Balancing Nuclear Equations
Balancing nuclear equations involves ensuring that the total number of protons and neutrons on both sides of the equation are equal, thus maintaining the conservation of mass and charge. This process is essential for accurately representing nuclear reactions, including radioactive decay and nuclear fission or fusion. It requires adjusting the coefficients of the reactants and products to achieve equilibrium.
6 Key excerpts on "Balancing Nuclear Equations"
- Kevin Reel, Derrick C. Wood, Scott A. Best(Authors)
- 2014(Publication Date)
- Research & Education Association(Publisher)
...Chapter 14 Nuclear Chemistry Fundamentals of Nuclear Chemistry Nuclear chemistry involves chemical reactions that deal with the loss and gain of particles found within the nucleus—protons and neutrons. In these reactions, the nucleus undergoes a fundamental change that alters its identity. In radioactive atoms, the nuclei are unstable and may emit particles or energy called radiation. You should know that only certain isotopes of elements are radioactive (radioisotopes). There is also a variety of decay processes that can occur that transform an unstable nucleus in a high energy state into a more stable nucleus. These decay processes result in a net release of energy and occur as first-order rate expressions. DID YOU KNOW? Without a doubt, you have a manmade radio-active material in your house: americium-241. Americium is used in smoke detectors to emit alpha particles, which, when blocked by smoke particles in the air, activates an alarm to warn occupants of danger. Types of Decay Processes Alpha Decay Alpha decay involves the emission of an alpha (α) particle, which is essentially a helium nucleus: or. This type of decay occurs for unstable radioisotopes having an atomic number greater than 82. Alpha particles are relatively harmless and cannot even penetrate a piece of paper. EXAMPLE: Beta Decay Beta decay involves the emission of a beta (β) particle,. This type of decay process occurs when the proton-to-neutron ratio within the nucleus is too low for the radioisotope. Beta particle emission involves the splitting of a neutron into a proton and high-speed electrons that are ejected from the nucleus. As a result, the emission of the beta particle will cause the number of neutrons to decrease and the number of protons to increase. Beta particles have more penetrating power than alpha particles, but are stopped readily by dense materials such as lead. EXAMPLE: For beta decay, the mass number will stay the same but the atomic number (# of protons) will increase...
eBook - ePub- Jeffrey Gaffney, Nancy Marley(Authors)
- 2017(Publication Date)
- Elsevier(Publisher)
...To understand why these two decay processes happen, it is useful to look at a plot of the number of protons (Z) versus the number or neutrons (A − Z) for the known isotopes as shown in Fig. 14.3. Fig. 14.3 The number of protons (Z) versus the number of neutrons (N) for the known isotopes. The radioactive decay process shown are: alpha decay (α), positive beta decay (β +), negative beta decay (β −), proton loss, neutron loss, and fission. Modified from Table isotopes.svg, Wikimedia Commons. There is a very narrow range of isotope stability represented by the narrow black area in Fig. 14.3. The nucleus of the stable isotopes below an atomic number of 20 has about the same number of protons and neutrons. Those above an atomic number of 20 have more neutrons than protons. The nuclides that lie outside of this narrow black zone of stability are unstable. The nuclides above the stable region are neutron-rich. They can achieve stability by neutron decay (β −) to yield a negative beta particle (e −) and a proton. Since the neutron and proton are about the same mass, the decay of a neutron to a proton does little to change the mass of the nucleus. However, the chemistry does change since β − decay results in a daughter isotope with one more proton. An example of a β − decay is nickel-63 decaying to the stable daughter isotope copper-63 by the emission of a negative beta particle. Nickel-63 has 28 protons and 35 neutrons. After β − decay, the copper-63 daughter has 29 protons and 34 neutrons. So, the total number of neutrons and protons does not change. A β − decay process will not change the mass number of the isotope, but it will move the daughter one place to the right on the periodic chart, in this case moving from element 28 (nickel) to element 29 (copper). The nuclides below the region of stability are neutron poor. They have two possible decay processes to reach stability...
- Rachel A. Powsner, Matthew R. Palmer, Edward R. Powsner(Authors)
- 2013(Publication Date)
- Wiley-Blackwell(Publisher)
...The work (energy) required to overcome the nuclear force, the work to remove a nucleon from the nucleus, is called the nuclear binding energy. Typical binding energies are in the range of 6 million to 9 million electron volts (MeV) (approximately one thousand to one million times the electron binding force). The magnitude of the binding energy is related to another fact of nature: the measured mass of a nucleus is always less than the mass expected from the sum of the masses of its neutrons and protons. The “missing” mass is called the mass defect, the energy equivalent of which is equal to the nuclear binding energy. This interchangeability of mass and energy was immortalized in Einstein’s equation E = mc 2. Figure 1.9 The nuclear binding force is strong enough to overcome the electrical repulsion between the positively charged protons. The Stable Nucleus: Not all elements have stable nuclei; however, they do exist for most of the light and mid-weight elements, i.e., those with atomic numbers (number of protons) up to and including bismuth (Z = 83). The exceptions are technetium (Z = 43) and promethium (Z = 61). All those with atomic numbers higher than 83, such as radium (Z = 88) and uranium (Z = 92), are inherently unstable because of their large size. For those nuclei with a stable state, there is an optimal ratio of neutrons to protons. For the lighter elements, this ratio is approximately 1 : 1; for increasing atomic weights, the number of neutrons exceeds the number of protons. A plot depicting the number of neutrons as a function of the number of protons is called the line of stability (Figure 1.10). Figure 1.10 The combinations of neutrons and protons that can coexist in a stable nuclear configuration all lie within the gray-shaded regions. Stability Strictly speaking, stability is a relative term. We call a nuclide stable when its half-life is so long as to be practically immeasurable—say, greater than 10 8 years...
eBook - ePubPrimer on Radiation Oncology Physics
Video Tutorials with Textbook and Problems
- Eric Ford(Author)
- 2020(Publication Date)
- CRC Press(Publisher)
...Consider the case of 235 uranium fission. Here a neutron combines with the 235 U to form 236 U, and this undergoes spontaneous decay as follows: U 235 + n → U 236 → B 141 a + K 92 r + 3 n + energy In this fission process energy is produced. Another way to extract energy is the fusion of low- Z nuclei which moves up the curve of binding energy from a light element to a heavier element. An example of this is the main reaction in the hot dense core of the sun: H 2 + H 2 → H 2 e + n + energy (3.27 MeV) Here 2 H is the deuterium nucleus, i.e. hydrogen with the addition of one neutron. 2.1.4 Nomenclature for Isotopes Nuclei can have different numbers of protons and neutrons as illustrated in Figure 2.1.3. Two isotopes of helium are shown: H 2 4 e and H 2 3 e. These are isotopes of one another, i.e. they have the same number of protons (Z = 2), and are therefore the same element, but they differ in the numbers of neutrons. Shown: H 2 4 e and H 1 3 are isotones of one another, i.e. they have the same number of neutrons. FIGURE 2.1.3 Example of nuclei with different numbers of protons and neutrons. 2.2 Nuclear Decay Schemes As noted above, some nuclei in Figure 2.1.1 are stable (black squares) but some are not stable and will decay. There are different decay modes depending on whether a nucleus is relatively neutron-rich (blue area in Figure 2.1.1) vs. neutron-poor (orange area). The following sections describe these various decays modes, the products that emerge after decay, and their energies. 2.2.1 Beta-Minus Decay This section describes beta-minus decay which occurs in neutron-rich nuclei. Figure 2.2.1 shows an example beta-minus decay of P 15 32 into S 16 32. A beta-minus particle is produced. Note that a “beta-minus” particle is the same as an electron. Also produced is an anti-neutrino (υ ¯), which, for medical physics purposes, is not important. Energy is released as well, denoted with the symbol E. In this decay we started with 15 protons and ended with 16 protons...
eBook - ePubChemistry
Concepts and Problems, A Self-Teaching Guide
- Richard Post, Chad Snyder, Clifford C. Houk(Authors)
- 2020(Publication Date)
- Jossey-Bass(Publisher)
...6 Chemical Equations Now that you are familiar with atoms, symbols, molecules, formulas, and nomenclature, let's look at what happens when we mix substances together. The most important result of your efforts with this chapter will be your ability to write a balanced chemical equation that represents the reaction between two or more different substances that produces at least one new substance. Chemical equations are the chemist's shorthand. They show at a glance what substances have been mixed together and what new substance(s) have been produced. Chemists are able to predict the products of a mixture of substances even though they may never have actually mixed the substances in the laboratory. This is very important to research chemists trying to prepare new products that are useful and beneficial to mankind. You will learn how to complete and balance several kinds of chemical equations and how chemists recognize whether or not a reaction does indeed occur when substances are mixed. You will discover that some things remain unchanged during a chemical reaction...
eBook - ePub- Kevin R. Reel(Author)
- 2013(Publication Date)
- Research & Education Association(Publisher)
...CHAPTER 5 Equations and Stoichiometry CHAPTER 5 EQUATIONS AND STOICHIOMETRY BALANCING CHEMICAL EQUATIONS • Balanced chemical reactions are an artifact of the Law of Conservation of Mass. Matter is neither created nor destroyed during a chemical change. • The same number of like atoms must exist both as reactants and products. • To balance reactions, alter only the coefficients; do not change the formula. • In general, when balancing reactions, balance the hydrogen and oxygen atoms last. • For a change in oxidation number that is not a combustion reaction, see the next section, “Balancing Oxidation-Reduction reactions.” • When faced with an unbalanced reaction that involves the combustion of a hydrocarbon, look for the following: 1. First match the number of carbons in the hydrocarbon in the reactants by adjusting the coefficient of carbon dioxide in the product. Example: Three carbon atoms on both sides. 2. Then balance the hydrogen atoms by adding a coefficient to water in the product so that the hydrogen atoms equal the number of hydrogen atoms in the reactant hydrocarbon. Example: Eight hydrogen atoms on both sides. 3. Finally, add up the oxygen atoms in the products (carbon dioxide and water) and adjust the coefficient for the oxygen molecule in the reactants so that all oxygen atoms balance. Example: Ten oxygen atoms on both sides. BALANCING OXIDATION-REDUCTION REACTIONS • Balancing oxidation-reduction reactions requires balancing both mass and charge. Not only must the same number of each kind of atom exist on both sides of the yield sign, but the number of electrons lost in oxidation must equal the number of electrons gained during reduction. • Balancing oxidation-reduction reactions is NOT the result of trial-and-error. The following is a well-established method that allows you to get the right answer the first time, every time. 1. Establish oxidation numbers for each atom. 2...
