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Chapter 13: Problem 90

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 period. (Hint: Radioactive decays obey first-order kinetics.)

Short Answer

Expert verified
After 10 half-lives, only 0.09765625% of the original radioactive sample remains.

Step by step solution

01

Understand the Concept

Each half-life reduces the amount of radioactivity by half. Hence, after one half-life, the fraction of the radioactive sample that remains is 1/2 or 0.5. After two half-lives, the remaining fraction is \((0.5)^2 =0.25\) or a quarter, and so on. Essentially, each half-life reduces the radioactivity level to half of the previous level.
02

Apply the Concept to 10 Half-lives

This means that after 10 half-lives, the fraction of the radioactive sample that remains is \((0.5)^{10} =0.0009765625\). This result is obtained by multiplying the remaining fraction after each half-life.
03

Convert to Percentage

In order to express this fraction as a percentage, multiply the fraction by 100, i.e. \(0.0009765625 \times 100 = 0.09765625\% \). Thus, after ten half-lives, only 0.09765625% of the original radioactive sample remains, which is considered relatively harmless in the nuclear industry.

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Most popular questions from this chapter

Chlorine oxide (ClO), which plays an important role in the depletion of ozone (see Problem 13.101 ), decays rapidly at room temperature according to the equation $$ 2 \mathrm{ClO}(g) \longrightarrow \mathrm{Cl}_{2}(g)+\mathrm{O}_{2}(g) $$ From the following data, determine the reaction order and calculate the rate constant of the reaction. $$ \begin{array}{ll} \hline \text { Time (s) } & {[\mathrm{ClO}](M)} \\ \hline 0.12 \times 10^{-3} & 8.49 \times 10^{-6} \\ 0.96 \times 10^{-3} & 7.10 \times 10^{-6} \\ 2.24 \times 10^{-3} & 5.79 \times 10^{-6} \\ 3.20 \times 10^{-3} & 5.20 \times 10^{-6} \\ 4.00 \times 10^{-3} & 4.77 \times 10^{-6} \\ \hline \end{array} $$

The following scheme in which A is converted to B, which is then converted to \(\mathrm{C}\) is known as a consecutive reaction. $$ \mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} $$ Assuming that both steps are first order, sketch on the same graph the variations of [A], [B], and [C] with time.

What are the characteristics of a catalyst?

What are the units for the rate constants of zeroorder, first-order, and second-order reactions?

The rate constant for the second-order reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ is \(0.80 / M \cdot \mathrm{s}\) at \(10^{\circ} \mathrm{C}\). (a) Starting with a concentration of \(0.086 M,\) calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when \([\mathrm{NOBr}]_{0}=0.072 M\) and \([\mathrm{NOBr}]_{0}=0.054 \mathrm{M}\)
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