---
title: Homework 1
allDay: false
startTime: 11:30
endTime: 16:00
date: 2024-09-03
completed: null
type: single
---
# Preamble
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](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/)
	*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:**
$$^{18}\text{N} \rightarrow ^{18}\text{O}$$
$$^{83}\text{Y} \rightarrow ^{83}\text{Sr} \rightarrow^{83}\text{Rb} \rightarrow ^{82}\text{Kr}$$
$$^{219}\text{Rn} \rightarrow ^{215}\text{Po} \rightarrow ^{211}\text{Pb} \rightarrow ^{211}\text{Bi} \rightarrow ^{207}\text{Ti} \rightarrow ^{207}\text{Pb} \rightarrow ^{203}\text{Hg} \rightarrow ^{203}\text{Tl}$$