/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 Neutron activation analysis was ... [FREE SOLUTION] | 91Ó°ÊÓ

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Neutron activation analysis was performed on small pieces of hair that had been taken from the exiled Napoleon after he died on the island of St. Helena in 1821\. This procedure involves exposing the samples to a source of neutrons. Some (stable) arsenic nuclei, if present in the sample, will absorb a neutron. In Napoleon's case the samples did contain abnormally high levels of arsenic, which supported the theory that his death was not due to natural causes. (a) These results came from studying beta emissions of the resulting \({ }^{76} \mathrm{As},\) nucleus. Write the nuclear equation for the neutron absorption and use it to determine the arsenic isotope initially present in the hair. (b) Write the nuclear equation for the subsequent beta decay of \({ }^{76} \mathrm{As}\). Use it to determine the nucleus after this decay.

Short Answer

Expert verified
(a) Initial arsenic isotope: \(^ {75} \mathrm{As}\). (b) After beta decay: \(^ {76} \mathrm{Se}\).

Step by step solution

01

Understanding Neutron Absorption

Neutron absorption involves a stable arsenic isotope in the hair absorbing a neutron and forming another isotope. The general reaction for neutron capture by an element can be expressed as \( ^A_ZX + ^1_0n \rightarrow {^{A+1}}_ZX \). Our goal is to identify the original arsenic isotope before neutron absorption. In the case of arsenic, if \( ^{76}\mathrm{As} \) is the product, the reaction is: \( ^{75}_{33}\mathrm{As} + ^1_0n \rightarrow ^{76}_{33}\mathrm{As} \). Hence, the initial isotope present is \( ^{75}\mathrm{As} \).
02

Writing the Beta Decay Equation

Beta decay involves the conversion of a neutron in the nucleus into a proton, emitting a beta particle. The beta decay equation for \( ^{76}_{33}\mathrm{As} \) would be \( ^{76}_{33}\mathrm{As} \rightarrow ^{76}_{34}Se + ^0_{-1}\beta \), where \(^0_{-1}\beta\) is the beta particle emitted, and \(^{76}_{34}Se\) is the resulting selenium nucleus.

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Key Concepts

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

Nuclear Equations
Nuclear equations are essential tools in representing the reactions that occur during nuclear processes. These equations depict the particles involved in a reaction, showing both initial and final states. This is particularly helpful when identifying isotopes and determining their changes through nuclear reactions.

For example, in the neutron activation analysis exercise, we begin with a stable isotope: arsenic-75 \( ^{75}_{33}\mathrm{As} \). During neutron absorption, this absorbs a neutron \( ^1_0n \), creating the isotope \( ^{76}_{33}\mathrm{As} \). Therefore, the nuclear equation is: \[ ^{75}_{33}\mathrm{As} + ^1_0n \rightarrow ^{76}_{33}\mathrm{As} \]

This equation provides insight into the type of reaction occurring, the isotopes involved, and what elements they transform into. Such nuclear equations are a cornerstone in understanding the processes of nuclear activation and decay.
Beta Decay
Beta decay is a fascinating nuclear process where a neutron in an atom's nucleus transforms into a proton. This results in the emission of a beta particle, which is essentially a high-energy electron.

Understanding beta decay is crucial for explaining how elements change into different elements during nuclear reactions. When arsenic-76 \( ^{76}_{33}\mathrm{As} \) undergoes beta decay, it converts into selenium-76 \( ^{76}_{34}\mathrm{Se} \), accompanied by the emission of a beta particle \( ^0_{-1}\beta \). Thus, the nuclear equation is:

\[ ^{76}_{33}\mathrm{As} \rightarrow ^{76}_{34}\mathrm{Se} + ^0_{-1}\beta \]

This transformation results in an increase in the atomic number by one, reflecting the increase of a proton in the nucleus. As the emitted beta particle has a negligible mass but carries a -1 charge, it helps maintain the balance of charge in the equation. This process is integral in transmuting elements in neutron activation analysis.
Arsenic Isotopes
Arsenic isotopes are pivotal in understanding various nuclear processes and analyses, such as neutron activation analysis. Isotopes are different forms of the same element that contain an identical number of protons but different numbers of neutrons.

In the given analysis, arsenic-75 \( ^{75}\mathrm{As} \) is the stable and naturally occurring isotope present in the sample. Upon absorbing a neutron, it transforms into arsenic-76 \( ^{76}\mathrm{As} \), an unstable isotope. This newly formed isotope is central to conducting further analyses due to its radioactivity.
  • The presence of isotopes like arsenic-76 allows researchers to track emissions, such as beta particles, to study a sample's composition.
  • These isotopes help determine potential exposure to elements or chemical processes, which was instrumental in the case of analyzing samples from Napoleon's hair.
Understanding arsenic isotopes forms an essential aspect of nuclear science and its applications in real-world scenarios.

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Most popular questions from this chapter

High-energy gamma ray photons can remove nucleons from nuclei in a process called the photonuclear effect. This is a process analogous to the photoelectric effect in atoms (see Section 27.2). (a) If you wanted to remove a single neutron from a nucleus, which of the following nuclei would likely require the higher energy photon: (1) \({ }^{12} \mathrm{C} ;\) (2) \({ }^{13} \mathrm{C} ;\) or (3) the energies would be about the same? Explain your reasoning. (b) Calculate the minimum energy of a photon required to eject a neutron from each of the two isotopes in part (a), neglecting any kinetic energy of the neutron or resulting nucleus. (c) Determine the wavelengths of the light associated (d) Explain why the actual with the photons in part (b). minimum energy in part (b) is higher than your result. [Hint: The initial photon contains linear momentum that must be conserved.]

Nitrogen-13, with a half-life of \(10.0 \mathrm{~min}\), decays by beta emission. (a) Write down the decay equation to determine the daughter product and whether the beta particle is a positron or electron. (b) If a sample of pure \({ }^{13} \mathrm{~N}\) has a mass of \(1.50 \mathrm{~g}\) at a certain time, what is the activity 35.0 min later? (c) What percentage of the sample is \({ }^{13} \mathrm{~N}\) at this time? (d) What alternative process could have happened to the nitrogen-13? Write down its decay equation and determine the daughter product for this process.

Carbon- 14 dating is used to determine the age of some unearthed bones. (a) If the activity of bone \(\mathrm{A}\) is higher than that of bone \(\mathrm{B}\), then bone \(\mathrm{A}\) is (1) older than, (2) younger than, (3) the same age as bone B. Explain your reasoning. (b) A sample of bone \(\mathrm{A}\) is found to have 4.0 beta decays/min per gram of carbon, while bone \(\mathrm{B}^{\prime} \mathrm{s}\) activity is only 1.0 beta decay \(/\) min for each gram of carbon. What is the age difference between the two bones?

Polonium-214 can decay by alpha decay. (a) The product of its decay has how many fewer protons than polonium- 214:(1) zero, (2) one, (3) two, or (4) four? (b) Write the nuclear equation for this decay and determine the daughter nucleus.

Write the nuclear equations expressing (a) the beta decay of \({ }^{60} \mathrm{Co}\) and \((\mathrm{b})\) the alpha decay of \({ }^{222} \mathrm{Rn}\)

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