From 041b3f2af0f12695c73cf97241ed476623c0739c Mon Sep 17 00:00:00 2001 From: Dane Sabo Date: Tue, 3 Sep 2024 16:06:43 -0400 Subject: [PATCH] vault backup: 2024-09-03 16:06:43 --- .../2024-09-03 Homework 1.md | 10 ++++++++-- 1 file changed, 8 insertions(+), 2 deletions(-) diff --git a/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md b/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md index cf0432b3..fc55c5ed 100644 --- a/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md +++ b/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md @@ -23,11 +23,14 @@ Complete the problems below being sure to show your work. If you need to lookup | $^{135}\text{Xe}$ | 54 | 81 | | $^{209}\text{Bi}$ | 83 | 126 | | $^{222}\text{Rn}$ | 86 | 136 | + **2. The atomic weight of $^{59}\text{Co}$ is 58.93319. How many times heavier is $^{12}\text{C}$? ** $\frac{ ^{59}\text{Co}}{^{12}\text{C}} = \frac{58.93319}{12.00000} = 4.91110 \text{ times larger}$ + **3. How many atoms are there in 10g of $^{12}\text{C}$?** $10 \text{g} \times \frac{1 \text{ mol } ^{12}\text{C}}{12 \text{g}} \times \frac{0.6022045 \times 10^{24} \text{ atoms}}{1 \text{ mol } ^{12}\text{C}} = 5.0184 \times 10^{23} \text{ atoms of } ^{12} \text{C}$ + **4. A beaker contains 50 g of ordinary water.** *a. How many moles of water are present?* $50 \text{g} \times \frac{1 \text{ mol } H_2O}{18.01528 \text{g}} = 2.77542 \text{ moles of water}$ @@ -35,17 +38,20 @@ $10 \text{g} \times \frac{1 \text{ mol } ^{12}\text{C}}{12 \text{g}} \times \fra $2.77542 \text{ moles of water } \times \frac{2 \text{mol} H}{1 \text{mol} H_2O} \times \frac{0.6022045 \times 10^{24} \text{ atoms}}{1 \text{ mol }H} = 3.34274 \times 10^{24} \text{ H atoms}$ *c. How many deuterium atoms?* $3.34274 \times 10^{24} \text{ H atoms} \times \frac{0.0156 ^2H}{1 H} = 5.21468 \times 10^{22} \text{ deuterium atoms}$ + **5. Find the mass of an atom of $^{235}\text{U}$** *a. in amu;* [235.043928 amu](https://ciaaw.org/uranium.htm) *b. in grams.* $1 \text{ atom } ^{235}U \times \frac{1 \text{ mol } ^{235}U}{0.6022045 \times 10^{24} \text{ atoms}} \times \frac{235.043928 \text{ g }}{1 \text{ mol } ^{235}U} = 3.90306 \times 10^{-20} \text{ g }$ + **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) $E = mc^2$ $E = 0.001 \text{ kg } \left(299792458 \text{ m/s }\right)^2$ $E = 8.98755 \times 10^{13} \text{ J }$ $\left(8.98755 \times 10^{13} \text{ J } \right) \times \frac{1 \text{ kg }}{3 \times 10^7 \text{ J }} = 2995850 \text{ kg of coal }$ + **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?* [Helium-3](https://people.physics.anu.edu.au/~ecs103/chart/) @@ -58,7 +64,7 @@ $\left(8.98755 \times 10^{13} \text{ J } \right) \times \frac{1 \text{ kg }}{3 \ $$\text{Ratio} = \frac{1 \text{ mCi}}{\lambda} = \frac{3.7 \times 10^{10} \frac{\text{decay}}{\text{s}}}{1.79241 \times 10^{-9} \frac{\text{decay}}{\text{s}}} = 2.06426 \times 10^{19}$$ Then we know we need this many atoms to decay (on average) at the mean activity. We now can convert to grams: $$\left(2.06426 \times 10^{19}\right) \times \frac{1 \text{ mol }^3H}{0.6022045 \times 10^{24} \text{atoms}} \times \frac{3.01605 \text{ g}}{1 \text{ mol } ^3H} = 1.03385 \times 10^{-4} \text{ g}$$ - + **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)?** First let's find the number of cobalt atoms: $$1 \text{g} \times \frac{1 \text{ mol } ^{60}\text{Co}}{59.934 \text{ g}} = 1.66850 \times 10^{-2} \text{ mol } ^{60}\text{Co}$$ @@ -70,4 +76,4 @@ $$ \left(1.66850 \times 10^{-2} \text{ mol } ^{60}\text{Co} \right) \times \frac **9. Using the chart of the nuclides, complete the following reactions. If a daughter nucleus is radioactive, indicate the complete decay chain:** $$^{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 $$ \ No newline at end of file +$$^{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}$$ \ No newline at end of file