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5
300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md
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@ -65,6 +65,9 @@ $$1 \text{g} \times \frac{1 \text{ mol } ^{60}\text{Co}}{59.934 \text{ g}} = 1.
Now we can find how much more strontium we need: Now we can find how much more strontium we need:
$$\frac{28.8 \text{ years}}{5.26 \text{ years}} = 5.47528$$ $$\frac{28.8 \text{ years}}{5.26 \text{ years}} = 5.47528$$
Finally we multiply this number by the moles of cobalt, and convert back to mass for strontium-90: Finally we multiply this number by the moles of cobalt, and convert back to mass for strontium-90:
$$5.47528 \times \frac{1.66850 \times 10^{-2} \text{ mol } ^{60}\text{Co}}{}$$ $$ \left(1.66850 \times 10^{-2} \text{ mol } ^{60}\text{Co} \right) \times \frac{5.47528 \text{ mol } ^{90}\text{Sr}}{1 \text{ mol } ^{60}\text{Co}} \frac{89.90773 \text{ g }}{1 \text{ mol } ^{90}\text{Sr}}= 82.13525 \text{ g } ^{90}\text{Sr}$$
**9. Using the chart of the nuclides, complete the following reactions. If a daughter nucleus is radioactive, indicate the complete decay chain:** **9. Using the chart of the nuclides, complete the following reactions. If a daughter nucleus is radioactive, indicate the complete decay chain:**
$$^{18}\text{N} \rightarrow ^{18}\text{O}$$
$$^{83}\text{Y} \rightarrow ^{83}\text{Sr} \rightarrow^{83}\text{Rb} \rightarrow ^{82}\text{Kr}$$
$$^{219}\text{Rn} \rightarrow $$