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It has been suggested that strontium-90 (generated by nuclear testing) deposited in the hot desert will undergo radioactive decay more rapidly because it will be exposed to much higher average temperatures. (a) Is this a reasonable suggestion? (b) Does the process of radioactive decay have an activation energy, like the Arrhenius behavior of many chemical reactions (Section 14.5)?

Short Answer

Expert verified
(a) It is not reasonable to suggest that strontium-90 will decay more rapidly at higher temperatures in a hot desert, as radioactive decay is not influenced by temperature, pressure, or presence of other substances. The decay rate depends solely on its intrinsic half-life and nature. (b) The process of radioactive decay does not have an activation energy, as it does not follow the Arrhenius behavior seen in many chemical reactions. Radioactive decay is a random process that is not affected by external factors like temperature, pressure, or presence of other substances.

Step by step solution

01

Understand radioactive decay

Radioactive decay is a random process where unstable atomic nuclei lose energy by emitting radiation or particles. Unlike chemical reactions, radioactive decay is not influenced by temperature, pressure, or the presence of other substances. The decay rate of a radioactive substance depends only on the nature of the substance and its half-life, which is the time required for half of its atoms to decay.
02

Answer part (a) considering radioactive decay

Now that we know that radioactive decay is not influenced by temperature, we can conclude that it is not reasonable to suggest that strontium-90 will decay more rapidly at higher temperatures in a hot desert. The decay rate depends solely on its intrinsic half-life and nature.
03

Understand Arrhenius behavior and activation energy

Activation energy is the energy barrier that must be overcome for a chemical reaction to occur, and it is affected by temperature. The Arrhenius equation describes the temperature dependence of reaction rates as: \[k = A~\mathrm{exp}\left(-\frac{E_\mathrm{a}}{RT}\right)\] where k is the rate constant, A is the pre-exponential factor, \(E_\mathrm{a}\) is the activation energy, R is the gas constant, and T is the temperature in Kelvin. As a result, chemical reactions that follow the Arrhenius behavior have their reaction rate modified by temperature due to the exponential temperature term in the equation.
04

Answer part (b) considering activation energy and Arrhenius behavior

Radioactive decay, as explained in Step 1, does not depend on external factors like temperature, pressure, or presence of other substances. This behavior contrasts with chemical reactions that are affected by these factors and typically follow Arrhenius behavior, with an associated activation energy. Since radioactive decay is not influenced by temperature, it does not have an activation energy, like the Arrhenius behavior of many chemical reactions. In conclusion: (a) It is not reasonable to suggest that strontium-90 will decay more rapidly at higher temperatures in a hot desert. (b) The process of radioactive decay does not have an activation energy, as it does not follow the Arrhenius behavior seen in many chemical reactions.

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Key Concepts

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

Strontium-90
Strontium-90 ((90)Sr) is a radioactive isotope that poses significant health concerns due to its beta-emission properties and its similarity to calcium, which allows it to accumulate in bones when ingested or inhaled. It is produced in nuclear fission and found in waste from nuclear reactors and fallout from nuclear testing. The biological half-life of strontium-90—the time it takes for half of the strontium-90 to be expelled from the body—is approximately 18 years, making it a long-term health risk.

Strontium-90 undergoes radioactive decay, a process that is not subject to external factors such as temperature or pressure. This means that regardless of the environment, whether it's a cold Arctic tundra or a hot desert, the decay rate for strontium-90 remains constant, solely determined by its own nuclear properties.
Half-life
The concept of half-life is fundamental in understanding radioactive decay, as it helps us determine how quickly a radioactive material will decrease in activity. The half-life is defined as the time required for half of the atoms of a radioactive substance to decay. For strontium-90, its half-life is approximately 28.8 years. This means that after 28.8 years, only half of the original amount of strontium-90 will remain; the other half would have decayed into a different element, yttrium-90, through beta decay.

It's essential to recognize that half-life is an intrinsic property of the radioactive isotope and is not influenced by environmental factors such as temperature. This is a key concept that leads us to understand why the decay rate of strontium-90 would not accelerate in a hot desert climate.
Activation Energy
Activation energy is a term often found in discussions about chemical reactions. It represents the minimum amount of energy that is required for reactants to transform into products during a chemical reaction. The higher the activation energy, the slower the reaction tends to be, as fewer molecules have sufficient energy to surpass this threshold at a given temperature.

Different from chemical reactions, radioactive decay is a nuclear process that does not require the surmounting of an energy barrier to proceed. It is governed by the principles of nuclear physics rather than chemical kinetics. Therefore, radioactive decay does not feature an activation energy—making this concept inapplicable to the decay of strontium-90 or any other radioactive isotope.
Arrhenius Equation
The Arrhenius equation is a critical expression in chemical kinetics, offering a quantitative way to understand how the rate of a chemical reaction depends on temperature and activation energy. Mathematically, it establishes a relationship between the rate constant ((k)), temperature (T), activation energy ((E_a)), and the Arrhenius constant (A).

The equation is given by: \[\begin{equation}k = A~\mathrm{exp}\left(-\frac{E_\mathrm{a}}{RT}\right)\end{equation}\]where R is the universal gas constant. The exp(-E_\mathrm{a}/RT) term shows an exponential decrease in reaction rate with increasing activation energy or decreasing temperature. The Arrhenius equation does not apply to strontium-90's radioactive decay since there is no dependency on temperature and no activation energy barrier to overcome.

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Most popular questions from this chapter

Calculations Involving Radioactive Decay and Time If we start with \(1.000 \mathrm{~g}\) of strontium- \(90,0.953 \mathrm{~g}\) will remain after \(2.00 \mathrm{yr}\). (a) What is the half-life of strontium- 90 ? (b) How much strontium- 90 will remain after \(5.00 \mathrm{yr}\) ? (c) What is the initial activity of the sample in becquerels and curies? SOLUTION (a) Analyze We are asked to calculate a half-life, \(t_{1 / 2}\) based on data that tell us how much of a radioactive nucleus has decayed in a time interval \(t=2.00 \mathrm{yr}\) and the information \(N_{0}=1.000 \mathrm{~g}, N_{t}=0.953 \mathrm{~g}\). Plan We first calculate the rate constant for the decay, \(k\), and then use that to compute \(t_{1 / 2}\). Solve Equation \(21.19\) is solved for the decay constant, \(k\), and then Equation \(21.20\) is used to calculate half-life, \(t_{1 / 2}\) : $$ \begin{aligned} k &=-\frac{1}{t} \ln \frac{N_{\mathrm{f}}}{N_{0}}=-\frac{1}{2.00 \mathrm{yr}} \ln \frac{0.953 \mathrm{~g}}{1.000 \mathrm{~g}} \\ &=-\frac{1}{2.00 \mathrm{yr}}(-0.0481)=0.0241 \mathrm{yr}^{-1} \\ t_{1 / 2} &=\frac{0.693}{k}=\frac{0.693}{0.0241 \mathrm{yr}^{-1}}=28.8 \mathrm{yr} \end{aligned} $$ (b) Analyze We are asked to calculate the amount of a radionuclide remaining after a given period of time. Plan We need to calculate \(N_{b}\) the amount of strontium present at time \(t\), using the initial quantity, \(N_{0}\), and the rate constant for decay, \(k\), calculated in part (a). Solve Again using Equation \(21.19\), with \(k=0.0241 \mathrm{yr}^{-1}\), we have $$ \ln \frac{N_{t}}{N_{0}}=-k t=-\left(0.0241 \mathrm{yr}^{-1}\right)(5.00 \mathrm{yr})=-0.120 $$ \(N_{t} / N_{0}\) is calculated from \(\ln \left(N_{t} / N_{0}\right)=-0.120\) using the \(e^{x}\) or INV LN function of a calculator: $$ \frac{N_{t}}{N_{0}}=e^{-1.120}=0.887 $$ Because \(N_{0}=1.000 \mathrm{~g}\), we have $$ N_{t}=(0.887) N_{0}=(0.887)(1.000 \mathrm{~g})=0.887 \mathrm{~g} $$ (c) Analyze We are asked to calculate the activity of the sample in becquerels and curies. Plan We must calculate the number of disintegrations per atom per second and then multiply by the number of atoms in the sample. Solve The number of disintegrations per atom per second is given by the decay constant, \(k\) : $$ k=\left(\frac{0.0241}{\mathrm{yr}}\right)\left(\frac{1 \mathrm{yr}}{365 \text { days }}\right)\left(\frac{1 \text { day }}{24 \mathrm{~h}}\right)\left(\frac{\mathrm{lh}}{3600 \mathrm{~s}}\right)=7.64 \times 10^{-10} \mathrm{~s}^{-1} $$ To obtain the total number of disintegrations per second, we calculate the number of atoms in the sample. We multiply this quantity by \(k\), where we express \(k\) as the number of disintegrations per atom per second, to obtain the number of disintegrations per second: $$ \begin{aligned} \left(1.000 \mathrm{~g}^{90} \mathrm{Sr}\right)\left(\frac{1 \mathrm{~mol}^{90} \mathrm{Sr}}{90 \mathrm{~g}^{90} \mathrm{Sr}}\right)\left(\frac{6.022 \times 10^{23} \mathrm{atoms} \mathrm{Sr}}{1 \mathrm{~mol}^{90} \mathrm{Sr}}\right)=6.7 \times 10^{21} \text { atoms }^{90} \mathrm{Sr} \\ \text { Total disintegrations/s } &=\left(\frac{7.64 \times 10^{-10} \text { disintegrations }}{\text { atom }{ }^{2} \mathrm{~s}}\right)\left(6.7 \times 10^{21} \text { atoms }\right) \\ &=5.1 \times 10^{12} \text { disintegrations/s } \end{aligned} $$ Because \(1 \mathrm{~Bq}\) is one disintegration per second, the activity is \(5.1 \times 10^{12} \mathrm{~Bq}\). The activity in curies is given by $$ \left(5.1 \times 10^{12} \text { disintegrations/s }\right)\left(\frac{1 \mathrm{Ci}}{3.7 \times 10^{10} \text { disintegrations/s }}\right)=1.4 \times 10^{2} \mathrm{Ci} $$ We have used only two significant figures in products of these calculations because we do not know the atomic weight of \({ }^{90} \mathrm{Sr}\) to more than two significant figures without looking it up in a special source.

Write balanced nuclear equations for the following processes: (a) rubidium-90 undergoes beta emission; (b) selenium-72 undergoes electron capture; (c) krypton- 76 undergoes positron emission; (d) radium-226 emits alpha radiation.

Why are nuclear transmutations involving neutrons generally easier to accomplish than those involving protons or alpha particles?

The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{~kJ} / \mathrm{min}\). (a) How much loss of mass from the Sun occurs in one day from just the encrgy falling on Farth? (b) If the energy released in the reaction $$ { }^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ \(\left({ }^{235} \mathrm{U}\right.\) nuclear mass, \(234.9935 \mathrm{amu} ;{ }^{141} \mathrm{Ba}\) nuclear mass, \(140.8833 \mathrm{amu} ;{ }^{92} \mathrm{Kr}\) nuclear mass, 91.9021 amu) is taken as typical of that occurring in a nuclear reactor, what mass of uranium-235 is required to equal \(0.10 \%\) of the solar energy that falls on Earth in \(1.0\) day?

According to current regulations, the maximum permissible dose of strontium- 90 in the body of an adult is \(1 \mu \mathrm{Ci}\left(1 \times 10^{-6} \mathrm{Ci}\right)\). Using the relationship rate \(=k N\), calculate the number of atoms of strontium- 90 to which this dose corresponds. To what mass of strontium- 90 does this correspond? The half-life for strontium- 90 is \(28.8 \mathrm{yr}\).

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