/*! 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 41 At the beginning of Section 43.7... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At the beginning of Section 43.7 the equation of a fission process is given in which \({ }^{235} \mathrm{U}\) is struck by a neutron and undergoes fission to produce \({ }^{144} \mathrm{Ba},{ }^{89} \mathrm{Kr},\) and three neutrons. The measured masses of these isotopes are \(235.043930 \mathrm{u}\left({ }^{235} \mathrm{U}\right), 143.922953 \mathrm{u}\left({ }^{144} \mathrm{Ba}\right),\) 88.917631 u \(\left({ }^{89} \mathrm{Kr}\right),\) and \(1.0086649 \mathrm{u}\) (neutron). (a) Calculate the energy (in \(\mathrm{MeV}\) ) released by each fission reaction. (b) Calculate the energy released per gram of \({ }^{235} \mathrm{U},\) in \(\mathrm{MeV} / \mathrm{g}\).

Short Answer

Expert verified
The calculated energy released in each fission process would be the result from Step 2. The energy released per gram of \(^{235} U\) is calculated in Step 3, which involves utilizing Avogadro's number and the mass of a mole of uranium.

Step by step solution

01

Calculate mass defect

To calculate the energy released during fission, we first need to find the 'mass defect'. Mass defect is the difference between the mass of the original nucleus (\(^{235} U\)) and the combined mass of its fission products (\(^{144} Ba\), \(^{89} Kr\) and three neutrons). Let's denote the mass of \(^{235} U\) as \(m_U\), \(^{144} Ba\) as \(m_Ba\), \(^{89} Kr\) as \(m_Kr\) and neutron as \(m_n\). So, Mass defect (\(∆m\)) = \(m_U - (m_Ba + m_Kr + 3m_n)\). After substituting the given values, we calculate the mass defect.
02

Convert mass defect into energy

The Energy (E) released can then be calculated using Einstein’s Mass-Energy equivalence principle, where \(E = ∆m*c^2\). Here, c is the speed of light. However, in atomic and nuclear processes, energy is usually measured in mega electron-volts (MeV). For these processes, the conversion factor from u (unified atomic mass units) to MeV is used, i.e., 1 u equals to about 931.5 MeV/c^2. Therefore, the energy released E (in MeV) can be obtained by multiplying the mass defect (∆m) by the conversion factor 931.5.
03

Calculate energy released per gram of Uranium

To calculate the energy per gram, we have to first convert the energy release per Uranium atom into per gram. Given that 1 mole of U-235 contains \(6.022*10^23\) atoms (Avogadro's number), and 1 mole of U-235 has a mass of 235 g. Then, multiplying the energy released per fission by Avogadro’s number and dividing by the atomic mass of U in grams, we get the energy released per gram of U-235.

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 91Ó°ÊÓ!

Key Concepts

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

Mass Defect
In the process of nuclear fission, the mass defect plays a crucial role in quantifying the energy release. Mass defect is defined as the difference between the mass of an original atomic nucleus and the combined mass of its resulting fission products. To understand why mass defect occurs, consider that the binding energy, which holds the nucleus together, leads to a loss of mass according to mass-energy equivalence. During fission, the nucleus of uranium (e.g., \(^{235}U\)) is split into smaller nuclei (e.g., \(^{144}Ba\) and \(^{89}Kr\)), along with several neutrons. The original uranium nucleus has a certain mass \(m_U\), while the fission products and additional neutrons have a total mass calculated as \(m_{Ba} + m_{Kr} + 3 m_n\) for three released neutrons. The mass defect \(\Delta m\) is then calculated as the mass of the original uranium nucleus minus the sum of the masses of fission products and neutrons: \[\Delta m = m_U - (m_{Ba} + m_{Kr} + 3 m_n)\] This mass defect can now be harnessed to calculate the energy output of the reaction.
Mass-Energy Equivalence
Mass-energy equivalence is a principle developed by Albert Einstein, which establishes that mass and energy are two forms of the same thing. This concept is eloquently summarized in the famous equation, \(E = mc^2\), where \(E\) is energy, \(m\) is mass, and \(c\) is the speed of light. In nuclear fission, the mass defect calculated previously can be transformed directly into energy. The difference in mass (mass defect), due to its conversion per Einstein's equation, results in the release of energy. In practical terms, especially in nuclear physics, energy is often expressed in mega electron volts (MeV). This requires converting the mass defect (usually given in unified atomic mass units or \(u\)) into energy using a proportional conversion factor. Specifically, \(1 \, u \, = \, 931.5 \, \text{MeV}/c^2\). Therefore, the energy released \(E\) from the fission process is calculated by multiplying the mass defect by this conversion factor: \[E = \Delta m \, \times \, 931.5\] This conversion allows easy comparison of energy outputs from fission reactions, aiding in understanding their potential energy yields.
Unified Atomic Mass Unit
The unified atomic mass unit, symbolized as \(u\), is a standard unit of mass used to express atomic and molecular weights. One \(u\) is defined as one twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear ground state. This unit provides a convenient way to express and compare very small masses like those of individual atoms or particles. In nuclear physics, \(u\) helps standardize measurements across calculations involving nuclear reactions like fission. When tackling exercises related to nuclear fission, using the unified atomic mass unit simplifies calculations, like those of mass defects, since the measured atomic masses are presented in this unit. It uniformly allows easy conversion using the well-established relationship in \(E = mc^2\), where \(1 \, u \, = \, 931.5 \, \text{MeV}/c^2\). Thus, unified atomic mass units offer a precise and practical framework for scientists to measure and communicate findings about atomic structures and reactions in nuclear science initiatives.

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

In the reaction that produces \({ }^{123} \mathrm{I}\), is there a minimum kinetic energy the protons need to make the reaction go? (a) No, because the proton has a smaller mass than the neutron. (b) No, because the total initial mass is smaller than the total final mass. (c) Yes, because the proton has a smaller mass than the neutron. (d) Yes, because the total initial mass is smaller than the total final mass.

Radioactive Fallout. One of the problems of in-air testing of nuclear weapons (or, even worse, the use of such weapons!) is the danger of radioactive fallout. One of the most problematic nuclides in such fallout is strontium-90 \(\left({ }^{90} \mathrm{Sr}\right)\), which breaks down by \(\beta\) decay with a half-life of 28 years. It is chemically similar to calcium and the can be incorporated into bones and teeth, where, due to its rather long halflife, it remains for years as an internal source of radiation. (a) What is the daughter nucleus of the \({ }^{90} \mathrm{Sr}\) decay? (b) What percentage of the original level of \({ }^{90} \mathrm{Sr}\) is left after 56 years? (c) How long would you have to wait for the original level to be reduced to \(6.25 \%\) of its original value?

The isotope \({ }_{39}^{90} \mathrm{Y}\) undergoes \(\beta^{-}\) decay with a half-life of \(64.0 \mathrm{~h}\). You measure an activity of \(8.0 \times 10^{16}\) Bq. (a) How many \({ }_{39}^{90} \mathrm{Y}\) nuclei are present in the sample at the time you make this measurement? (b) How many will be present after 12.0 days?

A sample of radioactive nuclei has \(N_{0}\) nuclei at time \(t=0 .\) The half- life of the decay is \(T_{1 / 2}\). In terms of \(N_{0}\), how many decays occur in the time period between \(t=0\) and \(t=0.500 T_{1 / 2} ?\)

Consider the nuclear reaction $$ { }_{14}^{28} \mathrm{Si}+\gamma \rightarrow{ }_{12}^{24} \mathrm{Mg}+\mathrm{X} $$where \(X\) is a nuclide. (a) What are \(Z\) and \(A\) for the nuclide \(X ?\) (b) Ignoring the effects of recoil, what minimum energy must the photon have for this reaction to occur? The mass of a \({ }_{14}^{28} \mathrm{Si}\) atom is \(27.976927 \mathrm{u}\), and the mass of a \({ }_{12}^{24} \mathrm{Mg}\) atom is \(23.985042 \mathrm{u}\).

See all solutions

Recommended explanations on Physics Textbooks

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.