{"id":13696,"date":"2014-02-13T23:27:50","date_gmt":"2014-02-13T23:27:50","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=13696"},"modified":"2026-03-25T10:52:53","modified_gmt":"2026-03-25T15:52:53","slug":"reading-nuclear-equations","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/reading-nuclear-equations\/","title":{"rendered":"Reading Nuclear Equations"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_3CgoACBOAd4\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_3CgoACBOAd4\" data-source-videoID=\"3CgoACBOAd4\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Reading Nuclear Equations Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Reading Nuclear Equations\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_3CgoACBOAd4:hover {cursor:pointer;} img#videoThumbnailImage_3CgoACBOAd4 {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/1502-thumb-final-2.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_3CgoACBOAd4\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_3CgoACBOAd4\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Reading Nuclear Equations\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_3CgoACBOAd4\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_3CgoACBOAd4\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_3CgoACBOAd4\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction YwE_Function() {\n  var x = document.getElementById(\"YwE\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"YwE_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"YwE\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Nuclear_Reactions_vs_Chemical_Reactions\" class=\"smooth-scroll\">Nuclear Reactions vs. Chemical Reactions<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Balancing_Nuclear_Equations\" class=\"smooth-scroll\">Balancing Nuclear Equations<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Condensed_Notation_for_Nuclear_Transmutation_Reactions\" class=\"smooth-scroll\">Condensed Notation for Nuclear Transmutation Reactions<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Common_Nuclear_Emissions_and_Abbreviations\" class=\"smooth-scroll\">Common Nuclear Emissions and Abbreviations<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Alpha_Decay_and_Alpha_Bombardment\" class=\"smooth-scroll\">Alpha Decay and Alpha Bombardment<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Beta_Particles_and_Positrons\" class=\"smooth-scroll\">Beta Particles and Positrons<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Review\" class=\"smooth-scroll\">Review<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this video on nuclear reactions!<\/p>\n<p>Today, we\u2019re going to delve into the notation used to represent them, and as we do that, we\u2019ll review their definition, how they usually take place, and the common species involved.<\/p>\n<h2><span id=\"Nuclear_Reactions_vs_Chemical_Reactions\" class=\"m-toc-anchor\"><\/span>Nuclear Reactions vs. Chemical Reactions<\/h2>\n<p>\nLet\u2019s begin with a quick refresher on nuclear reactions and how they differ from the chemical reactions that you\u2019re probably more familiar with.  <\/p>\n<p>In chemical reactions, electrons are redistributed between chemical species, breaking and forming new bonds, while the nuclei remain unchanged. This means that the elemental identity of the reactants doesn\u2019t change in the course of a chemical reaction. So for example, if we start with 4 atoms of hydrogen and 2 atoms of oxygen, we\u2019ll end with 4 atoms of hydrogen and 2 atoms of oxygen. <\/p>\n<div class=\"examplesentence\">\\(2\\text{H}_{2} + \\text{O}_{2}\\rightarrow 2\\text{H}_{2}\\text{O}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe atoms might be bonded differently, in new molecules, or have changed ionic states, but because the nucleons (the protons and neutrons) didn\u2019t participate in the reaction, the elemental identity of each atom goes unchanged.<\/p>\n<h3><span id=\"Atomic_Number,_Mass_Number,_and_Isotopes\" class=\"m-toc-anchor\"><\/span>Atomic Number, Mass Number, and Isotopes<\/h3>\n<p>\nRemember that all atoms of an element have the same atomic number, \\(Z\\), which is the number of protons. So if that number changes, the elemental identity of the atom changes as well. However, atoms of the same element can have different mass numbers (A), called isotopes, because the number of neutrons can vary without changing the identity of the atom. <\/p>\n<p>In nuclear reactions, the nucleons <em>are<\/em> involved in the reaction and thus, new elements and isotopes are formed in the process. So to accurately represent a nuclear reaction, we must include specific information about the nucleons in our notation.<\/p>\n<p>For example, consider this nuclear fusion reaction where two heavy isotopes of hydrogen, deuterium (hydrogen-2), and tritium (hydrogen-3), fuse to form helium and a highly energetic neutron.<\/p>\n<div class=\"examplesentence\">\\(\\frac{2}{1}\\text{H}+\\frac{3}{1}\\text{H}\\rightarrow \\frac{4}{2}\\text{He}+\\frac{1}{0}\\text{n}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe have maintained the same standard format used for a chemical equation with the reactants on the left side of the arrow and products on the right side of the arrow. But for each species, along with the atomic symbol, we have now included the mass number and atomic number.<\/p>\n<p>Because the mass number is the total nucleons, we get an indirect count of the neutrons as well, which is important because different isotopes can significantly impact the energetics of the reaction. <\/p>\n<p>Including this information also helps us check the mass and charge balance of our nuclear reaction. Remember that even though new elements are created, nuclear reactions still need to follow basic principles, like the law of mass conservation. <\/p>\n<p>This becomes slightly more complicated than in chemical reactions because appreciable mass-energy conversion takes place. Consequently, like in this reaction, the amount of energy released will often be included with the products to show the balanced equation.<\/p>\n<div class=\"examplesentence\">\\(\\frac{2}{1}\\text{H}+\\frac{3}{1}\\text{H}\\rightarrow \\frac{4}{2}\\text{He}+\\frac{1}{0}\\text{n}+17.5\\text{ MeV}\\)<\/div>\n<p>\n&nbsp;<br \/>\nHowever, more simply, we must conserve the number of nucleons in the reaction. So, even though nucleons are rearranging, we have to account for all of our neutrons and protons, which also ensures that we conserve charge throughout the reaction.  <\/p>\n<h2><span id=\"Balancing_Nuclear_Equations\" class=\"m-toc-anchor\"><\/span>Balancing Nuclear Equations<\/h2>\n<p>\nLet\u2019s consider our initial example. Are the nucleons balanced? <\/p>\n<div class=\"examplesentence\">\\(\\frac{2}{1}\\text{H}+\\frac{3}{1}\\text{H}\\rightarrow \\frac{4}{2}\\text{He}+\\frac{1}{0}\\text{n}\\)<\/div>\n<p>\n&nbsp;<br \/>\nBy adding the atomic numbers on the reactant side, we get \\(1+1=2\\) protons and on the product side, we get \\(2+0=2\\). So we started with 2 protons and ended with 2 protons, which is what we want! <\/p>\n<p>Next, add the mass numbers to confirm that we haven\u2019t lost any neutrons. From the reactant side, \\(2+3=5\\) and from the product side, \\(4+1=5\\). So we started with 5 nucleons and ended with 5 nucleons. Perfect! <\/p>\n<p>This is also helpful for predicting products. For example, if we have an unknown product, we can solve for its identity by balancing the reaction. Pause the video and work out the other product, \\(_{Z}^{A}\\textrm{X}\\), in the following radioactive decay of radium:<\/p>\n<div class=\"examplesentence\">\\(_{88}^{226}\\textrm{Ra}\\rightarrow _{Z}^{A}\\textrm{X}+_{4}^{2}\\textrm{He}\\)<\/div>\n<p>\n&nbsp;<br \/>\nGot it? Let\u2019s check. To solve for , set up the following simple equations.<\/p>\n<p>The sum of the mass numbers of the reactants is equal to the sum of the mass numbers of the products,<\/p>\n<div class=\"examplesentence\">\\(\\sum \\left ( A_{\\text{reactants}} \\right )=\\sum \\left ( A_{\\text{products}} \\right )\\)<br \/>\n\\(226=A+4\\)<\/div>\n<p>\n&nbsp;<br \/>\ntherefore <\/p>\n<div class=\"examplesentence\">\\(A = 222\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe also need to set the sum of the atomic numbers of the reactants equal to the sum of the atomic numbers of the products: <\/p>\n<div class=\"examplesentence\">\\(Z_{\\text{reactants}}=Z_{\\text{products}}\\)<br \/>\n\\(88 = Z + 2\\)<br \/>\n\\(Z = 86\\)<\/div>\n<p>\n&nbsp;<br \/>\nKnowing \\(Z\\) is 86 reveals radium decayed by emitting a helium-4 nuclei and leaving behind a radon atom.  <\/p>\n<div class=\"examplesentence\">\\(_{88}^{226}\\textrm{Ra}\\rightarrow _{A}^{Z}\\textrm{X}+\\frac{4}{2}\\text{He}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn these nuclear decay reactions, you\u2019ll also hear the reactant, in this case radium, called the parent and the larger product called the daughter, which in this case is radon. <\/p>\n<h2><span id=\"Condensed_Notation_for_Nuclear_Transmutation_Reactions\" class=\"m-toc-anchor\"><\/span>Condensed Notation for Nuclear Transmutation Reactions<\/h2>\n<p>\nYou may have noticed that in our notation, some of the information is redundant. For instance, given the atomic symbol and mass number, we can figure out the number of neutrons; consequently, you may see the atomic numbers dropped. <\/p>\n<p>More significantly, you\u2019ll sometimes see the reaction collapsed down into this format: <\/p>\n<div class=\"examplesentence\">\\(T\\left ( P,x \\right )R\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis notation is used for nuclear transmutation reactions, in which scientists induce the elemental conversion of a nucleus. In the experimental setup, there is a target species, \\(T\\), which is bombarded with a projectile, \\(P\\). This produces an emitted particle, \\(x\\), and a residual nucleus, \\(R\\).<\/p>\n<p>Using the notation of a standard chemical reaction, this is the same as saying <\/p>\n<div class=\"examplesentence\">\\(T+P\\rightarrow x+R\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, along with being shorter, this notation provides additional information about the experiment and the energy of the particles. Remember that to execute nuclear reactions, we often have to force positively charged particles together, which requires large amounts of energy, and typically produces lots of energy, so knowing which particles are high-energy is relevant to understanding the reaction.<\/p>\n<p>Let\u2019s consider another example. <\/p>\n<p>Beryllium-9 reacts with helium to produce carbon-12 and a neutron. <\/p>\n<div class=\"examplesentence\">\\(_{9}^{4}\\textrm{Be}+_{2}^{4}\\textrm{He}\\rightarrow _{6}^{12}\\textrm{C}+_{1}^{0}\\textrm{n}\\)<\/div>\n<p>\n&nbsp;<br \/>\nUsing the condensed notation, we write this reaction like this: <\/p>\n<div class=\"examplesentence\">\\(^{9}\\textrm{Be}\\left (^{4}\\textrm{He}, \\text{n} \\right )^{12}\\textrm{C}\\)<\/div>\n<p>\n&nbsp;<br \/>\nwhich now includes the information that beryllium was bombarded with high-energy helium particles to produce carbon and high-energy neutrons. <\/p>\n<h2><span id=\"Common_Nuclear_Emissions_and_Abbreviations\" class=\"m-toc-anchor\"><\/span>Common Nuclear Emissions and Abbreviations<\/h2>\n<p>\nApart from the condensed notation for the reaction, some particles are so common to nuclear reactions, particularly as emissions in radioactive decay, that scientists have given them their own name. Consequently, to quickly read nuclear reactions, it\u2019s nice to be comfortable with these terms as well. <\/p>\n<p>In this notation, the superscript and subscript prior to the emission symbol is the mass and charge, respectively. <\/p>\n<table class=\"ATable\" style=\"margin: auto; width: 100%;\">\n<caption style=\"text-align: center; font-weight: bold\">Common Nuclear Emissions and Abbreviations<\/caption>\n<thead>\n<tr>\n<th><strong>Particle<\/strong><\/th>\n<th><strong>Abbreviation<\/strong><\/th>\n<th><strong>Charge<\/strong><\/th>\n<th><strong>Mass (amu)<\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Alpha particle <\/td>\n<td>\\(_{4}^{2}\\textrm{He}_{4}^{2}\\textrm{He}^{2+}, \\text{ }_{4}^{2}\\textrm{a},_{4}^{2}\\textrm{a}^{2+}, \\text{a}\\)<\/td>\n<td>+2<\/td>\n<td>4.001506<\/td>\n<\/tr>\n<tr>\n<td>Beta Particle<\/td>\n<td>\\(_{-}\\frac{0}{1}\\beta ,\\beta ,\\beta &#8211;\\)<\/td>\n<td>-1<\/td>\n<td>0.000549<\/td>\n<\/tr>\n<tr>\n<td>Positron<\/td>\n<td>\\(_{+}\\frac{0}{1}\\beta ,\\beta+\\)<\/td>\n<td>+1<\/td>\n<td>0.000549<\/td>\n<\/tr>\n<tr>\n<td>Proton<\/td>\n<td>p,\\(\\frac{1}{1}p,\\frac{1}{1}\\text{H}\\)<\/td>\n<td>+1<\/td>\n<td>1.007276<\/td>\n<\/tr>\n<tr>\n<td>Neutron<\/td>\n<td>\\(\\text{n}, \\frac{1}{0}\\)<\/td>\n<td>0<\/td>\n<td>1.0008665<\/td>\n<\/tr>\n<tr>\n<td>Photon, gamma ray<\/td>\n<td>Y, \\(\\text{ }_{0}^{0}\\textrm{Y}\\)<\/td>\n<td>&#8211;<\/td>\n<td>&#8211;<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<\/p>\n<h2><span id=\"Alpha_Decay_and_Alpha_Bombardment\" class=\"m-toc-anchor\"><\/span>Alpha Decay and Alpha Bombardment<\/h2>\n<p>\nAlpha particles are high-energy helium-4 nuclei, two protons, and two neutrons. They have been stripped of their electrons, so they carry a +2 charge. They can be written as \\(_{4}^{2}\\textrm{He}^{2+}\\) , but are often abbreviated to simply \u03b1. <\/p>\n<p>Thus, returning to our previous examples, <\/p>\n<div class=\"examplesentence\">\\(^{9}\\text{Be}(^{4}\\text{He}, \\text{n})^{12}\\text{C and}_{88}^{226}\\textrm{Ra}\\rightarrow _{88}^{226}\\textrm{Rn}+_{2}^{4}\\textrm{He}\\)<\/div>\n<p>\n&nbsp;<br \/>\nwe now know that \\(^{4}_{2}\\text{H}\\) is an alpha particle. So the first reaction would be written like this: <\/p>\n<div class=\"examplesentence\">\\(_{88}^{226}\\textrm{Ra}\\rightarrow _{86}^{222}\\textrm{Rn}+a\\)<\/div>\n<p>\n&nbsp;<br \/>\nand is commonly referred to as the alpha decay of radium.<\/p>\n<p>The second reaction would be written like this: <\/p>\n<div class=\"examplesentence\">\\(^{9}\\text{Be}\\left ( \\text{a}, \\text{n} \\right )^{12}\\text{C}\\)<\/div>\n<p>\n&nbsp;<br \/>\nand read as, beryllium was bombarded with alpha particles to produce carbon and high-energy neutrons.<\/p>\n<h2><span id=\"Beta_Particles_and_Positrons\" class=\"m-toc-anchor\"><\/span>Beta Particles and Positrons<\/h2>\n<p>\nBeta particles are high-energy electrons emitted from the nucleus when a neutron is converted into a proton and denoted as \\(_{-1}^{0}\\beta\\) or \\(\\text{ }\\beta ^{-}\\). <\/p>\n<p>They are physically equivalent to electrons, e-, but are produced in a specific way, so it\u2019s best not to interchange the notation.<\/p>\n<div class=\"examplesentence\">\\(_{0}^{1}\\textrm{n}\\rightarrow _{1}^{1}\\textrm{p}+_{-1}^{0}\\beta\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Positrons are positively charged electrons or the antiparticle to electrons and are denoted as \\(_{+1}^{0}\\beta\\) or \\(\\text{ }\\beta ^{+}\\). <\/p>\n<p>They are formed when a proton in the nucleus is converted to a neutron, which then expels a positron. <\/p>\n<div class=\"examplesentence\">\\(_{1}^{1}\\textrm{p}\\rightarrow _{1}^{0}\\textrm{n}+_{+1}^{0}\\beta\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd gamma rays are high-energy photons with no mass or charge.<\/p>\n<p>Also remember that n is a neutron and p is a proton. <\/p>\n<p>Alright, we\u2019ve covered the basics of writing nuclear reactions, both the notation and the common abbreviations. You should now have the tools to read and write most nuclear reactions, but we\u2019ve really only scratched the surface of the theory, so don\u2019t worry if you still feel a little overwhelmed. There\u2019s a lot to learn here! <\/p>\n<hr>\n<h2><span id=\"Review\" class=\"m-toc-anchor\"><\/span>Review<\/h2>\n<p>\nFinally, let\u2019s wrap up with some practice (if you don\u2019t already, get a periodic table handy!).<\/p>\n<p>Pause the video and write the equations for the following nuclear reactions. For the second question, write the full equation as well as the shorthand notation for transmutation reactions. <\/p>\n<p>1. Naturally occurring thorium-232 undergoes \u03b1 decay.<\/p>\n<div style=\"text-align: center; margin-bottom: 20px;\">\n  <button class=\"buttontranscript\" onClick=\"toggle('Answer1')\">Show Answer<\/button>\n<\/div>\n<div id=\"Answer1\" class=\"showanswer\">\n  <strong><\/strong><\/p>\n<p style=\"text-align: left;\">First, write out what you know- Thorium decays into something plus an alpha particle <\/p>\n<p style=\"text-align: center;\">\\(_{90}^{232}\\textrm{Th}\\rightarrow _{Z}^{A}\\textrm{X}+_{2}^{4}\\textrm{a}\\)<\/p>\n<p>Now, solve for the unknown particle using the following equations<\/p>\n<p style=\"text-align: center;\">\\(232=A+4,A=228\\)<\/p>\n<p style=\"text-align: center;\">\\(90=Z+2,Z=88\\)<\/p>\n<p style=\"text-align: center;\">\\(\\text{ Therefore,}\\text{ X}=_{88}^{228}\\text{Ra}+_{2}^{4}\\alpha\\)<\/p>\n<p style=\"text-align: center;\">\\(_{90}^{232}\\textrm{Th}\\rightarrow _{Z}^{A}\\textrm{X}+_{2}^{4}\\textrm{a}\\)<\/p>\n<\/div>\n<p>\n&nbsp;<br \/>\n2. When aluminum foil was bombarded with \u03b1 particles, phosphorus-30 and neutrons were formed.<\/p>\n<div style=\"text-align: center; margin-bottom: 20px;\">\n  <button class=\"buttontranscript\" onClick=\"toggle('Answer2')\">Show Answer<\/button>\n<\/div>\n<div id=\"Answer2\" class=\"showanswer\">\n  <strong><\/strong><\/p>\n<p style=\"text-align: left;\">Again, write out what you know- aluminum was bombarded with alpha particles to form phosphorous-30 and neutrons.<br \/>\n&nbsp;<\/p>\n<p style=\"text-align: center;\">\\(_{13}^{A}\\textrm{Al}+_{2}^{4}\\textrm{}\\alpha \\to _{15}^{30}\\textrm{P}+_{0}^{1}\\textrm{n}\\)<\/p>\n<p>\n&nbsp;<br \/>\nNote that we don\u2019t know the mass number of aluminum, but can quickly solve for it. <br \/>\n&nbsp;<\/p>\n<p style=\"text-align: center;\">\\(A+4=30+1\\)<br \/>\n&nbsp;<br \/>\n\\(A=27\\)<br \/>\n&nbsp;<br \/>\n\\(_{13}^{27}\\textrm{Al}+_{2}^{4}\\textrm{}\\alpha\\to _{15}^{30}\\textrm{P}+_{0}^{1}\\textrm{n}\\)<\/p>\n<p>\n&nbsp;<br \/>\nAnd the shorthand notation is \\(^{27}Al\\)(\u03b1,n)\\(^{30}P\\)\n<\/p>\n<\/div>\n<p>\n&nbsp;<br \/>\nThat\u2019s it for this review on nuclear reactions! 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