80 lines
5.0 KiB
Markdown
80 lines
5.0 KiB
Markdown
---
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title: Homework 1
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allDay: false
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startTime: 11:30
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endTime: 16:00
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date: 2024-09-03
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completed: null
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type: single
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---
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# Preamble
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#Homework
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Dane Sabo
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Dane.Sabo@pitt.edu
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September 3rd, 2024
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# Instructions
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Complete the problems below being sure to show your work. If you need to lookup nuclear data from an external source please reference the source in your solutions.
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# Problems
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**1. How many neutrons and protons are there in the nuclei of the following atoms:**
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| Atom | Protons | Neutrons |
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| ----------------- | ------- | -------- |
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| $^7\text{Li}$ | 3 | 4 |
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| $^{24} \text{Mg}$ | 12 | 12 |
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| $^{135}\text{Xe}$ | 54 | 81 |
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| $^{209}\text{Bi}$ | 83 | 126 |
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| $^{222}\text{Rn}$ | 86 | 136 |
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**2. The atomic weight of $^{59}\text{Co}$ is 58.93319. How many times heavier is $^{12}\text{C}$? **
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$\frac{ ^{59}\text{Co}}{^{12}\text{C}} = \frac{58.93319}{12.00000} = 4.91110 \text{ times larger}$
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**3. How many atoms are there in 10g of $^{12}\text{C}$?**
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$10 \text{g} \times \frac{1 \text{ mol } ^{12}\text{C}}{12 \text{g}} \times \frac{0.6022045 \times 10^{24} \text{ atoms}}{1 \text{ mol } ^{12}\text{C}} = 5.0184 \times 10^{23} \text{ atoms of } ^{12} \text{C}$
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**4. A beaker contains 50 g of ordinary water.**
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*a. How many moles of water are present?*
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$50 \text{g} \times \frac{1 \text{ mol } H_2O}{18.01528 \text{g}} = 2.77542 \text{ moles of water}$
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*b. How many hydrogen atoms?*
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$2.77542 \text{ moles of water } \times \frac{2 \text{mol} H}{1 \text{mol} H_2O} \times \frac{0.6022045 \times 10^{24} \text{ atoms}}{1 \text{ mol }H} = 3.34274 \times 10^{24} \text{ H atoms}$
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*c. How many deuterium atoms?*
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$3.34274 \times 10^{24} \text{ H atoms} \times \frac{0.0156 ^2H}{1 H} = 5.21468 \times 10^{22} \text{ deuterium atoms}$
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**5. Find the mass of an atom of $^{235}\text{U}$**
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*a. in amu;*
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[235.043928 amu](https://ciaaw.org/uranium.htm)
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*b. in grams.*
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$1 \text{ atom } ^{235}U \times \frac{1 \text{ mol } ^{235}U}{0.6022045 \times 10^{24} \text{ atoms}} \times \frac{235.043928 \text{ g }}{1 \text{ mol } ^{235}U} = 3.90306 \times 10^{-20} \text{ g }$
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**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?**
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[The speed of light is 299,792,458 m/s.](https://en.wikipedia.org/wiki/Speed_of_light)
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$E = mc^2$
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$E = 0.001 \text{ kg } \left(299792458 \text{ m/s }\right)^2$
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$E = 8.98755 \times 10^{13} \text{ J }$
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$\left(8.98755 \times 10^{13} \text{ J } \right) \times \frac{1 \text{ kg }}{3 \times 10^7 \text{ J }} = 2995850 \text{ kg of coal }$
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**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.**
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*a. To what nucleus does $^3\text{H}$ decay?*
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[Helium-3](https://people.physics.anu.edu.au/~ecs103/chart/)
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*b. What is the mass in grams of 1 mCi of tritium?*
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First, we need to find the decay constant of tritium:
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$\lambda = \frac{0.693 \text{ decay}}{12.26 \text{ years}} = 1.79241 \times 10^{-9} \frac{\text{decay}}{\text{s}}$
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And we also know that one millicurie is:
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$1 \text{ mCi} = 3.7 \times 10^{10} \frac{\text{decay}}{\text{s}}$
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Therefore we find multiple of the decay constant we need:
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$$\text{Ratio} = \frac{1 \text{ mCi}}{\lambda} = \frac{3.7 \times 10^{10} \frac{\text{decay}}{\text{s}}}{1.79241 \times 10^{-9} \frac{\text{decay}}{\text{s}}} = 2.06426 \times 10^{19}$$
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Then we know we need this many atoms to decay (on average) at the mean activity. We now can convert to grams:
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$$\left(2.06426 \times 10^{19}\right) \times \frac{1 \text{ mol }^3H}{0.6022045 \times 10^{24} \text{atoms}} \times \frac{3.01605 \text{ g}}{1 \text{ mol } ^3H} = 1.03385 \times 10^{-4} \text{ g}$$
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**8. Approximately what mass of $^{90}\text{Sr}$ (T-1/2 = 28.8 years) has the same activity as 1g of $^{60}\text{Co}$ (T-1/2 = 5.26 years)?**
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First let's find the number of cobalt atoms:
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$$1 \text{g} \times \frac{1 \text{ mol } ^{60}\text{Co}}{59.934 \text{ g}} = 1.66850 \times 10^{-2} \text{ mol } ^{60}\text{Co}$$
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Now we can find how much more strontium we need:
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$$\frac{28.8 \text{ years}}{5.26 \text{ years}} = 5.47528$$
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Finally we multiply this number by the moles of cobalt, and convert back to mass for strontium-90:
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$$ \left(1.66850 \times 10^{-2} \text{ mol } ^{60}\text{Co} \right) \times \frac{5.47528 \text{ mol } ^{90}\text{Sr}}{1 \text{ mol } ^{60}\text{Co}} \frac{89.90773 \text{ g }}{1 \text{ mol } ^{90}\text{Sr}}= 82.13525 \text{ g } ^{90}\text{Sr}$$
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**9. Using the chart of the nuclides, complete the following reactions. If a daughter nucleus is radioactive, indicate the complete decay chain:**
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$$^{18}\text{N} \rightarrow ^{18}\text{O}$$
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$$^{83}\text{Y} \rightarrow ^{83}\text{Sr} \rightarrow^{83}\text{Rb} \rightarrow ^{82}\text{Kr}$$
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$$^{219}\text{Rn} \rightarrow ^{215}\text{Po} \rightarrow ^{211}\text{Pb} \rightarrow ^{211}\text{Bi} \rightarrow ^{207}\text{Ti} \rightarrow ^{207}\text{Pb} \rightarrow ^{203}\text{Hg} \rightarrow ^{203}\text{Tl}$$ |