{"id":72,"date":"2017-08-18T08:55:26","date_gmt":"2017-08-18T08:55:26","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=72"},"modified":"2017-09-18T18:44:35","modified_gmt":"2017-09-18T18:44:35","slug":"the-nucleus-and-nuclear-reactions","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/the-nucleus-and-nuclear-reactions\/","title":{"rendered":"The Nucleus and Nuclear Reactions"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/quantum-mechanics-part-ii\">\u2b05 Previous\u00a0Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/atomic-structure-periodicity-and-matter\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/review-2\">Next Lesson\u27a1<\/a><\/p>\n<\/div>\n<p><!-- UPDATE NEXT\/PREVIOUS ABOVE --><\/p>\n<p><!-- CONTENT STARTS HERE --><\/p>\n<h1 id=\"title\">Atomic Structure, Periodicity, and Matter:The Nucleus and Nuclear Reactions<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson we will review nuclear structure and stability. We will also review balancing nuclear\u00a0equations.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>Niels Bohr developed a planetary model of the hydrogen atom with the electron circling the nucleus like a planet\u00a0circling the sun<\/li>\n<li>Bohr\u2019s equation predicted the energy of an electron in the <em>n<\/em><sup>th<\/sup> energy level of a hydrogen atom.<\/li>\n<li>Erwin Schr\u00f6dinger used de Broglie\u2019s work to develop a theory proposing that the electron in an atom acted as a\u00a0wave.<\/li>\n<li>The Pauli exclusion principle specifies that each electron in an atom must have a unique set of\u00a0four quantum numbers.<\/li>\n<li>The aufbau (\u201cbuilding up\u201d) rule assumes that as electrons are added to an atom they fill the lowest energy\u00a0orbitals first.<\/li>\n<\/ul>\n<section>\n<h3>Nuclear Forces, Fusion, Fission, and Nuclear Stability<\/h3>\n<p>All atoms are composed of three subatomic particles, protons, neutrons, and electrons, the masses\u00a0and charges of which are given in Table 1. The protons and neutrons, called <abbr title=\" The protons and neutrons that make up the nucleus\">nucleons<\/abbr>, are located in the\u00a0nucleus, at the center of the atom. The electrons occupy the region around the nucleus. A typical atom has a radius\u00a0of about 1\u00d710<sup>\u221210<\/sup> m, but the radius of a nucleus is of the order of 1\u00d710<sup>15<\/sup> m. This means that\u00a0while the protons and neutrons in the nucleus make up more than 99.9 percent of the mass of a typical atom, they\u00a0occupy only 1 part in 10<sup>\u2212<\/sup><sup>15<\/sup> of its volume!<\/p>\n<table class=\"gas_law_table\" width=\"509\" cellspacing=\"0\" cellpadding=\"0\" align=\"center\">\n<thead>\n<tr>\n<td valign=\"top\" width=\"109\"><strong>Particle<\/strong><\/td>\n<td valign=\"top\" width=\"118\">\n<p class=\"center\"><strong>Symbol<\/strong><\/p>\n<\/td>\n<td valign=\"top\" width=\"123\">\n<p class=\"center\"><strong>Charge* <\/strong><\/p>\n<\/td>\n<td valign=\"top\" width=\"157\">\n<p class=\"center\"><strong>Mass (kg)<\/strong><\/p>\n<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td valign=\"top\" width=\"109\">Electron<\/td>\n<td valign=\"top\" width=\"118\">\n<p class=\"center\"><em>e<sup>\u2013<\/sup><\/em><\/p>\n<\/td>\n<td valign=\"top\" width=\"123\">\n<p class=\"center\">1\u2013<\/p>\n<\/td>\n<td valign=\"top\" width=\"157\">\n<p class=\"center\">9.10939 \u00d710<sup>\u201331<\/sup><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"109\">Proton<\/td>\n<td valign=\"top\" width=\"118\">\n<p class=\"center\"><em>p <\/em><\/p>\n<\/td>\n<td valign=\"top\" width=\"123\">\n<p class=\"center\">1+<\/p>\n<\/td>\n<td valign=\"top\" width=\"157\">\n<p class=\"center\">1.672623 \u00d7 10<sup>\u201327<\/sup><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"109\">Neutron<\/td>\n<td valign=\"top\" width=\"118\">\n<p class=\"center\"><em>n <\/em><\/p>\n<\/td>\n<td valign=\"top\" width=\"123\">\n<p class=\"center\">0<\/p>\n<\/td>\n<td valign=\"top\" width=\"157\">\n<p class=\"center\">1.674929 \u00d7 10<sup>\u201327<\/sup><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td class=\"figcaption\" colspan=\"4\" valign=\"top\">*The charge is given relative to the magnitude of the charge on<br \/>\nan electron, 1.602\u00d710<sup>\u201319<\/sup> C (Coulomb, the SI unit for charge).<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Objects with opposite electrical charges attract each other; objects with like charges repel each other. Coulomb\u2019s\u00a0electrical force law, <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_003.gif\" width=\"73\" height=\"38\" \/>, where <em>q<\/em><sub>1<\/sub> and <em>q<\/em><sub>2<\/sub> are the charges and <em>r<\/em> the distance between the charges,\u00a0describes the magnitude of the electric force between two point-like charges, and predicts that the repulsive force\u00a0between positively charged protons increases as the distance between them decreases. The very tiny dimensions of the\u00a0nucleus means that the positively charged protons in the nucleus are exceedingly close to one another, and so the\u00a0electrical force pushing them apart must be extremely large.<\/p>\n<p>An attractive force, the <abbr title=\"The force that exists between (or among) nucleons that holds the nucleus together\">strong nuclear\u00a0force<\/abbr>, is responsible for holding the nucleus together. While this force is sufficiently strong to\u00a0hold protons and neutrons together in the nucleus, it is a short-range force and has essentially zero magnitude at\u00a0separations greater than the dimensions of the nucleus. Neutrons, which have no electrical charge, contribute to\u00a0nuclear binding, and as the number of protons in the nucleus increases, the number of neutrons also increases. For\u00a0the first 20 elements, the neutron-proton ratio is very close to 1:1; for elements with more than 20 protons, the\u00a0ratio is greater than 1:1 and increases with increasing atomic number.<\/p>\n<p>Bringing protons and neutrons together to form a nucleus always releases energy\u2014the nucleus has a lower energy than\u00a0the uncombined protons and neutrons\u2014and the nucleus formed always has less mass than that of the protons and\u00a0neutrons that make it up. During such a reaction the lost mass, or <abbr title=\"The difference between the mass of a nucleus and the sum of the masses of the nucleons that make up the nucleus\">mass defect<\/abbr>, is converted into energy: the <abbr title=\"The energy released when a nucleus is formed from protons and neutrons\">nuclear binding energy<\/abbr>.\u00a0The nuclear binding energy can be calculated using Einstein\u2019s mass-energy relationship, \u2206<em>E <\/em>=\u00a0\u2206<em>mc<\/em><sup>2<\/sup>, where \u2206<em>E<\/em> is the binding energy, \u2206<em>m<\/em> is the mass defect, and <em>c<\/em> is\u00a0the speed of light.<\/p>\n<p>If we divide the binding energy by the number of nucleons in the nucleus and plot this ratio vs. the atomic number\u00a0(see graph below), we find that the most stable nucleus is the one with the largest amount of energy released per\u00a0nucleon, Fe\u201356. (Since formation of nuclei from nucleons is exothermic, binding energies are negative.)<\/p>\n<p>Nuclei with atomic numbers smaller than 26 (the atomic number of Fe\u201356) can undergo <abbr title=\"The process in which nuclei combine to form heavier nuclei\">fusion<\/abbr> to form heavier, more\u00a0stable nuclei with a release of energy. Nuclei with atomic numbers larger than 26 can undergo <abbr title=\"The process in which nuclei split apart to form lighter nuclei\">fission<\/abbr> to form\u00a0lighter, more stable nuclei, also with a release of energy.<\/p>\n<p><center><img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/nuclearchemistry3.AvgBindingEnergyPerNucleon1.gif\" alt=\"Ratio of binding energy to the number of nucleons in the nucleus vs. the atomic number\" \/><\/center><\/p>\n<p class=\"figcaption\">Ratio of binding energy to the number of nucleons in the nucleus vs. the atomic number<\/p>\n<p>For example, when a He\u20134 nucleus fuses with a C\u201312 nucleus to form O\u201316, 69.1\u00d710<sup>7<\/sup> kJ\/mol of energy is\u00a0released, and when a U\u2013238 nucleus splits into a Th\u2013234 nucleus and an alpha particle, 83.5\u00d710<sup>7<\/sup> kJ\/mol of\u00a0energy is released.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Is fission of Cs\u201355 into Fe\u201326 and Cu\u201329 energetically favorable?<\/p>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The correct answer is yes. Both Fe\u201326 and Cu\u201329 are more stable than\u00a0Cs\u201355, so energy would be\u00a0released.<\/p>\n<\/div>\n<\/section>\n<h3>Nuclear Synthesis<\/h3>\n<p>For nuclei to form from protons and lighter nuclei, reactant particles must collide with sufficient energy to\u00a0overcome repulsive electrical forces and approach each other closely enough for the strong nuclear force to bind\u00a0them together. At extremely high temperatures and pressures, collisions become energetic and frequent enough for\u00a0fusion to occur. Such conditions existed for a few minutes after the Big Bang, during which nuclei as heavy as\u00a0helium were formed, and they exist in stars and supernovas, in which heavier elements are formed. The very heaviest\u00a0elements in the Universe, such as actinium, thorium, and uranium are created in explosions of supernovas.<\/p>\n<h3>Radioactivity\u2014Alpha, Beta, and Gamma Decay<\/h3>\n<p>As of 2004 there were nearly 3,000 known <abbr title=\" A particular nuclear species, such as 14N or 15O\">nuclides<\/abbr>.\u00a0Most are man-made and undergo\u00a0<abbr title=\"Atoms with the same atomic number Z, but different masses, such as or , are called isotopes of each other.\">radioactive decay<\/abbr>. Of the 287 nuclides found in nature, 266 are stable; the remaining 21 are\u00a0radioactive.\u00a0There are a variety of radioactive decay pathways by which unstable, radioactive parent nuclei are transformed into\u00a0one or more daughter nuclei. These decay processes involve emission of subatomic particles and\/or radiation. While\u00a0there are a variety of subatomic particles involved in nuclear reactions, for our purposes we need only consider\u00a0protons, neutrons, electrons, and positrons. (Positrons are the antimatter counterpart of electrons, with the same\u00a0mass as electrons but a positive charge.)<\/p>\n<p>Decay of unstable nuclei is a random process\u2014the moment at which any particular nuclei in a sample will decay is\u00a0unpredictable and the number of radioactive nuclei in a sample diminishes exponentially with time. Each radioactive\u00a0species has a characteristic half-life \u2014 the time period during which the number of radioactive nuclei decreases by\u00a0half. Since the generally accepted age of the earth is 4.6 billion years, the only naturally occurring radioactive\u00a0<abbr title=\"Atoms of the same element that have the same atomic number but different atomic masses due to a different number of neutrons\">isotopes<\/abbr>,\u00a0or radioisotopes, will be those in one of the following three classes:<\/p>\n<ul>\n<li>Nuclides that have half-lives of at least 10<sup>9<\/sup> years, such as U\u2013238, or K\u201340, so detectable amounts\u00a0remain<\/li>\n<li>Nuclides that are products of decay of long-lived radioactive isotopes, such as Th\u2013234 produced by <em>\u03b1\u00a0<\/em>(alpha) decay of U\u2013238, or Ar\u201340 from \u03b2 (beta) decay of K\u201340<\/li>\n<li>Nuclides such as C\u201314, continually formed in the atmosphere by bombardment of N\u201314 by cosmic neutrons, <img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_005.gif\" width=\"144\" height=\"25\" \/><\/li>\n<\/ul>\n<h3>Alpha Decay<\/h3>\n<p>Alpha decay, the emission of a helium nucleus (or <em>\u03b1<\/em> particle) is a major decay path for elements with atomic\u00a0numbers 83 or higher, and a minor decay pathway for lighter nuclei. All nuclei with atomic numbers greater than 83\u00a0are unstable with respect to alpha decay; the mass of the parent nucleus is greater than the combined mass of the\u00a0daughter nucleus and alpha particle. A noteworthy example of alpha decay is the conversion of polonium\u2013210, to\u00a0lead\u2013206.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_010.gif\" width=\"42\" height=\"25\" align=\"absmiddle\" \/> \u2192\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_012.gif\" width=\"42\" height=\"25\" align=\"absmiddle\" \/> +\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_013.gif\" width=\"33\" height=\"25\" align=\"absmiddle\" \/><br \/>\n<em>\u03b1<\/em> decay<\/p>\n<p>Alpha particles can travel only a few centimeters in air and are stopped by a layer of tissue a few\u00a0cells thick. However, because of the large mass and charge of helium nuclei, alpha radiation causes extensive\u00a0ionization in living tissue and is the most damaging form of radioactivity, so if alpha sources are ingested,\u00a0radiation damage can be severe.<\/p>\n<h3>Beta Decay<\/h3>\n<p>Electron emission (or<em> \u03b2<\/em><sup>\u2013<\/sup> decay) in which a neutron inside the nucleus changes into a proton with\u00a0emission of an electron is the mode of beta decay most commonly discussed in introductory textbooks. There are two\u00a0other radioactive decay processes classified as beta decay, <em>\u03b2<\/em><sup>+<\/sup> decay, in which a proton inside a\u00a0nucleus changes into a neutron with the emission of a positron; and electron capture, in which a proton in the\u00a0nucleus changes into a neutron by capturing an inner shell atomic electron, usually a 1<em>s<\/em> electron. For each\u00a0of these nuclear processes, the mass number <em>A<\/em> remains constant, while <em>Z<\/em> changes by \u00b11.<\/p>\n<p>Beta decay is the principal decay path of radioisotopes, especially those with atomic numbers less than 83. An\u00a0example of <em>\u03b2<\/em><sup>\u2013<\/sup> decay is the decay of a C\u201314 nucleus into an N\u201314 nucleus, the basis of carbon\u201314\u00a0dating.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_014.gif\" width=\"28\" height=\"25\" align=\"absmiddle\" \/> \u2192\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_016.gif\" width=\"28\" height=\"25\" align=\"absmiddle\" \/> +\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_017.gif\" width=\"25\" height=\"25\" align=\"absmiddle\" \/><br \/>\n<em>\u03b2 <\/em>decay<\/p>\n<p>Beta particles can penetrate through the skin to depths of a few millimeters and damage living cells in their\u00a0path.<\/p>\n<h3>Gamma Decay<\/h3>\n<p>In \u03b3 (gamma) decay, a nucleus in an excited or metastable state decays to a lower energy state by emission of a high\u00a0energy photon, or gamma ray. Gamma ray emission usually follows alpha or beta decay, as a product daughter nucleus\u00a0decays to a lower energy state by emission of one or more \u03b3 rays. An example of a nuclear process involving gamma\u00a0rays is the decay of Co\u201360 to Ni\u201360. The first step in this reaction is the beta decay of Co\u201360 to a metastable,\u00a0excited Ni\u201360 nucleus.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_019.gif\" width=\"38\" height=\"25\" align=\"absmiddle\" \/> \u2192\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_021.gif\" width=\"41\" height=\"25\" align=\"absmiddle\" \/>+\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_017.gif\" width=\"25\" height=\"25\" align=\"absmiddle\" \/><\/p>\n<p>This is followed by emission of two gamma rays in succession.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_021.gif\" width=\"41\" height=\"25\" align=\"absmiddle\" \/> \u2192\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_025.gif\" width=\"32\" height=\"25\" align=\"absmiddle\" \/> + \u03b3<sub>1<\/sub> + \u03b3<sub>2<\/sub><br \/>\n\u03b3 decay<\/p>\n<p>Gamma radiation can pass completely through the body, leaving a trail of damaged cells in its wake. Focused gamma\u00a0radiation is used in radiation therapy to destroy cancerous tumors.<\/p>\n<h3>Nuclear Fission<\/h3>\n<p>There are two modes of radioactive decay: spontaneous fission or natural radioactivity, and induced fission or\u00a0artificial radioactivity. Radioactive isotopes, such as Th\u2013233, decay into lighter nuclides spontaneously.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_026.gif\" width=\"41\" height=\"25\" align=\"absmiddle\" \/> \u2192\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_028.gif\" width=\"42\" height=\"25\" align=\"absmiddle\" \/> +\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_029.gif\" width=\"25\" height=\"25\" align=\"absmiddle\" \/><\/p>\n<p>Induced fission, or artificial radioactivity, was first observed by Irene Curie and her husband, Pierre Joliot, in\u00a01934. When they bombarded aluminum foil with <em>\u03b1<\/em> particles produced by the decay of polonium, they discovered\u00a0that a radioactive isotope of phosphorus was formed:<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_030.gif\" width=\"160\" height=\"25\" \/><\/p>\n<p>Radioactive P\u201330 then decays by positron emission:<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_035.gif\" width=\"114\" height=\"25\" \/><\/p>\n<p>Today there are more than 2,000 artificial radionuclides that have been synthesized.<\/p>\n<h3>Balancing Nuclear Reactions<\/h3>\n<p>Nuclear reactions produce both particles and energy. In a nuclear reaction, charge, mass number (<em>A<\/em>), and\u00a0atomic number (<em>Z<\/em>) are all conserved.<\/p>\n<p>Reactants and products may include:<\/p>\n<ul>\n<li>alpha particles (<em>\u03b1<\/em>), which are fast moving helium nuclei and may be written as\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_039.gif\" width=\"33\" height=\"25\" align=\"absmiddle\" \/><\/li>\n<li>protons, <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_040.gif\" width=\"21\" height=\"25\" align=\"absmiddle\" \/><\/li>\n<li>beta particles <em>(\u03b2<\/em><sup>\u2013)<\/sup>,which are fast moving electrons, and may be written <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_041.gif\" width=\"22\" height=\"25\" align=\"absmiddle\" \/><\/li>\n<li>positrons (<em>\u03b2<\/em> +), which have the same mass as electrons but a positive charge, and may be written\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_042.gif\" width=\"24\" height=\"25\" align=\"absmiddle\" \/>, and<\/li>\n<li>neutrons, <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_043.gif\" width=\"22\" height=\"25\" align=\"absmiddle\" \/><\/li>\n<\/ul>\n<p>There are two conventions used in writing nuclear reactions you should keep in mind. The mass number of electrons and\u00a0positrons is zero, and the charge on a nucleus is usually not included. Two worked examples of balancing nuclear\u00a0reactions are given below, the first for a fission reaction, and the second for a fusion reaction.<\/p>\n<h3>Balancing a Fission Reaction<\/h3>\n<p>A Fe\u201360 nucleus can decay to form two beta particles and a new nucleus:\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_044.gif\" width=\"126\" height=\"25\" \/><\/p>\n<p>What is the identity of this new nucleus <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_048.gif\" width=\"26\" height=\"25\" align=\"absmiddle\" \/>?<\/p>\n<p>Conservation of mass numbers means the sum of the mass numbers of reactants must equal the sum of the mass numbers\u00a0of the products. Thus, for this reaction the mass number of <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_049.gif\" width=\"37\" height=\"25\" align=\"absmiddle\" \/> is equal to the mass number of\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_048.gif\" width=\"26\" height=\"25\" align=\"absmiddle\" \/> plus that\u00a0of 2 electrons, so<\/p>\n<p class=\"center\">60 = <em>A<sub>X<\/sub><\/em>+2(0)<br \/>\n<em>A<sub>X<\/sub><\/em> = 60 \u2013 0 = 60<\/p>\n<p>Conservation of atomic numbers means the sum of the atomic numbers of reactants must equal the sum of the atomic\u00a0numbers of the products, so<\/p>\n<p class=\"center\">26 = 2(\u20131) + <em>Z<sub>X<\/sub><sub><br \/>\n<\/sub><\/em><em>Z<sub>X<\/sub><\/em> = 26 \u2013 (\u2013 2) = 28<\/p>\n<p align=\"left\">Thus, <em>X<\/em> is a nickel nucleus. Our balanced equation is:\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_053.gif\" width=\"133\" height=\"25\" \/><\/p>\n<p>Check the answer. The sum of mass numbers is 60 (60 = 0 + 60), and the sum of atomic numbers is 26 (26 = 28 \u2013 2) for\u00a0both products and reactants.<\/p>\n<h3>Balancing a Fusion Reaction<\/h3>\n<p>In stars like our Sun, protons can combine to form helium nuclei. Balance the following skeleton reaction for this\u00a0process.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_057.gif\" width=\"110\" height=\"25\" \/><\/p>\n<p>In this unbalanced equation the sum of reactant mass numbers is 1 and that of products is 4. We need 4 protons to\u00a0balance mass numbers, so our equation becomes<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_058.gif\" width=\"121\" height=\"25\" \/><\/p>\n<p>At this point the sum of reactant atomic numbers is 4 and that of products is 3, so we need 1 more positron to\u00a0balance our equation.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_059.gif\" width=\"130\" height=\"25\" \/><\/p>\n<p>Check the answer. The sum of mass numbers is 4, (4 = 4 + 0), and the sum of atomic numbers is 4,\u00a0(4<strong>\u00b7<\/strong>1 = 2 + 2<strong>\u00b7<\/strong>1) for both products and reactants.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Balance the following nuclear reactions and identify each as either a fusion or a fission reaction.<\/p>\n<ol>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_060.gif\" width=\"166\" height=\"25\" align=\"absmiddle\" \/><\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_062.gif\" width=\"177\" height=\"25\" align=\"absmiddle\" \/><\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>a) <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_061.gif\" width=\"170\" height=\"25\" align=\"absmiddle\" \/>\u00a0(fusion)\u00a0Balancing mass numbers of reactants and products gives us\u00a0<em>A<\/em> + 3 = 4 + 2<strong>\u00b7<\/strong>1 = 6, so <em>A<\/em>\u00a0= 3<\/p>\n<p>Balancing atomic numbers of reactants and products gives us\u00a0<em>Z<\/em> + 2 = 2 + 2<strong>\u00b7<\/strong>1 = 4, so <em>Z<\/em> = 2<\/p>\n<p>b) <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_063.gif\" width=\"185\" height=\"25\" align=\"absmiddle\" \/>\u00a0(fission)\u00a0Balancing mass numbers of reactants and products gives us\u00a0235 = 139 + <em>A <\/em>+ 2<strong>\u00b7<\/strong>1, so <em>A<\/em>\u00a0= 94<\/p>\n<p>Balancing atomic numbers of reactants and products gives us\u00a092 = 56 + <em>Z<\/em> + 2 <strong>\u00b7<\/strong> 0, so <em>Z<\/em> = 36<\/p>\n<\/div>\n<\/section>\n<h3>Energy and mass \u2013 comparing nuclear and chemical reactions<\/h3>\n<p>It is a general result of Einstein\u2019s theory of special relativity that the mass of a system of particles held\u00a0together by attractive forces is less than that of the separated particles. This is true for chemical as well as\u00a0nuclear reactions, and means that in exothermic reactions of any kind the mass of the products is less that that of\u00a0the reactants. Conversely, in endothermic reactions the mass of products is greater than that of reactants.<\/p>\n<p>Can we measure these changes in mass? We can compare the predicted mass change for a <em>hypothetical<\/em> reaction\u00a0of hydrogen atoms to form helium atoms (in an actual nuclear reaction, reactants are nuclei, not atoms) with the\u00a0predicted mass change for an actual chemical reaction, hydrogen atoms combine to form hydrogen molecules.<\/p>\n<p>The balanced equation for the formation of helium atoms from hydrogen atoms is\u00a04\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_090.gif\" width=\"21\" height=\"25\" align=\"absmiddle\" \/>\u00a0(4 protons + 4 electrons) \u2192\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_105.gif\" width=\"33\" height=\"25\" align=\"absmiddle\" \/>(2 protons +\u00a02 electrons + 2 neutrons)<\/p>\n<p>In this reaction two of the four protons combine with two of the four electrons to form two neutrons; these combine\u00a0with the two remaining protons to form the helium nucleus. A helium atom, with two protons, two neutrons, and two\u00a0electrons, has a smaller mass than the four hydrogen atoms from which it is formed, and a hydrogen molecule has a\u00a0smaller mass than the two hydrogen atoms from which it is formed. Let\u2019s compare the mass loss when 1 mole of\u00a0hydrogen atoms is converted to helium by nuclear fusion to that when 1 mole of hydrogen atoms is converted to\u00a0hydrogen molecules.<\/p>\n<p>When 1 mole of hydrogen atoms is converted to 1 mole of helium, about 6.2\u00d710<sup>11<\/sup> J of energy is released.\u00a0When 1\u00a0mole of hydrogen atoms is converted to hydrogen molecules, about 2.2\u00d710<sup>5<\/sup> J of energy is\u00a0released. We can use Einstein\u2019s mass-energy relationship to calculate the mass changes for each of these reactions.<\/p>\n<p>Mass loss in fusion of 1 mole of hydrogen atoms to form 1 mole of helium:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_106.gif\" width=\"412\" height=\"46\" \/><\/center>Mass loss in reaction of 1 mole of hydrogen atoms to form hydrogen molecules:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_107.gif\" width=\"417\" height=\"46\" \/><\/center>\u2206<em>m<\/em> and \u2206<em>E<\/em> are negative for both of these reactions, so products have less mass and less energy than\u00a0reactants (energy is released to the surroundings). In the fusion reaction the mass change is easily measured with a\u00a0laboratory balance, but the mass change for the chemical reaction is too small to detect. This means that in\u00a0chemical reactions, masses of products and reactants are <em>equal<\/em> to within the precision of possible\u00a0measurements of mass; but with nuclear reactions, product and reactant masses are measurably different.<\/p>\n<h3>Energy and mass &#8211; Calculating the energy released in nuclear reactions<\/h3>\n<p>In the preceding section we used Einstein\u2019s mass-energy relationship to determine mass changes from the amount of\u00a0energy released in a reaction. Given masses of reactants and products, we can do this calculation in reverse: from\u00a0the change in mass that results from a reaction, we can determine the amount of energy released. In the next two\u00a0examples we calculate the amount of energy released when a mole of Fe\u201360 nuclei decays to form a mole of Ni\u201360\u00a0nuclei and two moles of electrons (<em>\u03b2<\/em> particles), and when four moles of hydrogen nuclei (protons) fuse to\u00a0form a mole of helium nuclei and two moles of positrons (<em>\u03b2<\/em><sup>+<\/sup> particles).<\/p>\n<p>The balanced equation for the first reaction, in which two neutrons in the Fe\u201360 nucleus decay to form two protons\u00a0and two electrons, is<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_110.gif\" width=\"132\" height=\"25\" \/><\/center>The mass of one mole of Fe\u201360 nuclei is 0.05993408 kg, that of one mole of Ni\u201360 is 0.05993079 kg, and that of two\u00a0moles of electrons is 0.00000110 kg, so the change in mass for this reaction is \u20130.00000219 kg. The energy produced\u00a0by this reaction can be calculated using Einstein\u2019s relationship between mass and energy.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_114.gif\" width=\"446\" height=\"24\" \/><\/center>For the second reaction,<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_115.gif\" width=\"130\" height=\"25\" \/><\/center>The mass of four moles of H\u20131 nuclei is 0.00402911 kg, that of one mole of He\u20134 nuclei is 0.00401506 kg, and that of\u00a0two moles of positrons is 0.00000110 kg, so the change in mass for this reaction is \u20130.00000276 kg. The energy\u00a0produced by this reaction can be calculated using Einstein\u2019s relationship between mass and energy,<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_117.gif\" width=\"440\" height=\"24\" \/><\/center>In comparison with the energy produced by the conversion of one mole of <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_120.gif\" width=\"36\" height=\"25\" align=\"absmiddle\" \/> to one\u00a0mole of <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_121.gif\" width=\"32\" height=\"25\" align=\"absmiddle\" \/> and\u00a0two moles of electrons, one mole\u00a0of H\u20131 nuclei produces 6.2\u00d710<sup>11<\/sup> J, more than three times as much energy.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Is fission of Cs\u2013119 into Fe\u201356 and Cu\u201363 energetically favorable? To answer this question, calculate the\u00a0change in\u00a0mass that would occur.<\/p>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>One mole of Cs\u2013119 nuclei has a mass of 0.11889232 kg, a mole of Fe\u201356 nuclei has a mass of 0.05592068\u00a0kg, and a\u00a0mole of Cu\u201363 nuclei has a mass of 0.06292370 kg. The mass change is \u20130.00004795 kg. Since mass\u00a0decreases in\u00a0this reaction, it is possible.<\/p>\n<\/div>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Is the reaction <img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_122.gif\" width=\"128\" height=\"25\" align=\"absmiddle\" \/> possible? The mass of one\u00a0mole of Fe\u201360 is 0.05993408 kg, that of one mole of Co\u201360 is 0.05993382 kg, and that of a mole of electrons\u00a0is\u00a00.00000055 kg.<\/p>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The mass change when one mole of Fe\u201360 reacts to form one mole of Co\u201360 and one mole of electrons is\u00a0+0.00000029 kg. Since the mass change is positive, this reaction is not possible.<\/p>\n<\/div>\n<\/section>\n<hr \/>\n<h3>Comparing fission and fusion<\/h3>\n<p>Since both nuclear fusion and nuclear fission reactions release energy, both have potential as energy sources,\u00a0although there are both practical and environmental issues with each.<\/p>\n<ul>\n<li>A gram of hydrogen produces about 6.5\u00d710<sup>11<\/sup> J of energy when converted into helium. A gram of\u00a0U\u2013235\u00a0produces about 7.0\u00d710<sup>10<\/sup> J of energy during spontaneous fission, or about 10% of that produced by\u00a0a\u00a0gram of hydrogen.<\/li>\n<li>Uranium is scarce and expensive to refine, while hydrogen is abundant and cheap to produce.<\/li>\n<li>Fission reactors produce large amounts of high-level, long-lasting radioactive waste while fusion reactors\u00a0would\u00a0not.<\/li>\n<\/ul>\n<p>Therefore, it is easy to understand the attractiveness of fusion as an abundant source of relatively safe energy.\u00a0The\u00a0problem with fusion is that though large amounts of time and money have been devoted to its development, no one\u00a0has\u00a0yet succeeded in achieving a sustained fusion reaction that produces more energy than it uses.<\/p>\n<h3>Conversion of nuclear energy into electrical energy<\/h3>\n<p>One of the products of nuclear reactions is heat. In a nuclear electrical power generating plant this heat is\u00a0used in\u00a0the same way as in oil, coal, or natural gas electrical power generating systems. It generates steam that drives\u00a0turbines connected to electrical generators. Presently, all nuclear power plants are based on energy derived\u00a0from a\u00a0sustained nuclear fission chain reaction, using either U\u2013235 or Pu\u2013239 as fuel.<\/p>\n<p>U\u2013235 and Pu\u2013239 are both what are termed fissile materials. When they absorb a neutron, they split into two\u00a0smaller<br \/>\nnuclei and release two or more neutrons on average. If conditions are such that one or more of these neutrons,\u00a0on\u00a0average, splits another uranium or plutonium nucleus, a sustained chain reaction can be achieved. There are a\u00a0great<br \/>\nnumber of possible fission pathways for each isotope. One example of neutron-induced fission of U\u2013235 is:<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_127.gif\" width=\"281\" height=\"25\" \/><\/p>\n<p>In a nuclear bomb, the nuclear chain reaction is uncontrolled and the result is a nuclear explosion; in a nuclear\u00a0generator, the chain reaction is controlled, and serves as a stable source of power.<\/p>\n<p>More than 99% of naturally occurring uranium is U\u2013238; less than 1% is U\u2013235. This is significant, because U\u2013238\u00a0is\u00a0not a fissile isotope, and cannot be used as a fuel for nuclear reactors. However, when U\u2013238 absorbs a neutron,\u00a0it\u00a0decays into Np\u2013239 which then decays to form Pu\u2013239 by a two-step process:<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_130.gif\" width=\"218\" height=\"25\" \/><\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/chemistry\/wp-content\/uploads\/sites\/3\/2017\/08\/s6_129.gif\" width=\"138\" height=\"25\" \/><\/p>\n<p>Non-fissile U\u2013238 can be converted into fissile Pu\u2013239 in breeder reactors, nuclear reactors that produce more fuel\u00a0than\u00a0they use. While the possibility of using breeder reactors to produce vast amounts of plutonium for use as fuel in\u00a0nuclear power plants has great appeal for some, the fact that plutonium is generally regarded as the most toxic\u00a0substance known and is a key component of nuclear weapons causes many others considerable concern.<\/p>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<p><!-- UPDATE NEXT\/PREVIOUS BELOW --><\/p>\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/quantum-mechanics-part-ii\">\u2b05Previous\u00a0Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/atomic-structure-periodicity-and-matter\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/review-2\">Next Lesson\u27a1<\/a><\/p>\n<\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous\u00a0Lesson\u00a0Workshop Index\u00a0Next Lesson\u27a1 Atomic Structure, Periodicity, and Matter:The Nucleus and Nuclear Reactions Objective In this lesson we will review nuclear structure and stability. We will also review balancing nuclear\u00a0equations. Previously we covered&#8230; Niels Bohr developed a planetary model of the hydrogen atom with the electron circling the nucleus like a planet\u00a0circling the sun Bohr\u2019s [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-72","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/72","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/comments?post=72"}],"version-history":[{"count":19,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/72\/revisions"}],"predecessor-version":[{"id":821,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/72\/revisions\/821"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=72"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}