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Chapter 42: Problem 92

High-mass radionuclides, which may be either alpha or beta emitters, belong to one of four decay chains, depending on whether their mass number \(A\) is of the form \(4 n, 4 n+1,4 n+2,\) or \(4 n+3\), where \(n\) is a positive integer. (a) Justify this statement and show that if a nuclide belongs to one of these families, all its decay products belong to the same family. Classify the following nuclides as to family: (b) \({ }^{235} \mathrm{U},\) (c) \({ }^{236} \mathrm{U},\) (d) \({ }^{238} \mathrm{U},\) (e) \({ }^{239} \mathrm{Pu},\) (f) \({ }^{240} \mathrm{Pu},(\mathrm{g}){ }^{245} \mathrm{Cm}\), (h) \({ }^{246} \mathrm{Cm},\) (i) \({ }^{249} \mathrm{Cf}\), and \((\mathrm{j}){ }^{253} \mathrm{Fm}\).

Short Answer

Expert verified
Each nuclide belongs to a decay chain family based on its mass number mod 4: \(^{235}U\) and \(^{239}Pu\) are in the \(4n+3\) family, \(^{236}U\) and \(^{240}Pu\) are in the \(4n\) family, \(^{238}U\) and \(^{246}Cm\) are in the \(4n+2\) family, and \(^{245}Cm\), \(^{249}Cf\), \(^{253}Fm\) are in the \(4n+1\) family.

Step by step solution

01

Understanding the Mass Number Forms

The given statement mentions four different decay chains based on the mass number \( A \) of the nuclide. These are \( 4n, 4n+1, 4n+2, \) and \( 4n+3 \). A radionuclide belongs to one of these forms if \( A \), its mass number, is in the form specified by each chain, where \( n \) is an integer. Depending on the form, the radionuclide undergoes a series of decays along one of these chains.
02

Justification of Decay Chain Consistency

Every decay, alpha or beta, changes the mass number \( A \) according to specific rules: Alpha decay reduces \( A \) by 4 (\( A-4 \)), preserving the modulo 4 property. Beta decay changes \( Z \) (the number of protons) but does not change \( A \) directly, preserving the residue mod 4. Thus, if a nuclide starts in a specific family, all its decay products remain in that family, maintaining the \( A \equiv n \mod 4 \).
03

Classifying Nuclides

By calculating \( A \mod 4 \), we can classify each nuclide into one of the four families:- (b) \(^{235}U\): \( 235 \mod 4 = 3 \), so it belongs to the \( 4n+3 \) family.- (c) \(^{236}U\): \( 236 \mod 4 = 0 \), so it belongs to the \( 4n \) family.- (d) \(^{238}U\): \( 238 \mod 4 = 2 \), so it belongs to the \( 4n+2 \) family.- (e) \(^{239}Pu\): \( 239 \mod 4 = 3 \), so it belongs to the \( 4n+3 \) family.- (f) \(^{240}Pu\): \( 240 \mod 4 = 0 \), so it belongs to the \( 4n \) family.- (g) \(^{245}Cm\): \( 245 \mod 4 = 1 \), so it belongs to the \( 4n+1 \) family.- (h) \(^{246}Cm\): \( 246 \mod 4 = 2 \), so it belongs to the \( 4n+2 \) family.- (i) \(^{249}Cf\): \( 249 \mod 4 = 1 \), so it belongs to the \( 4n+1 \) family.- (j) \(^{253}Fm\): \( 253 \mod 4 = 1 \), so it belongs to the \( 4n+1 \) family.

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Key Concepts

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

Alpha Decay
Alpha decay is a type of radioactive decay in which an unstable nucleus releases an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, much like a helium nucleus. This process decreases the atomic number by 2 and the mass number by 4.
During alpha decay, the heavy nucleus reduces in mass and typically becomes more stable. This is common among heavy nuclides with atomic numbers greater than 83.
  • Conservation in Alpha Decay: The total number of nucleons and charge are conserved. If a parent nucleus is of mass number \( A \), the daughter nucleus will be \( A-4 \).
  • Example: If a nuclide undergoes alpha decay, like \( ^{238}U \rightarrow ^{234}Th + ^{4}He \), the residual mass number and element both reflect this transformation.
This decay chain operation maintains the characteristic of the nuclear family, keeping the modulo 4 property unchanged.
Beta Decay
Beta decay is another form of radioactive decay. It involves the conversion of a neutron into a proton with the emission of a beta particle (which can be an electron or a positron).
Beta decay increases the atomic number by 1 without altering the mass number. This is because the conversion affects only the internal composition of the nucleus, changing particles but not adding or losing nucleons.
Beta decay is significant for:
  • Conservation of Mass Number: While \( A \), the mass number, remains constant during beta decay, the atomic number \( Z \) increases or decreases depending on the beta decay process.
  • Example: A typical beta decay process can be represented as, \( ^{14}C \rightarrow ^{14}N + e^- + \overline{u}_e \), where a neutron transforms to a proton, and an electron is emitted.
Beta decay ensures that nuclides remain in their original decay family according to modulo 4 properties.
Mass Number Modulo
The concept of mass number modulo 4 is crucial in the classification of decay chains. A mass number \( A \) can be classified into four groups depending on its remainder when divided by 4: \( 4n, 4n+1, 4n+2, \) and \( 4n+3 \). This is the modulo operation.
  • A radionuclide with \( A \equiv 0 \pmod{4} \) belongs to the Thorium series.
  • Radionuclides with \( A \equiv 1 \pmod{4} \) are part of the Neptunium series.
  • \( A \equiv 2 \pmod{4} \) nuclides belong to the Uranium series.
  • An \( A \equiv 3 \pmod{4} \) denotes the Actinium series.
This understanding helps predict how decay processes will unfold and helps keep track of a nuclide's "family" even after multiple transformations. In both alpha and beta decay, the mass number's modulo 4 property ensures consistency within these chains, maintaining the family's characteristic.
Nuclide Classification
Nuclides can be categorized into families based on their mass number when considering radioactive decay chains. Classification is essential for understanding a nuclide's pathway through different decay stages.
Nuclides are classified based on their mass number modulo 4. By determining \( A \mod 4 \), you can identify the family that a nuclide belongs to.
For example:
  • If \( ^{235}U \rightarrow 235 \mod 4 = 3 \), it is part of the Actinium family.
  • For \( ^{236}U \rightarrow 236 \mod 4 = 0 \), aligning it with the Thorium series.

Significance of Nuclide Classification

This categorization is particularly helpful in predicting decay products and ensuring the integrity of analytical processes involving radioactive materials. It allows scientists and researchers to interpret and manage nuclear reactions effectively.

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Most popular questions from this chapter

At the end of World War II, Dutch authorities arrested Dutch artist Hans van Meegeren for treason because, during the war, he had sold a masterpiece painting to the Nazi Hermann Goering. The painting, Christ and His Disciples at Emmaus by Dutch master Johannes Vermeer \((1632-1675),\) had been discovered in 1937 by van Meegeren, after it had been lost for almost 300 years. Soon after the discovery, art experts proclaimed that Emmaus was possibly the best Vermeer ever seen. Selling such a Dutch national treasure to the enemy was unthinkable treason. However, shortly after being imprisoned, van Meegeren suddenly announced that he, not Vermeer, had painted Emmaus. He explained that he had carefully mimicked Vermeer's style, using a 300 -year-old canvas and Vermeer's choice of pigments; he had then signed Vermeer's name to the work and baked the painting to give it an authentically old look. Was van Meegeren lying to avoid a conviction of treason, hoping to be convicted of only the lesser crime of fraud? To art experts, Emmaus certainly looked like a Vermeer but, at the time of van Meegeren's trial in 1947 , there was no scientific way to answer the question. However, in 1968 Bernard Keisch of Carnegie-Mellon University was able to answer the question with newly developed techniques of radioactive analysis. Specifically, he analyzed a small sample of white lead-bearing pigment removed from Emmaus. This pigment is refined from lead ore, in which the lead is produced by a long radioactive decay series that starts with unstable \({ }^{238} \mathrm{U}\) and ends with stable \({ }^{206} \mathrm{~Pb}\). To follow the spirit of Keisch's analysis, focus on the following abbreviated portion of that decay series, in which intermediate, relatively shortlived radionuclides have been omitted: $${ }^{230} \mathrm{Th} \frac{{ }_{75.4 \mathrm{ky}}}{ }^{226} \mathrm{Ra} \frac{{ }_{1.60 \mathrm{ky}}}{ }^{210} \mathrm{~Pb} \frac{{ }_{22.6 \mathrm{ky}}}{ }^{206} \mathrm{~Pb}$$ The longer and more important half-lives in this portion of the decay series are indicated. (a) Show that in a sample of lead ore, the rate at which the number of \({ }^{210} \mathrm{~Pb}\) nuclei changes is given by $$\frac{d N_{210}}{d t}=\lambda_{226} N_{226}-\lambda_{210} N_{210}$$ where \(N_{210}\) and \(N_{226}\) are the numbers of \({ }^{210} \mathrm{~Pb}\) nuclei and \({ }^{226} \mathrm{Ra}\) nuclei in the sample and \(\lambda_{210}\) and \(\lambda_{226}\) are the corresponding disintegration constants. Because the decay series has been active for billions of years and because the half-life of \({ }^{210} \mathrm{~Pb}\) is much less than that of \({ }^{226} \mathrm{Ra}\), the nuclides \({ }^{226} \mathrm{Ra}\) and \({ }^{210} \mathrm{~Pb}\) are in equilibrium; that is, the numbers of these nuclides (and thus their concentrations) in the sample do not change. (b) What is the ratio \(R_{226} / R_{210}\) of the activities of these nuclides in the sample of lead ore? (c) What is the ratio \(N_{226} / N_{210}\) of their numbers? When lead pigment is refined from the ore, most of the \(226 \mathrm{Ra}\) is eliminated. Assume that only \(1.00 \%\) remains. Just after the pigment is produced, what are the ratios (d) \(R_{226} / R_{210}\) and (e) \(N_{226} / N_{210} ?\) Keisch realized that with time the ratio \(R_{226} / R_{210}\) of the pigment would gradually change from the value in freshly refined pigment back to the value in the ore, as equilibrium between the \({ }^{210} \mathrm{~Pb}\) and the remaining \({ }^{226} \mathrm{Ra}\) is established in the pigment. If Emmaus were painted by Vermeer and the sample of pigment taken from it were 300 years old when examined in \(1968,\) the ratio would be close to the answer of (b). If Emmaus were painted by van Meegeren in the 1930 s and the sample were only about 30 years old, the ratio would be close to the answer of (d). Keisch found a ratio of \(0.09 .\) (f) Is Emmaus a Vermeer?

Cancer cells are more vulnerable to \(\mathrm{x}\) and gamma radiation than are healthy cells. In the past, the standard source for radiation therapy was radioactive \({ }^{60} \mathrm{Co},\) which decays, with a half-life of \(5.27 \mathrm{y},\) into an excited nuclear state of \({ }^{60} \mathrm{Ni}\). That nickel isotope then immediately emits two gamma-ray photons, each with an approximate energy of \(1.2 \mathrm{MeV}\). How many radioactive \({ }^{60} \mathrm{Co}\) nuclei are present in a \(6000 \mathrm{Ci}\) source of the type used in hospitals? (Energetic particles from linear accelerators are now used in radiation therapy.)

What is the binding energy per nucleon of the americium isotope \({ }_{95}^{244} \mathrm{Am} ?\) Here are some atomic masses and the neutron mass. $$\begin{array}{lr}{ }_{95}^{244} \mathrm{Am} & 244.064279 \mathrm{u} \\\\\mathrm{n} & 1.008665 \mathrm{u}\end{array} { }^{1} \mathrm{H} \quad 1.007825 \mathrm{u}$$

A \(1.00 \mathrm{~g}\) sample of samarium emits alpha particles at a rate of 120 particles/s. The responsible isotope is \({ }^{147} \mathrm{Sm},\) whose natural abundance in bulk samarium is \(15.0 \%\). Calculate the half-life.

The radionuclide \({ }^{56} \mathrm{Mn}\) has a half-life of \(2.58 \mathrm{~h}\) and is produced in a cyclotron by bombarding a manganese target with deuterons. The target contains only the stable manganese isotope \({ }^{55} \mathrm{Mn},\) and the manganese \(-\) deuteron reaction that produces \({ }^{56} \mathrm{Mn}\) is $${ }^{55} \mathrm{Mn}+\mathrm{d} \rightarrow{ }^{56} \mathrm{Mn}+\mathrm{p}$$ If the bombardment lasts much longer than the half-life of \({ }^{56} \mathrm{Mn}\), the activity of the \({ }^{56} \mathrm{Mn}\) produced in the target reaches a final value of \(8.88 \times 10^{10}\) Bq. (a) At what rate is \({ }^{56}\) Mn being produced? (b) How many \({ }^{56}\) Mn nuclei are then in the target? (c) What is their total mass?
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