/*! 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} Q44PE (a) Calculate the energy release... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Calculate the energy released in the neutron-induced fission reaction\(n{ + ^{235}}U{ \to ^{92}}Kr{ + ^{142}}Ba + 2n\),

Given \(m{(^{92}}Kr) = 91.926269{\rm{ }}u\) and

\(m{(^{142}}Ba) = 141.916361{\rm{ }}u\).

(b) Confirm that the total number of nucleons and total charge are conserved in this reaction.

Short Answer

Expert verified

(a) The energy released in the reaction\(n{ + ^{235}}U{ \to ^{92}}Kr{ + ^{142}}Ba + 2n\)is\(E = 179.4MeV\).

(b) It is confirmed that the total number of nucleons and total charge are conserved in \(n{ + ^{235}}U{ \to ^{92}}Kr{ + ^{142}}Ba + 2n\) reaction.

Step by step solution

01

Concept Introduction

The power kept in the centre of an atom is known as nuclear energy. All stuff in the cosmos is made up of tiny particles called atoms. The centre of an atom's nucleus is typically where most of its mass is concentrated. Neutrons and protons are the two subatomic particles that make up the nucleus. Atomic bonds that hold them together carry a lot of energy.

02

Energy Calculation

The energy released in a neutron-induced fission reaction is\(E = \Delta m{c^2}\),where\(\Delta m\)is the difference between the mass of the parent nucleus plus themass of the neutron, minus the sum of the masses of the products.

\(n = 1.008665u\), the difference in mass is –

\(\begin{align}{}\Delta m & = \left[ {m\left( {^{235}U} \right) + m(n)} \right] - \left[ {m\left( {^{92}Kr} \right) + m\left( {^{142}Ba} \right) + 2m(n)} \right]\\ & = [235.043924u + 1.008665u] - [91.926269u + 141.916361u + 2(1.008665)u]\\ & = 0.19262911\end{align}\)

Hence, the energy is –

\(\begin{align}{}E &= 0.192629n\frac{{931.5\frac{{McV}}{{{c^2}}}}}{u}{c^2}\\ &= 179.4MeV\end{align}\)

Therefore, the value for energy is obtained as \(E = 179.4MeV\).

03

Step 3:Number of nucleons and charge

The total number of nucleons and the total charge are conserved since,

The number of nucleons on the right side and on the left side of the reaction are equal –

\(\begin{align}{}A &= {(235 + 1)_{left{\rm{ }}}}\\ &= {(92 + 142 + 2)_{right{\rm{ }}}}\\ &= 236\end{align}\)

The total charge on the right side and on the left side of the reaction are equal –

\(\begin{align}{}Z & = {(92 + 0)_{left{\rm{ }}}}\\ &= {(36 + 56 + 0)_{right{\rm{ }}}}\\ &= 92\end{align}\)

Therefore, all the quantities are conserved.

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!

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

(a) Neutron activation of sodium, which is \({\rm{100\% }}\,{\,^{{\rm{23}}}}{\rm{Na}}\), produces\(^{{\rm{24}}}{\rm{Na}}\), which is used in some heart scans, as seen in Table 32.1. The equation for the reaction is \(^{23}Na + n{ \to ^{24}}Na + \gamma \). Find its energy output, given the mass of \(^{{\rm{24}}}{\rm{Na}}\) is \(23.990962\,u\).

(b) What mass of \(^{{\rm{24}}}{\rm{Na}}\) produces the needed \(5.0\)-mCi activity, given its half-life is \(15.0\,\;h\) ?

(a) Find the total energy released in \(MeV\) in each carbon cycle (elaborated in the above problem) including the annihilation energy.

(b) How does this compare with the proton-proton cycle output?

(a) Estimate the years that the deuterium fuel in the oceans could supply the energy needs of the world. Assume world energy consumption to be ten times that of the United States which is \(8 \times {10^9}J/y\) and that the deuterium in the oceans could be converted to energy with an efficiency of \(32\% \). You must estimate or look up the amount of water in the oceans and take the deuterium content to be \(0.015\% \) of natural hydrogen to find the mass of deuterium available. Note that approximate energy yield of deuterium is \(3.37 \times {10^{14}}J/kg\).

(b) Comment on how much time this is by any human measure. (It is not an unreasonable result, only an impressive one.)

The activities of \(^{131}I\) and \(^{123}I\) used in thyroid scans are given in Table \({\rm{32}}{\rm{.1}}\) to be 50 and \(70\mu Ci\), respectively. Find and compare the masses of \(^{131}I\)and \(^{123}I\) in such scans, given their respective half-lives are \({\rm{8}}{\rm{.04\;d}}\)and \({\rm{13}}{\rm{.2\;h}}\). The masses are so small that the radioiodine is usually mixed with stable iodine as a carrier to ensure normal chemistry and distribution in the body.

Why does the fusion of light nuclei into heavier nuclei release energy?

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.