Q63PE. - College Physics (Urone)

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.

Step-by-Step Solution

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.

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.