/*! 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} Q62PE. The ceramic glaze on a red-orang... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 31: Q62PE. (page 1117)

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.)

Short Answer

Expert verified
  1. The activity of the plate is \(6.19 \times {10^6}\,{\rm{Bq}}\).
  2. The total energy that will be released by the\(^{{\rm{238}}}{\rm{U}}\)decay is \(2.50 \times {10^{16}}\,{\rm{MeV}}\).
  3. The total energy cost \(13.3\,{\rm{cents }}\).

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 the activity of the plate

  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.

The number of atoms for \(^{{\rm{238}}}{\rm{U}}\) is

\(\begin{aligned}N & = \frac{{50\,{\rm{g}}}}{{238\,{\rm{g}}}}\left( {6.02 \times {{10}^{23}}} \right)\\ & = 1.26 \times {10^{23}}.\end{aligned}\)

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

\(\begin{aligned}R & = \frac{{0.693 \times 1.26 \times {{10}^{23}}}}{{\left( {4.468 \times {{10}^8}\,{\rm{y}}} \right)}}\\ & = \frac{{0.693 \times 1.26 \times {{10}^{23}}}}{{\left( {4.468 \times {{10}^8}\,{\rm{y}}} \right)\left( {3.16 \times {{10}^7}\,{\rm{s}}} \right)}}\\ & = 6.19 \times {10^6}\,{{\rm{s}}^{{\rm{ - 1}}}}\\ & = 6.19 \times {10^6}\,{\rm{Bq}}\\R & = 6.19 \times {10^6}\,{\rm{Bq}}\end{aligned}\)

Therefore, the activity of the plate is \(6.19 \times {10^6}\,{\rm{Bq}}\).

03

Calculate the total energy  

Assume that the amount of energy released by \(^{{\rm{238}}}{\rm{U}}\)per decay is\(4.27\,{\rm{MeV}}\). As a result, the total energy released will be

\(\begin{aligned}E & = \left( {6.19 \times {{10}^6}\,{\rm{Bq}}} \right)(4.27\,{\rm{MeV}})(30\,{\rm{y}})\\ & = \left( {6.19 \times {{10}^6}\,{{\rm{s}}^{{\rm{ - 1}}}}} \right)(4.27\,{\rm{MeV}})(30\,{\rm{y}})\left( {\frac{{3.16 \times {{10}^7}\,{\rm{s}}}}{{1\,{\rm{y}}}}} \right)\\ & = 2.50 \times {10^{16}}\,{\rm{MeV}}\\E & = 2.50 \times {10^{16}}\,{\rm{MeV}}\end{aligned}\)

Therefore, the total energy that will be released by the\(^{{\rm{238}}}{\rm{U}}\)decay is \(2.50 \times {10^{16}}\,{\rm{MeV}}\).

04

Calculate the total energy

  1. Convert MeV to \({\rm{kWs}}\) now.

\(\begin{aligned}E & = 2.50 \times {10^{16}}\,{\rm{MeV}}\left( {\frac{{1.60 \times {{10}^{ - 13}}\,{\rm{kWs}}}}{{1\,{\rm{MeV}}}}} \right)\\ & = 4.0 \times {10^3}\,{\rm{kWs}}\end{aligned}\)

Because the cost of electricity is 12 cents per kWh Thus, the total cost is:

\(\begin{aligned}Total{\rm{ }}\cos t & = 4.0 \times {10^3}\,{\rm{kWs}}\left( {\frac{{12\,{\rm{cent }}}}{{{\rm{kWh}}}}} \right)\left( {\frac{{1\;\,{\rm{h}}}}{{3600\,{\rm{s}}}}} \right)\\ & = 13.3\,{\rm{cents}}\\total{\rm{ }}\cos t{\rm{ }} & = 13.3\,{\rm{cents }}\end{aligned}\)

Therefore, the total energy cost \(13.3\,{\rm{cents }}\).

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

(a) Calculate the energy released in the\({\rm{\alpha }}\)decay of\({}^{{\rm{238}}}{\rm{U}}\). (b) What fraction of the mass of a single\({}^{{\rm{238}}}{\rm{U}}\)is destroyed in the decay? The mass of\({}^{{\rm{234}}}{\rm{Th}}\)is\(234.043593\,{\rm{u}}\). (c) Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?

Spontaneous radioactive decay occurs only when the decay products have less mass than the parent, and it tends to produce a daughter that is more stable than the parent. Explain how this is related to the fact that more tightly bound nuclei are more stable. (Consider the binding energy per nucleon.)

World War II aircraft had instruments with glowing radium-painted dials (see figure ). The activity of one such instrument was \(1.0 \times {10^5}\,{\rm{Bq}}\) when new.

(a) What mass of \(^{{\rm{226}}}{\rm{Ra}}\) was present?

(b) After some years, the phosphors on the dials deteriorated chemically, but the radium did not escape. What is the activity of this instrument \({\rm{57}}{\rm{.0}}\)years after it was made?

(a) Write the complete \({{\rm{\beta }}^{\rm{ - }}}\) decay equation for the neutron.

(b) Find the energy released in the decay.
  1. The \(^{{\rm{210}}}\) Po source used in a physics laboratory is labelled as having an activity of \(1.0\,{\rm{\mu Ci}}\) on the date it was prepared. A student measures the radioactivity of this source with a Geiger counter and observes 1500 counts per minute. She notices that the source was prepared 120 days before her lab. What fraction of the decays is she observing with her apparatus?
  2. Identify some of the reasons that only a fraction of the 伪s emitted are observed by the detector.
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Linear Momentum

Read Explanation

Quantum Physics

Read Explanation

Waves Physics

Read Explanation

Radiation

Read Explanation

Force

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.