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300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md
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@ -23,11 +23,14 @@ Complete the problems below being sure to show your work. If you need to lookup
| $^{135}\text{Xe}$ | 54 | 81 | | $^{135}\text{Xe}$ | 54 | 81 |
| $^{209}\text{Bi}$ | 83 | 126 | | $^{209}\text{Bi}$ | 83 | 126 |
| $^{222}\text{Rn}$ | 86 | 136 | | $^{222}\text{Rn}$ | 86 | 136 |
**2. The atomic weight of $^{59}\text{Co}$ is 58.93319. How many times heavier is $^{12}\text{C}$? ** **2. The atomic weight of $^{59}\text{Co}$ is 58.93319. How many times heavier is $^{12}\text{C}$? **
$\frac{ ^{59}\text{Co}}{^{12}\text{C}} = \frac{58.93319}{12.00000} = 4.91110 \text{ times larger}$ $\frac{ ^{59}\text{Co}}{^{12}\text{C}} = \frac{58.93319}{12.00000} = 4.91110 \text{ times larger}$
**3. How many atoms are there in 10g of $^{12}\text{C}$?** **3. How many atoms are there in 10g of $^{12}\text{C}$?**
$10 \text{g} \times \frac{1 \text{ mol } ^{12}\text{C}}{12 \text{g}} \times \frac{0.6022045 \times 10^{24} \text{ atoms}}{1 \text{ mol } ^{12}\text{C}} = 5.0184 \times 10^{23} \text{ atoms of } ^{12} \text{C}$ $10 \text{g} \times \frac{1 \text{ mol } ^{12}\text{C}}{12 \text{g}} \times \frac{0.6022045 \times 10^{24} \text{ atoms}}{1 \text{ mol } ^{12}\text{C}} = 5.0184 \times 10^{23} \text{ atoms of } ^{12} \text{C}$
**4. A beaker contains 50 g of ordinary water.** **4. A beaker contains 50 g of ordinary water.**
*a. How many moles of water are present?* *a. How many moles of water are present?*
$50 \text{g} \times \frac{1 \text{ mol } H_2O}{18.01528 \text{g}} = 2.77542 \text{ moles of water}$ $50 \text{g} \times \frac{1 \text{ mol } H_2O}{18.01528 \text{g}} = 2.77542 \text{ moles of water}$
@ -35,17 +38,20 @@ $10 \text{g} \times \frac{1 \text{ mol } ^{12}\text{C}}{12 \text{g}} \times \fra
$2.77542 \text{ moles of water } \times \frac{2 \text{mol} H}{1 \text{mol} H_2O} \times \frac{0.6022045 \times 10^{24} \text{ atoms}}{1 \text{ mol }H} = 3.34274 \times 10^{24} \text{ H atoms}$ $2.77542 \text{ moles of water } \times \frac{2 \text{mol} H}{1 \text{mol} H_2O} \times \frac{0.6022045 \times 10^{24} \text{ atoms}}{1 \text{ mol }H} = 3.34274 \times 10^{24} \text{ H atoms}$
*c. How many deuterium atoms?* *c. How many deuterium atoms?*
$3.34274 \times 10^{24} \text{ H atoms} \times \frac{0.0156 ^2H}{1 H} = 5.21468 \times 10^{22} \text{ deuterium atoms}$ $3.34274 \times 10^{24} \text{ H atoms} \times \frac{0.0156 ^2H}{1 H} = 5.21468 \times 10^{22} \text{ deuterium atoms}$
**5. Find the mass of an atom of $^{235}\text{U}$** **5. Find the mass of an atom of $^{235}\text{U}$**
*a. in amu;* *a. in amu;*
[235.043928 amu](https://ciaaw.org/uranium.htm) [235.043928 amu](https://ciaaw.org/uranium.htm)
*b. in grams.* *b. in grams.*
$1 \text{ atom } ^{235}U \times \frac{1 \text{ mol } ^{235}U}{0.6022045 \times 10^{24} \text{ atoms}} \times \frac{235.043928 \text{ g }}{1 \text{ mol } ^{235}U} = 3.90306 \times 10^{-20} \text{ g }$ $1 \text{ atom } ^{235}U \times \frac{1 \text{ mol } ^{235}U}{0.6022045 \times 10^{24} \text{ atoms}} \times \frac{235.043928 \text{ g }}{1 \text{ mol } ^{235}U} = 3.90306 \times 10^{-20} \text{ g }$
**6. The complete combustion of 1 kg of bituminous coal releases about $3\times 10^7 \text{J}$ in heat energy. The conversion of 1 g of mass into energy is equivalent to the burning of how much coal?** **6. The complete combustion of 1 kg of bituminous coal releases about $3\times 10^7 \text{J}$ in heat energy. The conversion of 1 g of mass into energy is equivalent to the burning of how much coal?**
[The speed of light is 299,792,458 m/s.](https://en.wikipedia.org/wiki/Speed_of_light) [The speed of light is 299,792,458 m/s.](https://en.wikipedia.org/wiki/Speed_of_light)
$E = mc^2$ $E = mc^2$
$E = 0.001 \text{ kg } \left(299792458 \text{ m/s }\right)^2$ $E = 0.001 \text{ kg } \left(299792458 \text{ m/s }\right)^2$
$E = 8.98755 \times 10^{13} \text{ J }$ $E = 8.98755 \times 10^{13} \text{ J }$
$\left(8.98755 \times 10^{13} \text{ J } \right) \times \frac{1 \text{ kg }}{3 \times 10^7 \text{ J }} = 2995850 \text{ kg of coal }$ $\left(8.98755 \times 10^{13} \text{ J } \right) \times \frac{1 \text{ kg }}{3 \times 10^7 \text{ J }} = 2995850 \text{ kg of coal }$
**7. Tritium ($^3\text{H}$) decays by negative beta decay with a half-life of 12.26 years. The atomic weight of $^3\text{H}$ is 3.016.** **7. Tritium ($^3\text{H}$) decays by negative beta decay with a half-life of 12.26 years. The atomic weight of $^3\text{H}$ is 3.016.**
*a. To what nucleus does $^3\text{H}$ decay?* *a. To what nucleus does $^3\text{H}$ decay?*
[Helium-3](https://people.physics.anu.edu.au/~ecs103/chart/) [Helium-3](https://people.physics.anu.edu.au/~ecs103/chart/)
@ -70,4 +76,4 @@ $$ \left(1.66850 \times 10^{-2} \text{ mol } ^{60}\text{Co} \right) \times \frac
**9. Using the chart of the nuclides, complete the following reactions. If a daughter nucleus is radioactive, indicate the complete decay chain:** **9. Using the chart of the nuclides, complete the following reactions. If a daughter nucleus is radioactive, indicate the complete decay chain:**
$$^{18}\text{N} \rightarrow ^{18}\text{O}$$ $$^{18}\text{N} \rightarrow ^{18}\text{O}$$
$$^{83}\text{Y} \rightarrow ^{83}\text{Sr} \rightarrow^{83}\text{Rb} \rightarrow ^{82}\text{Kr}$$ $$^{83}\text{Y} \rightarrow ^{83}\text{Sr} \rightarrow^{83}\text{Rb} \rightarrow ^{82}\text{Kr}$$
$$^{219}\text{Rn} \rightarrow $$ $$^{219}\text{Rn} \rightarrow ^{215}\text{Po} \rightarrow ^{211}\text{Pb} \rightarrow ^{211}\text{Bi} \rightarrow ^{207}\text{Ti} \rightarrow ^{207}\text{Pb} \rightarrow ^{203}\text{Hg} \rightarrow ^{203}\text{Tl}$$