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Chapter 42: Problem 28

The plutonium isotope \({ }^{239} \mathrm{Pu}\) is produced as a by-product in nuclear reactors and hence is accumulating in our environment. It is radioactive, decaying with a half-life of \(2.41 \times 10^{4} \mathrm{y} .\) (a) How many nuclei of Pu constitute a chemically lethal dose of \(2.00 \mathrm{mg} ?(\mathrm{~b})\) What is the decay rate of this amount?

Short Answer

Expert verified
(a) The number of nuclei is approximately \( 5.04 \times 10^{18} \). (b) The decay rate is approximately \(1.45 \times 10^{11}\) decays per year.

Step by step solution

01

Calculate the number of moles in 2.00 mg

First, determine the number of moles in the 2.00 mg of \(^{239} \text{Pu}\). The molar mass of \(^{239} \text{Pu}\) is 239 g/mol. Use the formula to find moles: \[\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\]\. Convert the mass from mg to g: \[\text{mass} = 2.00 \text{ mg} = 0.00200 \text{ g}\] Thus, the number of moles is: \[\text{moles} = \frac{0.00200 \text{ g}}{239 \text{ g/mol}}\].
02

Determine number of nuclei

Use Avogadro's number (\(6.022 \times 10^{23}\)) to find the number of nuclei in the sample. Multiply the moles by Avogadro's number: \[\text{nuclei} = \text{moles} \times 6.022 \times 10^{23}\].
03

Find the decay rate

Use the formula for the decay rate, which is \(\lambda N\), where \(\lambda\) is the decay constant and \(N\) is the number of nuclei. The decay constant \(\lambda\) can be calculated using the half-life: \[\lambda = \frac{0.693}{\text{Half-life}}\]\. Thus, \[\lambda = \frac{0.693}{2.41 \times 10^4 \text{ y}}\].\Now, use the decay rate formula: \[\text{Decay Rate} = \lambda \times N\].

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Key Concepts

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

Radioactive Decay
Radioactive decay is a fundamental concept in nuclear physics where unstable atomic nuclei lose energy by emitting radiation. This process transforms the original nucleus into a different element or isotope. Radioactive decay can occur through various types of emissions, such as:
  • Alpha particles (helium nuclei)
  • Beta particles (electrons or positrons)
  • Gamma rays (high-energy photons)
In our exercise, we deal with plutonium-239 ( 239Pu), which undergoes alpha decay. This means it emits alpha particles, thus decreasing both its atomic mass and atomic number. The nature of radioactive decay is random and spontaneous, influencing how it is mathematically modeled in physics. Understanding this randomness is crucial when assessing the safety and impact of radioactive materials.
Half-Life Calculation
The concept of half-life is essential to understanding radioactive decay. The half-life is the time it takes for half of a given amount of a radioactive substance to decay. This constant is unique to each isotope.In the case of 239Pu, with a half-life of 24,100 years, understanding half-life allows us to predict how long it stays radioactive. To calculate half-life, the following considerations are important:- Half-life is independent of the initial quantity of the isotope.- Half-life helps in determining the rate of decay and longevity of a substance's radioactivity.To use the half-life in calculations, we often need the decay constant (denoted by \( \lambda \)). The relationship between half-life and the decay constant is given by the formula: \(\lambda = \frac{0.693}{\text{Half-life}}\)This equation allows us to convert the half-life into a quantitative measure of how rapidly the substance decays.
Decay Constant
The decay constant, \( \lambda \), is a critical parameter in radioactive decay equations. It is a measure of the probability of decay of a nucleus per unit time. The decay constant links the half-life of a radioactive isotope to the rate at which its nuclei decay. It can be calculated using the half-life formula as mentioned earlier: \( \lambda = \frac{0.693}{\text{Half-life}} \).This constant plays a vital role in determining the decay rate, using the formula:\[ \text{Decay Rate} = \lambda \times N \]where:
  • \( \lambda \) is the decay constant
  • \( N \) is the number of radioactive nuclei present
This mathematical relationship helps quantify the activity of a radioactive sample, which is crucial in applications such as dating ancient artifacts or assessing the safety of nuclear materials. Understanding \( \lambda \) is key to predicting how quickly a radioactive substance will diminish.

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Most popular questions from this chapter

\(.\) co An \(\alpha\) particle \(\left({ }^{4} \mathrm{He}\right.\) nucleus \()\) is to be taken apart in the following steps. Give the energy (work) required for each step: (a) remove a proton, (b) remove a neutron, and (c) separate the remaining proton and neutron. For an \(\alpha\) particle, what are (d) the total binding energy and (e) the binding energy per nucleon? (f) Does either match an answer to (a), (b), or (c)? Here are some atomic masses and the neutron mass. \(\begin{array}{llll}{ }^{4} \mathrm{He} & 4.00260 \mathrm{u} & { }^{2} \mathrm{H} & 2.01410 \mathrm{u} \\ { }^{3} \mathrm{H} & 3.01605 \mathrm{u} & { }^{1} \mathrm{H} & 1.00783 \mathrm{u} \\ \mathrm{n} & 1.00867 \mathrm{u} & & \end{array}\)

A penny has a mass of \(3.0 \mathrm{~g} .\) Calculate the energy that would be required to separate all the neutrons and protons in this coin from one another. For simplicity, assume that the penny is made entirely of \({ }^{63} \mathrm{Cu}\) atoms (of mass \(62.92960 \mathrm{u}\) ). The masses of the protonplus-electron and the neutron are \(1.00783 \mathrm{u}\) and \(1.00866 \mathrm{u}\), respectively.

After a brief neutron irradiation of silver, two isotopes are present: \({ }^{108} \mathrm{Ag}\left(T_{1 / 2}=2.42 \mathrm{~min}\right)\) with an initial decay rate of \(3.1 \times\) \(10^{5} / \mathrm{s}\), and \({ }^{110} \mathrm{Ag}\left(T_{1 / 2}=24.6 \mathrm{~s}\right)\) with an initial decay rate of \(4.1 \times\) \(10^{6} / \mathrm{s}\). Make a semilog plot similar to Fig. 42-9 showing the total combined decay rate of the two isotopes as a function of time from \(t=0\) until \(t=10 \mathrm{~min}\). We used Fig. 42-9 to illustrate the extraction of the half-life for simple (one isotope) decays. Given only your plot of total decay rate for the two-isotope system here, suggest a way to analyze it in order to find the half-lives of both isotopes.

Generally, more massive nuclides tend to be more unstable to alpha decay. For example, the most stable isotope of uranium, \({ }^{238} \mathrm{U}\), has an alpha decay half-life of \(4.5 \times 10^{9} \mathrm{y} .\) The most stable isotope of plutonium is \({ }^{244} \mathrm{Pu}\) with an \(8.0 \times 10^{7} \mathrm{y}\) half-life, and for curium we have \({ }^{248} \mathrm{Cm}\) and \(3.4 \times 10^{5} \mathrm{y}\). When half of an original sample of \({ }^{238} \mathrm{U}\) has decayed, what fraction of the original sample of (a) plutonium and (b) curium is left?

Under certain rare circumstances, a nucleus can decay by emitting a particle more massive than an alpha particle. Consider the decays $$ { }^{223} \mathrm{Ra} \rightarrow{ }^{209} \mathrm{~Pb}+{ }^{14} \mathrm{C} \quad \text { and } \quad{ }^{223} \mathrm{Ra} \rightarrow{ }^{219} \mathrm{Rn}+{ }^{4} \mathrm{He} $$ Calculate the \(Q\) value for the (a) first and (b) second decay and determine that both are energetically possible. (c) The Coulomb barrier height for alpha-particle emission is \(30.0 \mathrm{MeV}\). What is the barrier height for \({ }^{14} \mathrm{C}\) emission? (Be careful about the nuclear radii.) The needed atomic masses are \(\begin{array}{ll}{ }^{223} \mathrm{Ra} & 223.01850 \mathrm{u} \\ { }^{209} \mathrm{~Pb} & 208.98107 \mathrm{u}\end{array}\) $$ { }^{14} \mathrm{C} \quad 14.00324 \mathrm{u} $$ \(14.00324 \mathrm{u}\) \(4.00260 \mathrm{u}\) \({ }^{4} \mathrm{He}\) $$ { }^{219} \mathrm{Rn} \quad 219.00948 \mathrm{u} $$
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