Q62PE. - College Physics (Urone)

Question

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.)

Step-by-Step Solution

Answer

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 \(13.3\,{\rm{cents }}\).

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

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 \(6.19 \times {10^6}\,{\rm{Bq}}\).

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 \(2.50 \times {10^{16}}\,{\rm{MeV}}\).

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 \(13.3\,{\rm{cents }}\).