/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 The radioisotope hydrogen-3 (tri... [FREE SOLUTION] | 91Ó°ÊÓ

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The radioisotope hydrogen-3 (tritium) is used in fusion reactors. It is a \(\beta\) -emitter with a half-life of \(12.33\) years. Calculate the fraction of a hydrogen-3 sample that will remain after \(50.0\) years.

Short Answer

Expert verified
After 50 years, about 5.9% of tritium remains.

Step by step solution

01

Understand the Half-Life Concept

The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. For tritium, this time is 12.33 years. This means every 12.33 years, the quantity of tritium remaining is halved.
02

Determine the Number of Half-Lives Passed

To calculate how many half-lives have passed in 50.0 years, divide the total time period by the half-life of tritium.\[\text{Number of half-lives} = \frac{50.0}{12.33}\]
03

Calculate the Number of Half-Lives

Calculate the value from the previous step.\[\text{Number of half-lives} = \frac{50.0}{12.33} \approx 4.057\]Approximately 4.057 half-lives have passed.
04

Calculate the Remaining Fraction of Tritium

The fraction of tritium remaining after a certain number of half-lives can be calculated using the formula:\[\text{Remaining fraction} = \left( \frac{1}{2} \right)^{\text{Number of half-lives}} = \left( \frac{1}{2} \right)^{4.057} \]
05

Compute the Remaining Fraction

Perform the calculation:\[\text{Remaining fraction} = \left( \frac{1}{2} \right)^{4.057} \approx 0.059\]This means approximately 5.9% of the original tritium remains after 50.0 years.

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Key Concepts

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

Understanding Half-Life
Half-life is a fundamental concept in the study of radioactive decay. It represents the time required for half of the radioactive atoms in a sample to break down, or decay, into another form. In simple terms, if you have a certain amount of a radioactive substance, in one half-life period, half of that substance will have transformed into another element or isotope. This concept is crucial for predicting how long a radioactive material will continue to exist in its reactive form.

For example, in the case of tritium, the half-life is 12.33 years. This means that if you start with 100 grams of tritium, in 12.33 years, you will have about 50 grams remaining; the rest will have decayed. After another 12.33 years, you will have approximately 25 grams.
  • The half-life remains constant, irrespective of the initial quantity.
  • Understanding half-life helps in calculating the remaining quantity of a radioactive isotope over time.
Whether you are studying chemistry, physics, or environmental science, understanding half-life aids in grasping how radioactivity affects materials and environments.
Tritium and Its Uses
Tritium is a radioactive isotope of hydrogen, known scientifically as hydrogen-3. It naturally occurs in very low quantities in the atmosphere from interactions with cosmic rays, but it is also produced in nuclear reactors. Since tritium can emit beta particles, which are electrons, it is crucial in various scientific and industrial applications.

One of the significant applications of tritium is in fusion reactors. Fusion reactors aim to replicate the sun’s energy production method, using isotopes like tritium and deuterium to produce energy through nuclear fusion. In essence, tritium acts as a fuel in these reactors:
  • Tritium combines with deuterium at high temperatures to produce helium, along with a significant amount of energy.
  • The energy from this fusion process has the potential to become a clean and nearly limitless energy source.
However, handling tritium requires caution due to its radioactive nature. Safety protocols are paramount to prevent leaks and manage its decay effectively.
Fusion Reactors: The Future of Energy
Fusion reactors are seen as a revolutionary step towards sustainable energy, given their potential to provide an almost infinite source of energy without the downsides of traditional nuclear power. Unlike fission reactors, which split heavy atomic nuclei, fusion reactors work by merging light nuclei, like tritium and deuterium, to form a heavier nucleus. This process releases a large amount of energy, akin to what fuels our sun.

Some significant advantages of fusion reactors include:
  • No greenhouse gas emissions, making them environmentally friendly.
  • They produce significantly less radioactive waste compared to fission reactors.
  • The fuel sources (like tritium and deuterium) are more abundant and easier to source, as they primarily come from water and lithium.
While the promise of fusion energy is tremendous, achieving it is incredibly challenging due to the extreme conditions needed to sustain the reactions, such as high temperatures and pressures. Research and development efforts worldwide continue to make advancements, edging closer to making fusion reactors a viable energy source for the future.

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Most popular questions from this chapter

Sulfuryl chloride decomposes according to the equation \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) Using the following initial-rate data, determine the order of the reaction with respect to \(\mathrm{SO}_{2} \mathrm{Cl}_{2}:\) \begin{tabular}{cc} \hline & \multicolumn{2}{l} { Initial rate of reaction of } \\ {\(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]_{0} / \mathrm{mol} \cdot \mathrm{L}^{-1}\)} & \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) / \mathrm{mol} \cdot \mathrm{L}^{-1} \cdot \mathrm{s}^{-1}\) \\ \hline \(0.10\) & \(2.2 \times 10^{-6}\) \\ \(0.20\) & \(4.4 \times 10^{-6}\) \\ \(0.30\) & \(6.6 \times 10^{-6}\) \\ \(0.40\) & \(8.8 \times 10^{-6}\) \\ \hline \end{tabular} Calculate the value of the rate constant.

Show that for a first-order reaction, the time required for \(99.9 \%\) of the reaction to take place is about 10 times that required for \(50 \%\) of the reaction to take place.

The reaction described by the equation \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HCl}(g)\) was studied at \(300 \mathrm{~K}\), and the following initial-rate data were collected: \begin{tabular}{ccc} \hline & & Initial rate of formation \\ Run & {\(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right]_{0} / \mathrm{M}\)} & of \(\mathbf{C}_{2} \mathrm{H}_{4}(g) / \mathrm{M} \cdot \mathrm{s}^{-1}\) \\ \hline 1 & \(0.93\) & \(2.40 \times 10^{-30}\) \\ 2 & \(0.55\) & \(4.00 \times 10^{-30}\) \\ 9 & \(0.90\) & \(6.54 \times 10^{-30}\) \\ \hline \end{tabular} Determine the rate law and the value of the rate constant for the reaction.

The following initial-rate data were obtained for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{3}(g)\) as described by the equation $$ \mathrm{N}_{2} \mathrm{O}_{3}(g) \rightarrow \mathrm{NO}(g)+\mathrm{NO}_{2}(g) $$ \begin{tabular}{ccc} \hline Run & Initial pressure of \(P_{\mathrm{N}_{2} \mathrm{O}_{3}} /\) Torr & Initial rate of formation of \(\mathrm{NO}_{2}(g) /\) Torr \(\cdot \mathrm{s}^{-1}\) \\ \hline 1 & \(0.91\) & \(5.5\) \\ 2 & \(1.4\) & \(8.4\) \\ 3 & \(2.1\) & 13 \\ \hline \end{tabular} Determine the rate law for the reaction, expressed in terms of \(P_{\mathrm{N}_{2} \mathrm{o}_{3}}\) rather than \(\left[\mathrm{N}_{2} \mathrm{O}_{3}\right] .\) Calculate the value of the rate constant for the reaction.

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