/*! 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 73 Production of a helium nucleus f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 73

Production of a helium nucleus from a heavy atom is referred to as decay.

Short Answer

Expert verified
The production of a helium nucleus from a heavy atom is referred to as alpha decay. This type of radioactive decay involves the emission of an alpha particle (or helium nucleus with 2 protons and 2 neutrons) and transforms the parent atom into a new element with a lower atomic number. The alpha decay process can be represented as \( ^A_ZX \rightarrow ^{A-4}_{Z-2}Y + ^4_2\alpha \), with conservation of mass and charge. The released energy (Q-value) is carried by the emitted alpha particle in the form of kinetic energy.

Step by step solution

01

Understanding Decay

Decay is a spontaneous process in which an unstable atomic nucleus loses energy by emitting particles (such as alpha particles, beta particles, or gamma rays) or radiation. In this case, we are focusing on the production of a helium nucleus, which occurs during alpha decay.
02

Describe Alpha Decay

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle (consisting of 2 protons and 2 neutrons) and transforms into a new element with a lower atomic number. The alpha particle produced is essentially a helium nucleus.
03

Illustrate the Alpha Decay Equation

Consider a heavy atom with atomic number Z and mass number A. During alpha decay, the atom loses 2 protons and 2 neutrons, forming a helium nucleus. The alpha decay can be represented by the following equation: \( ^A_ZX \rightarrow ^{A-4}_{Z-2}Y + ^4_2\alpha \) Where: - \( ^A_ZX \) represents the initial atom with atomic number Z and mass number A - \( ^{A-4}_{Z-2}Y \) represents the resulting atom after alpha decay, with atomic number (Z-2) and mass number (A-4) - \( ^4_2\alpha \) is the emitted alpha particle or helium nucleus
04

Conservation Laws in Alpha Decay

During alpha decay, both mass and charge are conserved. The conservation of mass means that the mass number of the parent atom (A) must equal the combined mass numbers of the daughter atom and the helium nucleus. The conservation of charge means that the atomic number of the parent atom (Z) must equal the combined atomic numbers of the daughter atom and the helium nucleus.
05

Energy Release During Alpha Decay

Since this decay process is spontaneous, the parent atom loses energy during alpha decay. The energy released can be calculated using the difference in the masses of the parent atom and the sum of the masses of the daughter atom and the helium nucleus. This released energy, known as the Q-value, is carried away by the emitted alpha particle in the form of kinetic energy.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Radioactive Decay
Radioactive decay is a natural process where unstable atomic nuclei lose energy to attain stability. This can occur through the emission of particles like alpha, beta, and gamma rays.
A primary example is alpha decay, in which an unstable nucleus emits an alpha particle. This process lowers its energy and transforms it into a different element.
  • Alpha particles consist of two protons and two neutrons, similar to a helium nucleus.
  • The nucleus of the atom changes, reducing its atomic number and mass.
Such transformations allow the nucleus to reach a more stable state, removing excess energy and shifting its identity based on the transformed element. Understanding decay helps us examine natural and artificial nuclear processes.
Helium Nucleus
A helium nucleus, also known as an alpha particle, is a small cluster of two protons and two neutrons. It is positive in charge, thanks to its protons.
When a heavy nucleus undergoes alpha decay, it ejects this alpha particle, which contributes to the change in the original element's structure.
  • Helium nucleus creation marks not just a shift in atomic structure but also mass loss.
  • It is incredibly stable, explaining why alpha particles are chosen by nature as emission candidates in radioactive decay.
  • After emission, it travels away, carrying energy as kinetic energy.
This behavior of the helium nucleus not only alters the initial atom but also helps the universe in maintaining atomic stability.
Conservation Laws
Conservation laws are crucial in understanding alpha decay. These laws ensure that certain physical quantities remain consistent before and after a reaction.
During alpha decay, both mass and charge need to be conserved.
  • The conservation of mass states that the mass number of the parent atom equals the total mass number of the daughter atom and the helium nucleus. This is why we see a reduction in mass number by four in alpha decay equations.
  • The conservation of charge requires the atomic number of the parent atom to match the total atomic number of the daughter and helium nucleus. In practice, this results in a reduction by two of the atomic number.
These laws are key to understanding the equation represented as:
\(^A_ZX \rightarrow ^{A-4}_{Z-2}Y + ^4_2\alpha\)

They provide us with a robust framework to predict the products of alpha decay and validate the principles of physics in nuclear reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Complete each of the following nuclear equations by supplying the missing particle. a. \({ }_{80}^{201} \mathrm{Hg}+{ }_{-1}^{0} \mathrm{e} \rightarrow ?\) \(\mathrm{b} . ? \rightarrow{ }_{83}^{210} \mathrm{Br}+{ }_{-1}^{0} \mathrm{e}\) c. \({ }_{83}^{210} \mathrm{Bi} \rightarrow ?_{-1}^{0} \mathrm{e}\)

Technetium-99 has been used as a radiographic agent in bone scans ( \({ }_{43}^{99} \mathrm{Tc}\) is absorbed by bones). If \({ }_{43}^{99} \mathrm{Tc}\) has a half-life of 6.0 hours, what fraction of an administered dose of \(100 \mu \mathrm{g}\) of \({ }_{43}^{99} \mathrm{Tc}\) remains in a patient's body after 2.0 days?

How have \({ }_{53}^{131} \mathrm{I}\) and \({ }_{81}^{201} \mathrm{Ti}\) been used in medical diagnosis? Why are these particular nuclides especially well suited for this purpose?

A certain radioactive nuclide has a half-life of 80.9 years. How long does it take for \(87.5 \%\) of a sample of this substance to decay?

During the research that led to production of the two atomic bombs used against Japan in World War II, different mechanisms for obtaining a supercritical mass of fissionable material were investigated. In one type of bomb, what is essentially a gun was used to shoot one piece of fissionable material into a cavity containing another piece of fissionable material. In the second type of bomb, the fissionable material was surrounded with a high explosive that, when detonated, compressed the fissionable material into a smaller volume. Discuss what is meant by critical mass, and explain why the ability to achieve a critical mass is essential to sustaining a nuclear reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.