/*! 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} Q63PE. Large amounts of depleted uraniu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 31: Q63PE. (page 1151)

Large amounts of depleted uranium \({{\rm{(}}^{{\rm{238}}}}{\rm{U)}}\)are available as a by-product of uranium processing for reactor fuel and weapons. Uranium is very dense and makes good counter weights for aircraft. Suppose you have a \(4000\,{\rm{kg}}\)block of\(^{{\rm{238}}}{\rm{U}}\).

  1. Find its activity.
  2. How many calories per day are generated by thermalization of the decay energy?
  3. Do you think you could detect this as heat? Explain.

Short Answer

Expert verified
  1. Its activity is \(R = 4.97 \times {10^{11}}\,{\rm{Bq}}\).
  2. Calories are generated by per day is \(E = 7.0 \times {10^3}\,{\rm{cal/day }}\).
  3. The level of energy is really high. As a result, the energy might be released as heat.

Step by step solution

01

Concept Introduction

For a given number of nuclei, the shorter the half-life, the more decays per unit time. As a result, activity R should be proportional to N, the number of radioactive nuclei, and inversely proportional to t1/2, their half-life. In reality, your instincts are spot on. It can be demonstrated that a source's activity is\(R = \frac{{0.693N}}{{{t_{1/2}}}}\).

02

Find its activity.

  1. The following is the relationship between activity, half-life, and the number of atoms:

\(R = \frac{{0.693N}}{{{t_{1/2}}}}\)

Where, \({t_{1/2}}\) is the half life, \(R\) is the activity and \(N\) is the number of atoms.

Number of atoms for \(^{{\rm{238}}}{\rm{U}}\)

\(\begin{aligned}N = \frac{{4000\;\,{\rm{kg}} \times 1000\;\,{\rm{g}}}}{{238\,{\rm{g}}}}\left( {6.02 \times {{10}^{23}}} \right)\\ = 1.01 \times {10^{28}}\end{aligned}\)

Now, in the previous equation, enter in the values of\(N\) and\({{\rm{t}}_{{\rm{1/2}}}}\)and solve for\(R\).

\(\begin{aligned}R = \frac{{0.693 \times 1.01 \times {{10}^{28}}}}{{\left( {4.468 \times {{10}^9}\,{\rm{y}}} \right)}}\\ = \frac{{0.693 \times 1.01 \times {{10}^{28}}}}{{\left( {4.468 \times {{10}^9}\,{\rm{y}}} \right)\left( {3.16 \times {{10}^7}\,{\rm{s}}} \right)}}\\ = 4.97 \times {10^{10}}\,{{\rm{s}}^{{\rm{ - 1}}}}\\ = 4.97 \times {10^{10}}\,{\rm{Bq}}\\R = 4.97 \times {10^{10}}\,{\rm{Bq}}\end{aligned}\)

Therefore, its activity is \(R = 4.97 \times {10^{10}}\,{\rm{Bq}}\).

03

Calories are generated by per day

  1. Assume that the amount of energy released by \(^{{\rm{238}}}{\rm{U}}\) each decay is \({\rm{4}}{\rm{.27MeV}}\). As a result, the total energy released will be

\(\begin{aligned}E = \left( {4.97 \times {{10}^{10}}\,{\rm{Bq}}} \right)(4.27\,{\rm{MeV}})\\ = \left( {4.97 \times {{10}^{10}}\,{{\rm{S}}^{{\rm{ - 1}}}}} \right)(4.27\,{\rm{MeV}})\\ = 2.12 \times {10^{11}}\,{\rm{MeV/s}}\end{aligned}\)

\(\begin{aligned}{\rm{So,}}E = \left( {2.12 \times {{10}^{11}}\,{\rm{MeV/s}}} \right)\left( {1.60 \times {{10}^{ - 13}}\,{\rm{joule }}} \right)(86400\;\,{\rm{s/day}})\\ = 2.93 \times {10^3}\;{\rm{J/day }}\end{aligned}\)

Now, energy is measured in calories per day.

\(\begin{aligned}E = \left( {2.93 \times {{10}^3}\;\,{\rm{J/day }}} \right)\left( {\frac{{1\,{\rm{cal}}}}{{4.184\;\,{\rm{J}}}}} \right)\\ = 7.0 \times {10^2}\,{\rm{cal/day}}\\{\rm{ }}E = 7.0 \times {10^2}\,{\rm{cal/day }}\end{aligned}\)

Therefore, calories are generated by per day is\(E = 7.0 \times {10^2}\,{\rm{cal/day }}\).

04

Write about heat energy

  1. The level of energy is really high. As a result, the energy might be released as heat.

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!

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

Is it possible for light emitted by a scintillator to be too low in frequency to be used in a photomultiplier tube? Explain.

The ceramic glaze on a red-orange Fiesta ware plate is \({{\rm{U}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}\)and contains \({\rm{50}}{\rm{.0}}\)grams of \(^{{\rm{238}}}{\rm{U}}\), but very little \(^{{\rm{235}}}{\rm{U}}\). (a)

  1. What is the activity of the plate?
  2. Calculate the total energy that will be released by the \(^{{\rm{238}}}{\rm{U}}\)decay.
  3. If energy is worth \({\rm{12}}{\rm{.0}}\)cents per\({\rm{kW \times h}}\), what is the monetary value of the energy emitted? (These plates went out of production some 30 years ago, but are still available as collectibles.)

(a) Calculate the radius of\(^{{\rm{58}}}{\rm{Ni}}\), one of the most tightly bound stable nuclei. (b) What is the ratio of the radius of\(^{{\rm{58}}}{\rm{Ni}}\)to that of\(^{{\rm{258}}}{\rm{Ha}}\), one of the largest nuclei ever made? Note that the radius of the largest nucleus is still much smaller than the size of an atom.

Find the length of a side of a cube having a mass of \(1.0\,{\rm{kg}}\) and the density of nuclear matter, taking this to be \(2.3 \times {10^{17}}\,{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\).

In a \({\rm{3 \times 1}}{{\rm{0}}^{\rm{9}}}\)-year-old rock that originally contained some\(^{{\rm{238}}}{\rm{U}}\)which has a half-life of \({\rm{4}}{\rm{.5 \times 1}}{{\rm{0}}^{\rm{9}}}\) years, we expect to find some \(^{{\rm{238}}}{\rm{U}}\)remaining in it. Why are \(^{{\rm{226}}}{\rm{Ra}}{{\rm{,}}^{{\rm{222}}}}{\rm{Rn, and}}{{\rm{ }}^{{\rm{210}}}}{\rm{Po}}\) also found in such a rock, even though they have much shorter half-lives (1600 years, 3.8 days, and 138 days, respectively)?
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Dynamics

Read Explanation

Scientific Method Physics

Read Explanation

Quantum Physics

Read Explanation

Fluids

Read Explanation

Astrophysics

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.