Chapter 4, Problem 4.42P

To determine

The activity of each Uranium isotope in 1g HEU in Bq and Ci in Highly enriched uranium (HEU) which contains 1.5% U-4, 93.5% U-5.0% U-8 as percentages by weight.

Answer to Problem 4.42P

Activity of U-4

  $\begin{array}{l}{A}_{234}=3.4355\times {\text{10}}^5\text{Bq}\\ A_{234}=0.9285\times {\text{10}}^{-5}\text{Ci}\end{array}$

Activity of U-5

  $\begin{array}{r}{A}_{235}=7.4568\times {\text{10}}^4\text{Bq}\\ A_{235}=2.0153\times {\text{10}}^{-6}\text{Ci}\end{array}$

Activity of U-8

  $\begin{array}{l}{A}_{238}=0.62\times {\text{10}}^3\text{Bq}\\ A_{238}=0.16756\times {\text{10}}^{-7}\text{Ci}\end{array}$

Explanation of Solution

Given:

  $\text{1}\text{.5\% U-4, 93}\text{.5\% U-5, 5}\text{.0\% U-8}\text{.}$ Percentages are by weight.

  $\text{Avogadro's Constant}\text{=}\text{6}\text{.022}\times {\text{10}}^{23}{\text{mol}}^{\text{-1}}$

Formula used:

  $\begin{array}{c}A=\lambda N\\ A=\text{ Activity in becquerel }\left(\text{Bq}\right)\\ N=\text{ The number of undecayed nuclei}\\ \\ \lambda ={\text{ Decay constant (s}}^{\text{-1}}\text{)}\\ \lambda =\frac{0.693}{Halflife}\\ N=\frac{\text{mass}}{\text{Atomic}\text{mass}}\times \text{avogadro's Constant}\\ \end{array}$

Calculation:

Decay Constant of each Uranium isotope.

  $\lambda =\frac{0.693}{Halflife}$

Half-life of ${\text{U}}_{\text{234}}\text{=246,000}\text{years}$

  $\begin{array}{l}{\lambda }_{234}=\frac{0.693}{246,000\times 365.25\times 24\times 3600s}\\ {\lambda }_{234}=\frac{0.693}{7.76\times {10}^{12}}\\ {\lambda }_{234}=0.089\times {10}^{-12}{\text{s}}^{-1}\end{array}$

Half-life of ${\text{U}}_{\text{235}}\text{=704}\text{million years}$

  $\begin{array}{l}{\lambda }_{235}=\frac{\text{0}\text{.693}}{\text{704,000,000×365}\text{.25×24×3600}\text{ s}}\\ {\lambda }_{235}=\frac{0.693}{2.22\times {10}^{16}}\\ {\lambda }_{235}=0.312\times {10}^{-16}{\text{s}}^{\text{-1}}\end{array}$

Half-life of ${\text{U}}_{\text{238}}\text{=4}\text{.46}\text{billion}\text{years}$

  $\begin{array}{l}{\lambda }_{238}=\frac{0.693}{4,460\times {10}^6\times 365.25\times 24\times 3600\text{s}}\\ {\lambda }_{238}=\frac{0.693}{1.407\times {10}^{17}}\\ {\lambda }_{238}=0.492\times {10}^{-17}{\text{s}}^{\text{-1}}\end{array}$

The number of undecayed nuclei

  $N=\frac{\text{mass}}{\text{Atomic}\text{mass}}\times \text{avogadro's Constant}$

  $\text{avogadro's Constant}\text{=}\text{6}\text{.022}\times {\text{10}}^{23}{\text{mol}}^{\text{-1}}$

  $\begin{array}{l}{N}_{234}\\ \text{Atomic}\text{mass}\text{=234}\text{g}{\text{mole}}^{-1}\end{array}$

  $\begin{array}{c}N=\frac{\text{mass}}{\text{Atomic}\text{mass}}\times \text{avogadro's Constant}\\ {\text{N}}_{234}=\frac{1g}{\text{234}\text{g}/\text{mole}}\times \frac{1.5}{100}\times \text{6}\text{.022}\times {\text{10}}^{23}{\text{mole}}^{-1}\\ {\text{N}}_{234}=0.0386\times {\text{10}}^{21}\end{array}$

  $\begin{array}{l}{N}_{235}\\ \text{Atomic}\text{mass}\text{=235}\text{g}{\text{mole}}^{-1}\end{array}$

  $\begin{array}{l}N=\frac{\text{mass}}{\text{Atomic}\text{mass}}\times \text{avogadro's Constant}\\ {\text{N}}_{235}=\frac{1g}{\text{235}\text{g}/\text{mole}}\times \frac{93.5}{100}\times \text{6}\text{.022}\times {\text{10}}^{23}{\text{mole}}^{\text{-1}}\\ {\text{N}}_{235}=0.0239\times {\text{10}}^{23}\\ {\text{N}}_{235}=23.9\times {\text{10}}^{20}\end{array}$

  $\begin{array}{l}{N}_{238}\\ \text{Atomic}\text{mass}\text{=238}\text{g}{\text{mole}}^{-1}\end{array}$

  $\begin{array}{c}N=\frac{\text{mass}}{\text{Atomic}\text{mass}}\times \text{avogadro's Constant}\\ {\text{N}}_{238}=\frac{\text{1g}}{\text{238}\text{g}/\text{mole}}\text{×}\frac{\text{5}}{\text{100}}\text{×6}{\text{.022×10}}^{\text{23}}{\text{mole}}^{\text{-1}}\\ {\text{N}}_{238}=0.00126\times {\text{10}}^{23}\\ {\text{N}}_{238}=1.26\times {\text{10}}^{20}\end{array}$

Activity of each Uranium isotope.

Activity of U-4

  $\begin{array}{l}{A}_{234}=\lambda N\\ A_{234}=0.089\times {10}^{-12}{\text{s}}^{-1}\times 0.0386\times {\text{10}}^{21}\\ A_{234}=0.0034\times {10}^9\\ A_{234}=3.4355\times {\text{10}}^6\text{decays}\text{per}\text{second}\\ A_{234}=3.4355\times {\text{10}}^6\text{Bq}\\ A_{234}=3.4355\times {\text{10}}^6\frac{\text{decays}}{\text{second}}\text{×}\frac{\text{1}\text{Curie}}{\text{3}{\text{.7×10}}^{\text{10}}\raisebox{1ex}{$\text{decays}$}\!\left/ \!\raisebox{-1ex}{$\text{second}$}\right.}\\ A_{234}=0.9285\times {\text{10}}^{\text{-4}}\text{Ci}\end{array}$

Activity of U-5

  $\begin{array}{l}{A}_{235}=\lambda N\\ A_{235}=0.312\times {10}^{-16}{\text{s}}^{-1}\times 23.9\times {\text{10}}^{20}\\ A_{235}=7.4568\times {\text{10}}^4\text{decays}\text{per}\text{second}\\ A_{235}=7.4568\times {\text{10}}^4\text{Bq}\\ \\ A_{235}=7.4568\times {\text{10}}^4\frac{\text{decays}}{\text{second}}\text{×}\frac{\text{1}\text{Curie}}{\text{3}{\text{.7×10}}^{\text{10}}\raisebox{1ex}{$\text{decays}$}\!\left/ \!\raisebox{-1ex}{$\text{second}$}\right.}\\ A_{235}=2.0153\times {\text{10}}^{\text{-6}}\text{Ci}\end{array}$

Activity of U-8

  $\begin{array}{l}{A}_{238}=\lambda N\\ A_{238}=0.492\times {10}^{-17}{\text{s}}^{-1}\times 1.26\times {\text{10}}^{20}\\ A_{238}=0.62\times {\text{10}}^3\text{decays}\text{per}\text{second}\\ A_{238}=0.62\times {\text{10}}^3\text{Bq}\\ \\ A_{238}=0.62\times {\text{10}}^{\text{3}}\frac{\text{decays}}{\text{second}}\text{×}\frac{\text{1}\text{Curie}}{\text{3}{\text{.7×10}}^{\text{10}}\raisebox{1ex}{$\text{decays}$}\!\left/ \!\raisebox{-1ex}{$\text{second}$}\right.}\\ A_{238}=0.16756\times {\text{10}}^{\text{-7}}\text{Ci}\end{array}$

Conclusion:

The activity of each Uranium isotope in 1g HEU in Bq and Ci is given below.

Activity of U-4

  $\begin{array}{l}{A}_{234}=3.4355\times {\text{10}}^5\text{Bq}\\ A_{234}=0.9285\times {\text{10}}^{-5}\text{Ci}\end{array}$

Activity of U-5

  $\begin{array}{r}{A}_{235}=7.4568\times {\text{10}}^4\text{Bq}\\ A_{235}=2.0153\times {\text{10}}^{-6}\text{Ci}\end{array}$

Activity of U-8

  $\begin{array}{l}{A}_{238}=0.62\times {\text{10}}^3\text{Bq}\\ A_{238}=0.16756\times {\text{10}}^{-7}\text{Ci}\end{array}$

(b)

To determine

The total activity of 1g of enriched Uranium, in Bq and in Ci.

Answer to Problem 4.42P

  $\begin{array}{c}\text{Total}\text{activity}=3.5\times {\text{10}}^6\text{Bq}\\ =0.95\times {\text{10}}^{-4}\text{Ci}\end{array}$

Explanation of Solution

Given:

Most prominent Uranium isotope in Highly enriched uranium (HEU) is U-5. It’s about 93.5% as percentages by weight.

Formula used:

  $\begin{array}{c}A=\lambda N\\ A=\text{ Activity in becquerel }\left(\text{Bq}\right)\\ N=\text{ The number of undecayed nuclei}\\ \\ \lambda ={\text{ Decay constant (s}}^{\text{-1}}\text{)}\\ \lambda =\frac{0.693}{Halflife}\\ N=\frac{\text{mass}}{\text{Atomic}\text{mass}}\times \text{avogadro's Constant}\\ \end{array}$

Calculation:

  $\begin{array}{c}\text{Total }\text{activity}\text{= Activity}\text{of}\text{U-4+U-5+U-8}\\ =3.4355\times {\text{10}}^{\text{6}}\text{Bq+7}{\text{.4568×10}}^{\text{4}}\text{Bq+0}{\text{.62×10}}^{\text{3}}\text{Bq}\\ =3.5\times {\text{10}}^6\text{Bq}\\ =3.5\times {\text{10}}^6\frac{\text{decays}}{\text{second}}\text{×}\frac{\text{1}\text{Curie}}{\text{3}{\text{.7×10}}^{\text{10}}\raisebox{1ex}{$\text{decays}$}\!\left/ \!\raisebox{-1ex}{$\text{second}$}\right.}\\ =0.95\times {\text{10}}^{-4}\text{Ci}\end{array}$

Conclusion:

The total activity of 1g of enriched Uranium, in Bq and in Ci is as given below.

  $\begin{array}{c}\text{Total}\text{activity}=3.5\times {\text{10}}^6\text{Bq}\\ =0.95\times {\text{10}}^{-4}\text{Ci}\end{array}$