing Remaining vs. Decayed Amount: The formula $N=N_0(\frac{1}{2}{)}^{(t/t_{½})}$ gives you the amount remaining. A question asking "how much has decayed" requires an extra step: Decayed Amount = $N_0-N$.

• Correction: Read the question carefully. If asked for the amount decayed, calculate $N$ first, then subtract it from $N_0$.

1. Logarithm Mistakes: When using logarithms to solve for $t$ or $t_{½}$, students often misuse the properties of logs.

• Correction: Recall that $\ln (\frac{N}{N_0})=\ln (N)-\ln (N_0)$ and $\ln (\frac{1}{2})=-\ln (2)$. Write each step clearly to avoid sign errors.

Summary

• The half-life ($t_{½}$) is the constant time for half of a radioactive sample to decay, modeled by the exponential decay formula $N=N_0(\frac{1}{2}{)}^{(t/t_{½})}$.
• You can manipulate this formula using logarithms to solve for any variable: remaining quantity $N$, elapsed time $t$, or the half-life $t_{½}$ itself.
• Carbon dating uses the known half-life of Carbon-14 (5,730 years) and the measured remaining fraction in an artifact to calculate its age.
• In medical isotope applications, half-life calculations ensure proper dosage for diagnostic imaging and safe handling based on decayed activity.
• Avoid common errors by carefully checking the exponent order, ensuring consistent units, distinguishing between remaining and decayed amounts, and applying logarithm rules correctly.