Question

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

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

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 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 \(R = 4.97 \times {10^{10}}\,{\rm{Bq}}\).

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\(E = 7.0 \times {10^2}\,{\rm{cal/day }}\).

Write about heat energy

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