Problem 65 - Physics for Scientists and Engineers with Modern Physics

Question

Tritium dating. The \({ }_{1}^{3} \mathrm{H}\) isotope of hydrogen, which is called tritium (because it contains three nucleons), has a half-life of 12.3 yr. It can be used to measure the age of objects up to about \(100 \mathrm{yr} .\) It is produced in the upper atmosphere by cosmic rays and brought to Earth by rain. As an application, determine approximately the age of a bottle of wine whose \({ }_{1}^{3} \mathbf{H}\) radiation is about \(\frac{1}{10}\) that present in new wine.

Step-by-Step Solution

Answer

The wine is approximately 41 years old.

Step 1: Understand the Half-Life Concept

The half-life of a radioactive material is the time it takes for half of the radioactive isotopes in a sample to decay. For tritium, the half-life is 12.3 years. This means that every 12.3 years, the amount of tritium is halved.

Step 2: Use the Half-Life Formula

The decay of radioactive material is modeled by the formula \( N(t) = N_0 \left( \frac{1}{2} \right)^{t/T} \), where \( N(t) \) is the amount of substance remaining after time \( t \), \( N_0 \) is the original amount of the substance (new wine), and \( T \) is the half-life.

Step 3: Setup the Equation

Given that the tritium radiation in the old wine is \( \frac{1}{10} \) of that in new wine, we set up the equation: \( \frac{N_0}{10} = N_0 \left( \frac{1}{2} \right)^{t/12.3} \). We want to solve for \( t \), the age of the wine.

Step 4: Solve for t

First, simplify the equation: \( \frac{1}{10} = \left( \frac{1}{2} \right)^{t/12.3} \). Take the logarithm of both sides to get: \( \log\left( \frac{1}{10} \right) = \frac{t}{12.3} \log\left( \frac{1}{2} \right) \). Solve for \( t \): \[ t = 12.3 \cdot \frac{\log\left( \frac{1}{10} \right)}{\log\left( \frac{1}{2} \right)} \]

Step 5: Calculate the Result

Using the logarithm values: \( \log\left( \frac{1}{10} \right) = -1 \) and \( \log\left( \frac{1}{2} \right) \approx -0.301 \), plug these into the equation: \[ t \approx 12.3 \times \frac{-1}{-0.301} \approx 40.9 \text{ years} \]

Key Concepts

Half-Life Concept

The half-life of a radioactive isotope is a fascinating concept in physics and chemistry. It is the time required for half of an isotope in a sample to undergo decay.
For example, if you start with 100 atoms of a radioactive substance, over one half-life, approximately 50 atoms would have decayed.
Each substance has a characteristic half-life that is constant and unaffected by external conditions.

• In the case of tritium (\( {}_1^3 \mathrm{H} \)), the half-life is 12.3 years.
• This means after 12.3 years, only half of the initial amount of tritium will remain, and it will continue to halve every subsequent 12.3 years.
• Understanding half-life helps us calculate the age of ancient objects, like the bottle of wine in our problem, by measuring how much of a specific isotope remains.

Radioactive Decay Formula

Radioactive decay is mathematically described using specific formulas that help us predict the amount of substance left after a certain period.
One commonly used formula is the half-life decay formula:
\[N(t) = N_0 \left( \frac{1}{2} \right)^{t/T} \]

• \(N(t)\) represents the amount of radioactive isotope remaining after time \(t\).
• \(N_0\) is the initial amount of radioactive isotope present.
• \(T\) is the half-life period of the isotope.

For the wine problem, the formula helps us set up the equation:
\[\frac{N_0}{10} = N_0 \left( \frac{1}{2} \right)^{t/12.3}\]This equation implies that in the old wine, the tritium content is \(\frac{1}{10}\)th of what was originally present in new wine, leading us to solve for the age \(t\).
This formula is crucial because it quantifies the decay process in a consistent way, providing a robust tool for age estimation.

Logarithm in Physics

Logarithms play a significant role in physics, especially when dealing with exponential relationships like radioactive decay.
They help in solving equations where the variable in question is an exponent, such as the decay formula.
To find\(t\) in our equation:

• Set \(\frac{1}{10} = \left( \frac{1}{2} \right)^{t/12.3}\).
• Take the logarithm of both sides to facilitate solving.

This leads to:\[\log\left( \frac{1}{10} \right) = \frac{t}{12.3} \log\left( \frac{1}{2} \right)\]By rearranging the formula, you can solve for \(t\):
\[ t = 12.3 \cdot \frac{\log\left( \frac{1}{10} \right)}{\log\left( \frac{1}{2} \right)}\]Logarithms thus simplify calculations by turning multiplicative relationships into additive ones.
As each side of the equation can now be isolated, it makes it easier to calculate an unknown exponential factor such as \(t\).