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Tritium has a half-life of \(12.33 \mathrm{yr}\). What percentage of the \({ }^{3} \mathrm{H}\) nuclei in a tritium sample will decay during a period of 5 years?

Short Answer

Expert verified
About 35.1% of the tritium nuclei will decay in a period of 5 years.

Step by step solution

01

Understand the Exponential Decay Model

The exponential decay model can be written as \(N=N_0 e^{-λt}\), where \(N\) is the final amount, \(N_0\) is the initial amount, \(λ\) is the decay constant, and \(t\) is time. Here, the decay constant can be estimated as \(λ = \ln(2)/T_{1/2}\), where \(T_{1/2}\) is the half-life period. This formula can be used to calculate the percentage of the decayed substance.
02

Calculate the Decay Constant λ

The decay constant \(λ\) can be calculated using the given half-life period of tritium. Substituting \(T_{1/2} = 12.33\) years into \(λ = \ln(2)/T_{1/2}\), the decay constant can be found.
03

Substitute the values

Substitute \(λ\) and the time \(t = 5\) years into the exponential decay formula \(N=N_0 e^{-λt}\). Since \(N_0\) is initial amount, hence 100%, one can calculate the remaining amount \(N\), which will be the percentage of tritium left after 5 years.
04

Calculate the percentage decayed

Since the question asks for the percentage decayed, subtract the percentage remaining from 100%. This will give the percentage of tritium that has decayed over 5 years.

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Key Concepts

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

Half-Life
The concept of half-life is fundamental to the comprehension of exponential decay, particularly when discussing radioactive substances like tritium. Half-life is the time needed for half of the radioactive nuclei in a sample to undergo decay. This period is a constant for each radioactive isotope, meaning it does not depend on the initial amount of the substance or external factors like temperature.

For example, if a substance has a half-life of 12.33 years, as in the case of tritium from our exercise, it will take that amount of time for half of the tritium atoms to decay. Let's visualize this concept: if you start with a 100 grams sample, after 12.33 years, you would have 50 grams of tritium left, and the other 50 grams would have decayed into other elements or isotopes.

Furthermore, understanding half-life is crucial for determining how much of a substance will remain after a certain period. By applying the half-life concept, one can also infer how much of a substance has decayed, which is exactly what the textbook exercise asked for. In simple terms, the half-life gives us a prediction tool for how the quantity of a radioactive sample will decrease over time.
Decay Constant
The decay constant, represented by the Greek letter lambda (\( \text{λ} \)), is an intrinsic property of a radioactive substance that indicates the rate at which the substance undergoes decay. It is a measurement of the probability of decay per unit time.

In mathematical terms, the decay constant connects to the half-life through the equation \[ \lambda = \frac{\ln(2)}{T_{1/2}} \.\] Using this equation, if we know the half-life of a substance, we can calculate the decay constant, which will then apply to the exponential decay formula. In the context of the textbook exercise, after calculating the decay constant for tritium, it becomes a simple task to determine the percentage of a sample that will decay over a given time span.

Exponential Decay Formula

The decay constant is an integral part of the exponential decay formula: \[ N = N_0 e^{-\lambda t} \.\] This formula allows us to calculate the remaining amount of a substance after time 't'. The decay constant helps us understand and predict how quickly or slowly a substance will decay over time, which is essential for both scientific research and practical applications such as medical treatments and archaeological dating.
Radioactive Decay
Radioactive decay is a random but predictable process where unstable atomic nuclei release energy in the form of particles or electromagnetic waves. This process leads to the transformation of original elements into different elements or isotopes. Radioactive decay is the principle behind phenomena such as radiometric dating and is extensively used in fields including medicine, archaeology, and nuclear energy.

The predictability of radioactive decay lies in the statistical pattern of a large number of nuclei. Even though we cannot predict the decay of a single atom, the overall decay rate of a sizable sample is remarkably consistent, governed by the substance's decay constant and half-life. The exercise tackled the concept by asking us to calculate the decay of tritium nuclei over five years, demonstrating the practical application of understanding radioactive decay.

  • Types of Radioactive Decay: There are three main types of radioactive decay: alpha, beta, and gamma decay, each involving different particles and energy release.
  • Alpha Decay: Involves the release of alpha particles, which are helium nuclei.
  • Beta Decay: Occurs when a neutron transforms into a proton, emitting a beta particle (electron).
  • Gamma Decay: Involves the release of gamma radiation, which is high-energy electromagnetic radiation.

The study of radioactive decay is pivotal in understanding the stability of atoms, predicting the behavior of radioactive materials, and managing safety concerns associated with radioactivity.

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Most popular questions from this chapter

(a) Use energy methods to calculate the distance of closest approach for a head-on collision between an alpha particle with an initial energy of \(0.5 \mathrm{MeV}\) and a gold nucleus \(\left({ }^{197} \mathrm{Au}\right)\) at rest. (Assume that the gold \(n u-\) cleus remains at rest during the collision.) (b) What minimum initial speed must the alpha particle have in order to approach to a distance of \(300 \mathrm{fm}\) ?

A laboratory stock solution is prepared with an initial activity due to \({ }^{24} \mathrm{Na}\) of \(2.5 \mathrm{mCi} / \mathrm{mL}\), and \(10 \mathrm{~mL}\) of the stock solution is diluted (at \(t_{0}=0\) ) to a working solution with a total volume of \(250 \mathrm{~mL}\). After \(48 \mathrm{~h}\), a \(5-\mathrm{mL}\) sample of the working solution is monitored with a counter. What is the measured activity? (Note: \(1 \mathrm{~mL}=\) 1 milliliter, and the half-life of \({ }^{24} \mathrm{Na}\) is \(15.0 \mathrm{~h}\).)

A freshly prepared sample of a certain radioactive isotope has an activity of \(10 \mathrm{mCi}\). After an elapsed time of \(4 \mathrm{~h}\), its activity is \(8 \mathrm{mCi}\). (a) Find the decay constant and half-life of the isotope. (b) How many atoms of the isotope were contained in the freshly prepared sample? (c) What is the sample's activity \(30 \mathrm{~h}\) after it is prepared?

During the manufacture of a steel engine component, radioactive iron \(\left({ }^{59} \mathrm{Fe}\right)\) is included in the total mass of \(0.2 \mathrm{~kg}\). The component is placed in a test engine when the activity due to this isotope is \(20 \mu\) Ci. After a 1000 -h test period, oil is removed from the engine and found to contain enough \({ }^{59} \mathrm{Fe}\) to produce 800 disintegrations \(/\) min per liter of oil. The total volume of oil in the engine is \(6.5 \mathrm{~L} .\) Calculate the total mass worn from the engine component per hour of operation. (The halflife for \({ }^{59} \mathrm{Fe}\) is \(45.1\) days.)

The concept of radioactive half-life was described in Section 13.4, and Equation \(13.11\) gives the relationship between \(T_{1 / 2}\) and \(\lambda\). Another parameter that is often useful in the description of radioactive processes is the mean life, \(\tau\). Although the half-life of a radioactive isotope is accurately known, it is not possible to predict the time when any individual atom will decay. The mean life is a measure of the average length of existence of all the atoms in a particular sample. Show that \(\tau=1 / \lambda\). (Hint: Remember that \(\tau\) is essentially an average value, and use the fact that the number of atoms that decay between \(t\) and \(t+d t\) is equal to \(d N\). Furthermore, note that these \(d N\) atoms have a finite time of existence, \(t\) )

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