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.)

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

Step-by-step solution

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