# FHSST Physics/Atomic Nucleus/Nuclear Reactions

Inside the Atomic Nucleus The Free High School Science Texts: A Textbook for High School Students Studying Physics Main Page - << Previous Chapter (Modern Physics) Composition - Nucleus - Nuclear Force - Binding Energy and Nuclear Masses - Radioactivity - Nuclear Reactions - Detectors - Nuclear Energy - Nuclear Reactors - Nuclear Fusion - Origin of the Universe Elementary Particles: Beta Decay - Particle Physics - Quarks and Leptons - Forces of Nature

# Nuclear Reactions

Those of you who studied chemistry, are familiar with the notion of chemical reaction, which, in essence, is just regrouping of atoms that constitute molecules. As a result, reagent chemical compounds are transformed into product compounds.

In the world of nuclear particles, similar processes are possible. When nuclei are close to each other, nucleons from one nucleus can jump into another one. This happens because there are attractive and repulsive forces between the nucleons. The complicated interplay of these forces may cause their regrouping. As a result, the reagent particles are transformed into product particles. Such processes are called nuclear reactions.

For example, when two isotopes ${\displaystyle {}_{2}^{3}}$He collide, the six nucleons constituting them, can rearrange in such a way that the isotope ${\displaystyle {}_{2}^{4}}$He is formed and two protons are liberated. Similarly to chemical reactions, this process is denoted as

 ${\displaystyle {\begin{matrix}{}_{2}^{3}{\rm {He}}+{}_{2}^{3}{\rm {He}}\longrightarrow {}_{2}^{4}{\rm {He}}+p+p+12.86\,{\rm {MeV}}\ \end{matrix}}}$ (15.3)

The same as in chemical reactions, nuclear reactions can also be either exothermic (i.e. releasing energy) or endothermic (i.e. requiring an energy input). The above reaction releases 12.86MeV of energy. This is because the total mass on the left hand side of Eq. (15.3) is in 12.86MeV greater than the total mass of the products on the right hand side (you can check this using Table 15.1).

Thus, when considering a particular nuclear reaction, we can always learn if it releases or absorbs energy. For this, we only need to compare total masses on the left and right hand sides of the equation. Now, you can understand why it is very convenient to express masses in the units of energy.

Composing equations like (15.3), we should always check the superscripts and subscripts of the nuclei in order to have the same number of nucleons and the same charge on both sides of the equation. In the above example, we have six nucleons and the charge +4 in both the initial and final states of the reaction. To make the checking of nucleon number and charge conservation easier, sometimes the proton and neutron are denoted with superscripts and subscripts as well, namely, ${\displaystyle {}_{1}^{1}p}$and ${\displaystyle {}_{0}^{1}n}$.

In this case, all we need is to check that sum of superscripts and sum of subscripts are the same on both sides of the equation.