Q61PE.
Question
Armor-piercing shells with depleted uranium cores are fired by aircraft at tanks. (The high density of the uranium makes them effective.) The uranium is called depleted because it has had its \(^{{\rm{235}}}{\rm{U}}\)removed for reactor use and is nearly pure\(^{{\rm{238}}}{\rm{U}}\). Depleted uranium has been erroneously called non-radioactive. To demonstrate that this is wrong:
(a) Calculate the activity of \(60.0\;\,{\rm{g}}\)of pure\(^{{\rm{238}}}{\rm{U}}\).
(b) Calculate the activity of \(60.0\;\,{\rm{g}}\)ofnatural uranium, neglecting the\(^{{\rm{234}}}{\rm{U}}\) and all daughter nuclides.
Step-by-Step Solution
Verifieda. The activity of pure \(^{{\rm{238}}}{\rm{U}}\)is\({\rm{2}}{\rm{.48 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{\;Bq}}\).
b. The activity of pure \(^{{\rm{238}}}{\rm{U}}\)neglecting the \(^{{\rm{234}}}{\rm{U}}\)and all daughter nuclides is\(7.49 \times {10^5}\,{\rm{Bq}}\).
Uranium: Uranium is a chemical element with the atomic number 92 and the symbol U. It's a silvery-grey metal from the periodic table's actinide series. There are 92 protons and 92 electrons in a uranium atom, with 6 valence electrons.
(a)
Considering the given information:
Mass of the pure \(^{{\rm{238}}}{\rm{U}}\) is
\(m = 60.0{\rm{\;}}\,{\rm{g}}\)
Apply the formula:
The activity is,
\(R=\frac{0.693N}{t_{1/2}}\)
The number of \(^{{\rm{238}}}{\rm{U}}\)in \(60.0{\rm{\;}}\,{\rm{g}}\) of uranium is:
\(\begin{aligned}N = \frac{{6.02 \times {{10}^{23}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}}}{{238\;\,{\rm{g/mol}}}} \times 60.0\;\,{\rm{g}}\\N = 1.5176 \times {10^{23}}\end{aligned}\)
The number of \(^{{\rm{238}}}{\rm{U}}\)in \(60.0{\rm{\;}}\,{\rm{g}}\) of uranium is:
\(\begin{aligned}N = \frac{{6.02 \times {{10}^{23}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}}}{{238\;\,{\rm{g/mol}}}} \times 60.0\;\,{\rm{g}}\\N = 1.5176 \times {10^{23}}\end{aligned}\)
The \(^{{\rm{238}}}{\rm{U}}\) half-life is,
\({t_{1/2}} = 1.41 \times {10^{17}}\,{\rm{sec}}\)
Putting the all values,
\(\begin{aligned}R = \frac{{0.693 \times 1.5176 \times {{10}^{23}}}}{{1.41 \times {{10}^{17}}\,{\rm{sec}}}}\\R = 7.459 \times {10^5}\,{\rm{Bq}}\end{aligned}\)
Therefore, the required activity of pure \(^{{\rm{238}}}{\rm{U}}\)is \(7.459 \times {10^5}\,{\rm{Bq}}\).
(b)
Considering the given information:
Mass of the pure \(^{{\rm{238}}}{\rm{U}}\) is
\(m = 60.0{\rm{\;}}\,{\rm{g}}\)
Apply the formula:
The activity is,
\(R=\frac{0.693N}{t_{1/2}}\)
\({\rm{0}}{\rm{.9928}}\) is the abundance of \(^{{\rm{238}}}{\rm{U}}\).
The activity of the \(^{{\rm{238}}}{\rm{U}}\)is
\(\begin{aligned}{l}R = \left( {7.459 \times {{10}^5}\,{\rm{Bq}}} \right) \times 0.9928\\R = 7.405 \times {10^5}\,{\rm{Bq}}\end{aligned}\)
\({\rm{0}}{\rm{.0072}}\)is the abundance of \(^{{\rm{234}}}{\rm{U}}\).
The number of \(^{{\rm{234}}}{\rm{U}}\)is,
\(\begin{aligned}{l}N = 1.5176 \times {10^{23}} \times 0.0072\\N = 1.0926 \times {10^{21}}\end{aligned}\)
The half-life of \(^{{\rm{234}}}{\rm{U}}\)is,
\({t_{1/2}} = 2.22 \times {10^{17}}\,{\rm{sec}}\)
Putting all values, the activity of \(^{{\rm{234}}}{\rm{U}}\)
\(\begin{aligned}{l}{R_{234}} = \frac{{0.693 \times 1.0926 \times {{10}^{21}}}}{{2.22 \times {{10}^{17}}\,\sec }}\\{R_{234}} = 3.4106 \times {10^3}\,{\rm{Bq}}\end{aligned}\)
As a result, the overall activity is:
\(\begin{aligned}{l}R = 7.459 \times {10^5}\,{\rm{Bq}} + 3.4106 \times {10^3}\,{\rm{Bq}}\\R = 7.49 \times {10^5}\,{\rm{Bq}}\end{aligned}\)
Therefore, the required activity of pure \(^{{\rm{238}}}{\rm{U}}\)neglecting the \(^{{\rm{234}}}{\rm{U}}\)and all daughter nuclides is\(7.49 \times {10^5}\,{\rm{Bq}}\).