Chapter 31, Problem 59GP

(a)

To determine

Total mass required to operate the plants for 1 year.

Answer to Problem 59GP

Total mass required to operate the plants for 1 year is $\mathrm{m}=1309\text{ kg}$

Explanation of Solution

Given:

Power generated, $\mathrm{P}=3400\text{ MW}=34\times {10}^8\text{W}$

Energy released per fission, $\mathrm{E}=200\text{MeV}=2\times {10}^8\text{eV}$

Formula used:

The work done is given by the formula,

  $\mathrm{W}=\mathrm{q}\mathrm{\Delta }\mathrm{V}$

Calculation:

Energy produced in one year can be given as:

  $\begin{array}{l}=\mathrm{P}\times \mathrm{t}\\ =\left(3400\times {10}^6\text{W}\right)\left(3.156\times {10}^7\text{s}\right)\\ =\left(3400\times {10}^6\text{J/s}\right)\left(3.156\times {10}^7\text{s}\right)\\ =1.073\times {10}^{17}\text{J}\end{array}$

Converting energy released into joules, we get:

  $\begin{array}{l}\mathrm{E}=200\text{MeV}\\ \mathrm{E}=200\times {10}^6\times 1.6\times {10}^{-19}\text{J}\\ \mathrm{E}=3.2\times {10}^{-11}\text{J}\end{array}$

Now, the number of fission reactions is given by:

  $\begin{array}{l}\mathrm{N}=\frac{1.073\times {10}^{17}\text{J}}{3.2\times {10}^{-11}\text{J}}\\ \mathrm{N}=3.35\times {10}^{27}\end{array}$

Hence, the total mass can be calculated as follows:

  $\begin{array}{l}\mathrm{m}=\frac{\left(1.073\times {10}^{17}\text{J}\right)\left(0.235\right)}{\left(3.2\times {10}^{-11}\text{J}\right)\left(6.022\times {10}^{23}\right)}\\ \mathrm{m}=1309\text{ kg}\end{array}$

Conclusion:

Hence, the total mass is $\mathrm{m}=1309\text{ kg}$

(b)

To determine

Total radioactivity of $\text{S}\text{r}$

Answer to Problem 59GP

The total radioactivity is $\text{4}\text{.1}\times {\text{10}}^6\text{Ci}$

Explanation of Solution

Given:

6% of fission produces $\text{S}\text{r}$

Half-life of ${\mathrm{\beta }}^{-}$ emitter is 29 yr.

Calculation:

To find the radioactivity, we need to find the number of Strontium atoms as follows:

  $\begin{array}{l}\mathrm{N}=\frac{\left(0.06\right)\left(1.073\times {10}^{17}\text{J}\right)}{\left(3.2\times {10}^{-11}\text{J}\right)}\\ \mathrm{N}=2.01\times {10}^{26}\end{array}$

Now, the radioactivity can be calculated as follows:

  $\begin{array}{l}=\frac{0.693\mathrm{N}}{{\mathrm{t}}_{\frac{1}{2}}}\\ =\frac{\left(0.693\right)\left(2.01\times {10}^{26}\right)}{\left(29\times 3.156\times {10}^7\text{s}\right)}\\ =1.522\text{ decay/s}\\ =\frac{\left(1.522\right)}{\left(3.7\times {10}^{10}\right)}\text{Ci}\\ \text{=4}\text{.1}\times {\text{10}}^6\text{Ci}\end{array}$

Conclusion:

The total radioactivity is $\text{4}\text{.1}\times {\text{10}}^6\text{Ci}$