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300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md

@@ -65,6 +65,9 @@ $$1 \text{g} \times \frac{1 \text{ mol } ^{60}\text{Co}}{59.934 \text{ g}} =  1.
 Now we can find how much more strontium we need:
 $$\frac{28.8 \text{ years}}{5.26 \text{ years}} = 5.47528$$
 Finally we multiply this number by the moles of cobalt, and convert back to mass for strontium-90:
-$$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:**
+$$^{18}\text{N} \rightarrow ^{18}\text{O}$$
+$$^{83}\text{Y} \rightarrow ^{83}\text{Sr} \rightarrow^{83}\text{Rb} \rightarrow ^{82}\text{Kr}$$
+$$^{219}\text{Rn} \rightarrow $$