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| title | allDay | startTime | endTime | date | completed | type |
|---|---|---|---|---|---|---|
| Homework 1 | false | 11:30 | 16:00 | 2024-09-03 | null | single |
Preamble
#Homework Dane Sabo Dane.Sabo@pitt.edu September 3rd, 2024
Instructions
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.
Problems
1. How many neutrons and protons are there in the nuclei of the following atoms:
| Atom | Protons | Neutrons |
|---|---|---|
^7\text{Li} |
3 | 4 |
^{24} \text{Mg} |
12 | 12 |
^{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}
b. How many hydrogen 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?
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
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.
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
b. What is the mass in grams of 1 mCi of tritium?
First, we need to find the decay constant of tritium:
\lambda = \frac{0.693 \text{ decay}}{12.26 \text{ years}} = 1.79241 \times 10^{-9} \frac{\text{decay}}{\text{s}}
And we also know that one millicurie is:
1 \text{ mCi} = 3.7 \times 10^{10} \frac{\text{decay}}{\text{s}}
Therefore we find multiple of the decay constant we need:
\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}
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:
\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 ^{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}