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In the nuclear industry, workers use a rule of thumb that the radioactivity from any sample will be relatively harmless after 10 half-lives. Calculate the fraction of a radioactive sample that remains after this time. (Hint: Radioactive decays obey first-order kinetics.)

Short Answer

Expert verified
After 10 half-lives, the fraction of the original sample remaining is \((1/2)^{10}\) or approximately 0.000977.

Step by step solution

01

Definition of Half-Life

Half-life is the time taken for half of the radioactive substance to decay. Radioactivity is a first order process which means that the rate of decay is proportional to the amount of radioactive substance. After every half-life, half of the substance is decayed, leaving half of the starting material intact.
02

Apply the Rule of Ten Half-Lives

The rule of thumb in the nuclear industry states that after 10 half-lives, the radioactivity from any sample will be relatively harmless. If we imagine starting with a sample of 1 unit of radioactive substance, after each half-life, half of the initial material would remain. Therefore, the fraction of the original sample remaining after 1 half-life is \(1/2\), after 2 half-lives it's \((1/2)^2\) or \(1/4\), after 3 it's \((1/2)^3\) or \(1/8\), and so on.
03

Calculation of Remaining Fraction after 10 Half-Lives

By the same reasoning, after 10 half-lives, the fraction of the original sample remaining would be \((1/2)^{10}\). To calculate this, one must raise 1/2 to the power of 10.

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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 crucial concept in the study of radioactive decay, particularly in the nuclear industry. It refers to the time required for half of a radioactive substance to undergo decay. This means that every specific interval known as the half-life, half of the radioactive atoms in a sample will have transformed into a different substance.

The practical importance of half-life can be seen in how we predict the behavior of radioactive materials. For instance, if a radioactive substance has a half-life of ten years, in a decade only 50% of the original radioactive atoms remain. Another ten years later, only 25% (or half of the remaining 50%) would still be radioactive. This predictable pattern helps scientists and those working in nuclear industries to estimate how long a substance will remain hazardous.

In everyday terms, you could think about it like the melting of ice—only instead of melting, the atoms transform into something else.
First-Order Kinetics Explaining Decay
Radioactive decay is governed by first-order kinetics. This means that the rate at which the material decays is directly proportional to the amount of material present. In simpler terms, more material means faster decay. As the material decreases, the decay rate slows down.

Mathematically, first-order kinetics can be expressed with a simple equation:
  • The rate of decay = -k[N], where - - [N] is the concentration of the radioactive substances, and - k is the rate constant, unique to each substance.
With this understanding, you can see that even as the material halves over each half-life, the decay process continues efficiently. First-order kinetics provide a reliable method to calculate how much radioactive substance remains over time and plays a critical role in managing radioactive materials safely.
Significance in the Nuclear Industry
Within the nuclear industry, the concepts of half-life and first-order kinetics are fundamental in managing the safety and longevity of radioactive materials. Because radioactive materials emit energy that can be dangerous, understanding these concepts helps workers estimate when a substance will become safe.

This safety guideline—that radioactivity becomes relatively harmless after 10 half-lives—offers an operational framework for predicting the decay timeline of a substance.
  • For instance, if a particular isotope's half-life is 30 years, it takes 300 years before the material is virtually safe.
  • Planning and waste management strategies heavily rely on these calculations.
This understanding allows the nuclear industry to forecast risks, schedule checks, and ensure proper containment and disposal of radioactive materials, thereby safeguarding the environment and communities.

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Most popular questions from this chapter

Consider the reaction $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \\\ & \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+\mathrm{H}^{+}(a q)+\mathrm{I}^{-}(a q) \end{aligned} $$ How could you follow the progress of the reaction by measuring the electrical conductance of the solution?

In recent years ozone in the stratosphere has been depleted at an alarmingly fast rate by chlorofluorocarbons (CFCs). A CFC molecule such as \(\mathrm{CFCl}_{3}\) is first decomposed by UV radiation: $$ \mathrm{CFCl}_{3} \longrightarrow \mathrm{CFCl}_{2}+\mathrm{Cl} $$ The chlorine radical then reacts with ozone as follows: $$ \begin{array}{c} \mathrm{Cl}+\mathrm{O}_{3} \longrightarrow \mathrm{ClO}+\mathrm{O}_{2} \\ \mathrm{ClO}+\mathrm{O} \longrightarrow \mathrm{Cl}+\mathrm{O}_{2} \end{array} $$ (a) Write the overall reaction for the last two steps. (b) What are the roles of \(\mathrm{Cl}\) and \(\mathrm{ClO} ?\) (c) Why is the fluorine radical not important in this mechanism? (d) One suggestion to reduce the concentration of chlorine radicals is to add hydrocarbons such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) to the stratosphere. How will this Work?

Define the half-life of a reaction. Write the equation relating the half-life of a first-order reaction to the rate constant.

The rate law for the reaction $$ \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ is given by rate \(=k\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{NO}_{2}^{-}\right]\). At \(25^{\circ} \mathrm{C},\) the rate constant is \(3.0 \times 10^{-4} / M \cdot\) s. Calculate the rate of the reaction at this temperature if \(\left[\mathrm{NH}_{4}^{+}\right]=0.26 \mathrm{M}\) and \(\left[\mathrm{NO}_{2}^{-}\right]=0.080 \mathrm{M}\)

The integrated rate law for the zero-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) is \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-k t .\) (a) Sketch the follow- ing plots: (i) rate versus \([\mathrm{A}]_{t}\) and (ii) \([\mathrm{A}]_{t}\) versus \(t\). (b) Derive an expression for the half-life of the reaction. (c) Calculate the time in half-lives when the integrated rate law is no longer valid, that is, when \([\mathrm{A}]_{t}=0\)

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