Fresh rainwater or surface water contains enough tritium \(\left({ }^{3} \mathrm{H}\right)\) to show \(5.5\) decay events per minute per \(100 . \mathrm{g}\) water. Tritium has a half-life of \(12.3\) years. You are asked to check a vintage wine that is claimed to have been produced in \(1946 .\) How many decay events per minute should you expect to observe in \(100 . \mathrm{g}\) of that wine?

Short Answer

Expert verified
You should expect to observe about 0.0596 decay events per minute in 100g of the vintage wine claimed to have been produced in 1946.

Step by step solution

01

Calculate the number of half-lives

First, we have to find out how many half-lives have passed since 1946. We can accomplish this by dividing the time elapsed (from 1946 to now) by the half-life of tritium. Let's assume it is the year 2022. The elapsed time will be: Elapsed time = Current year - Vintage year Elapsed time = 2022 - 1946 = 76 years Now, divide the elapsed time by the half-life of tritium to find the number of half-lives: Number of half-lives = Elapsed time / Half-life of tritium Number of half-lives = 76 years / 12.3 years
02

Calculate the decay events per minute in the wine

Now that we know the number of half-lives, we can use this information to determine the decay events per minute in 100g of the wine. We can use the half-life formula, with the initial decay events per minute in rainwater or surface water, and the number of half-lives: Decay events per minute = Initial decay events per minute x (1 / 2) ^ Number of half-lives Decay events per minute = 5.5 events per minute x (1 / 2) ^ (76 years / 12.3 years)
03

Calculate the final answer

Now, we will simplify the expression to find the decay events per minute in the wine: Decay events per minute ≈ 5.5 events per minute x (1 / 2) ^ (6.178861788619) Decay events per minute ≈ 5.5 events per minute x 0.0108422274 Decay events per minute ≈ 0.05963225 events per minute Therefore, you should expect to observe about 0.0596 decay events per minute in 100g of the vintage wine claimed to have been produced in 1946.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life of Radioactive Isotopes
Understanding the half-life of radioactive isotopes is crucial in various fields, from archaeology to medicine. Half-life is the time required for half the atoms in a sample of a radioactive isotope to decay.

For instance, if we start with 100 unstable atoms with a half-life of one year, we'd expect to have 50 left after one year, 25 after two years, and so on. In our example with tritium \( ^3H \), which has a half-life of 12.3 years, it takes that amount of time for half of the tritium atoms in the sample to transform into another element through the process of radioactive decay.

When examining vintage wine for authentication, knowing the half-life allows us to calculate the decrease in radioactivity over time. This calculation is based on the assumption that the initial amount of tritium was known and that its decrease follows a predictable pattern described by the half-life.
Nuclear Chemistry
Nuclear chemistry deals with the reactions and processes that occur within atomic nuclei. Radioactive decay, a spontaneous process by which an unstable atomic nucleus loses energy, is a key concept in this field.

Tritium decay—where the tritium nucleus \( ^3H \) decomposes into a stable helium isotope \( ^3He \)—is a prime example of beta decay, one type of radioactive decay.

Why is tritium decay significant?

Its presence in organic matter like wine allows us to date its production. This is due to the predictability of nuclear processes, which can be described using mathematical models. Thus, nuclear chemistry not only elucidates how these decays occur but also provides tools to measure time and verify historical claims, as is the goal in testing the vintage wine.
Scientific Problem-Solving
Scientific problem-solving involves using a systematic approach to understand and resolve complex issues. Here's how we applied it:

Firstly, we identified our primary objective: to determine the number of decay events per minute in the vintage wine. Secondly, we gathered necessary data, such as the half-life of tritium and the initial decay events in contemporary water samples.

Then, using a mathematical formula that incorporates the concept of half-life, we calculated the expected decay events. To achieve a clearer and more reliable outcome, we meticulously implemented each step, ensuring careful consideration of each variable. This approach exemplifies how precision and methodical procedures are essential to scientific inquiry and finding solutions to intricate problems like determining the age of a vintage wine based on radioactive isotope decay.

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Most popular questions from this chapter

The curie (Ci) is a commonly used unit for measuring nuclear radioactivity: 1 curie of radiation is equal to \(3.7 \times 10^{10}\) decay events per second (the number of decay events from \(1 \mathrm{~g}\) radium in \(1 \mathrm{~s}\) ). A \(1.7-\mathrm{mL}\) sample of water containing tritium was injected into a \(150-\mathrm{lb}\) person. The total activity of radiation injected was \(86.5 \mathrm{mCi}\). After some time to allow the tritium activity to equally distribute throughout the body, a sample of blood plasma containing \(2.0 \mathrm{~mL}\) water at an activity of \(3.6 \mu \mathrm{Ci}\) was removed. From these data, calculate the mass percent of water in this \(150-\mathrm{lb}\) person.

A rock contains \(0.688 \mathrm{mg}^{206} \mathrm{~Pb}\) for every \(1.000 \mathrm{mg}{ }^{238} \mathrm{U}\) present. Assuming that no lead was originally present, that all the \({ }^{206} \mathrm{~Pb}\) formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between \({ }^{238} \mathrm{U}\) and \({ }^{206} \mathrm{~Pb}\) is negligible, calculate the age of the rock. (For \({ }^{238} \mathrm{U}\), \(t_{1 / 2}=4.5 \times 10^{9}\) years. \()\)

Much of the research on controlled fusion focuses on the problem of how to contain the reacting material. Magnetic fields appear to be the most promising mode of containment. Why is containment such a problem? Why must one resort to magnetic fields for containment?

Calculate the amount of energy released per gram of hydrogen nuclei reacted for the following reaction. The atomic masses are \({ }^{1} \mathrm{H}, 1.00782 \mathrm{amu} ;{ }_{1}^{2} \mathrm{H}, 2.01410\) amu; and an electron, \(5.4858 \times\) \(10^{-4}\) amu. (Hint: Think carefully about how to account for the electron mass.) $${ }_{1}^{1} \mathrm{H}+{ }_{1}^{1} \mathrm{H} \longrightarrow{ }_{1}^{2} \mathrm{H}+{ }_{+1}^{0} \mathrm{e}$$

Strontium-90 and radon-222 both pose serious health risks. \({ }^{90} \mathrm{Sr}\) decays by \(\beta\) -particle production and has a relatively long half-life (28.9 years). Radon- 222 decays by \(\alpha\) -particle production and has a relatively short half-life (3.82 days). Explain why each decay process poses health risks.

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