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Chapter 14: Problem 74

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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Most popular questions from this chapter

The reaction \(2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow \mathrm{C}\) is first order with respect to \(\mathrm{A}\) and \(\mathrm{B}\). When the initial concentrations are \([\mathrm{A}]=1.6 \times 10^{-2} M\) and \([\mathrm{B}]=2.4 \times 10^{-3} M,\) the rate is \(4.1 \times 10^{-4} M / \mathrm{s} .\) Calculate the rate constant of the reaction.

Consider this mechanism for the enzyme-catalyzed reaction $$ \mathrm{E}+\mathrm{S} \stackrel{k_{1}}{\rightleftharpoons_{-1}} \mathrm{ES} \quad \text { (fast equilbrium) } $$ $$ \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \quad(\text { slow }) $$ Derive an expression for the rate law of the reaction in terms of the concentrations of \(\mathrm{E}\) and \(\mathrm{S}\). (Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.)

For the reaction $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ the frequency factor \(A\) is \(8.7 \times 10^{12} \mathrm{~s}^{-1}\) and the activation energy is \(63 \mathrm{~kJ} / \mathrm{mol}\). What is the rate constant for the reaction at \(75^{\circ} \mathrm{C} ?\)

The rate law for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g) $$ is given by rate \(=k[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]\). (a) What is the order of the reaction? (b) A mechanism involving these steps has been proposed for the reaction $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) & \longrightarrow \mathrm{NOCl}_{2}(g) \\ \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) \longrightarrow & 2 \mathrm{NOCl}(g) \end{aligned} $$ If this mechanism is correct, what does it imply about the relative rates of these two steps?

These data were collected for the reaction between hydrogen and nitric oxide at \(700^{\circ} \mathrm{C}\) : \(2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\) $$ \begin{array}{cllc} \text { Experiment } & {\left[\mathrm{H}_{2}\right]} & {[\mathrm{NO}]} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 0.010 & 0.025 & 2.4 \times 10^{-6} \\ 2 & 0.0050 & 0.025 & 1.2 \times 10^{-6} \\ 3 & 0.010 & 0.0125 & 0.60 \times 10^{-6} \end{array} $$ (a) Determine the order of the reaction. (b) Calculate the rate constant. (c) Suggest a plausible mechanism that is consistent with the rate law. (Hint: Assume the oxygen atom is the intermediate.)
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