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Alpha Decay

Alpha Decay

In alpha decay, a \({}^{4}\text{He}\) nucleus simply breaks away from the parent nucleus, leaving a daughter with two fewer protons and two fewer neutrons than the parent (see this figure). One example of \(\alpha \) decay is shown in this figure for \({}^{\text{238}}\text{U}\). Another nuclide that undergoes \(\alpha \) decay is \({}^{\text{239}}\text{Pu}\). The decay equations for these two nuclides are

\({}^{\text{238}}\text{U}\to {}^{\text{234}}{\text{Th}}_{\text{92}}^{\text{234}}+{}^{4}\text{He}\)

and

\({}^{\text{239}}\text{Pu}\to {}^{\text{235}}\text{U}+{}^{4}\text{He}.\)

If you examine the periodic table of the elements, you will find that Th has \(Z=\text{90}\), two fewer than U, which has \(Z=\text{92}\). Similarly, in the second decay equation, we see that U has two fewer protons than Pu, which has \(Z=\text{94}\). The general rule for \(\alpha \) decay is best written in the format \({}_{Z}^{A}{\text{X}}_{N}\). If a certain nuclide is known to \(\alpha \) decay (generally this information must be looked up in a table of isotopes, such as in this appendix), its \(\alpha \) decay equation is

\({}_{Z}^{A}{\text{X}}_{N}\to {}_{Z-2}^{A-4}{\text{Y}}_{N-2}+{}_{2}^{4}{\text{He}}_{2}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}(\alpha \phantom{\rule{0.25em}{0ex}}\text{decay})\)

where Y is the nuclide that has two fewer protons than X, such as Th having two fewer than U. So if you were told that \({}^{\text{239}}\text{Pu}\)\(\alpha \) decays and were asked to write the complete decay equation, you would first look up which element has two fewer protons (an atomic number two lower) and find that this is uranium. Then since four nucleons have broken away from the original 239, its atomic mass would be 235.

It is instructive to examine conservation laws related to \(\alpha \) decay. You can see from the equation \({}_{Z}^{A}{\text{X}}_{N}\to {}_{Z-2}^{A-4}{\text{Y}}_{N-2}+{}_{2}^{4}{\text{He}}_{2}\) that total charge is conserved. Linear and angular momentum are conserved, too. Although conserved angular momentum is not of great consequence in this type of decay, conservation of linear momentum has interesting consequences. If the nucleus is at rest when it decays, its momentum is zero.

In that case, the fragments must fly in opposite directions with equal-magnitude momenta so that total momentum remains zero. This results in the \(\alpha \) particle carrying away most of the energy, as a bullet from a heavy rifle carries away most of the energy of the powder burned to shoot it. Total mass–energy is also conserved: the energy produced in the decay comes from conversion of a fraction of the original mass. As discussed in Atomic Physics, the general relationship is

\(E\phantom{\rule{0.15em}{0ex}}=(\Delta m){c}^{2}.\)

Here, \(E\) is the nuclear reaction energy (the reaction can be nuclear decay or any other reaction), and \(\Delta m\) is the difference in mass between initial and final products. When the final products have less total mass, \(\Delta m\) is positive, and the reaction releases energy (is exothermic). When the products have greater total mass, the reaction is endothermic (\(\Delta m\) is negative) and must be induced with an energy input. For \(\alpha \) decay to be spontaneous, the decay products must have smaller mass than the parent.

Example: Alpha Decay Energy Found from Nuclear Masses

Find the energy emitted in the \(\alpha \) decay of \({}^{\text{239}}\text{Pu}\).

Strategy

Nuclear reaction energy, such as released in α decay, can be found using the equation \(E=(\Delta m){c}^{2}\). We must first find \(\Delta m\), the difference in mass between the parent nucleus and the products of the decay. This is easily done using masses given in this appendix.

Solution

The decay equation was given earlier for \({}^{\text{239}}\text{Pu}\) ; it is

\({}^{\text{239}}\text{Pu}\to {}^{\text{235}}\text{U}+{}^{4}\text{He}.\)

Thus the pertinent masses are those of \({}^{\text{239}}\text{Pu}\), \({}^{\text{235}}\text{U}\), and the \(\alpha \) particle or \({}^{4}\text{He}\), all of which are listed in this appendix. The initial mass was \(m{(}^{\text{239}}\text{Pu})=\text{239}\text{.}\text{052157 u}\). The final mass is the sum \(m{(}^{\text{235}}\text{U})\text{+}m{(}^{4}\text{He})\text{= 235}\text{.}\text{043924 u + 4.002602 u = 239.046526 u}\). Thus,

\(\begin{array}{lll}\Delta m& =& m{(}^{\text{239}}\text{Pu})-[m{(}^{\text{235}}\text{U})+m({}^{4}\text{He})]\\ & =& \text{}\text{239.052157 u}-\text{239.046526 u}\\ & =& \text{0.0005631 u.}\end{array}\)

Now we can find \(E\) by entering \(\Delta m\) into the equation:

\(E=(\Delta m){c}^{2}=(0\text{.005631 u}){c}^{2}.\)

We know \(\text{1 u}=\text{931.5 MeV/}{c}^{2}\), and so

\(E=(0\text{.}\text{005631})(\text{931.5 MeV}/{c}^{2})({c}^{2})=\text{5.25 MeV}.\)

Discussion

The energy released in this \(\alpha \) decay is in the \(\text{MeV}\) range, about \({\text{10}}^{6}\) times as great as typical chemical reaction energies, consistent with many previous discussions. Most of this energy becomes kinetic energy of the \(\alpha \) particle (or \({}^{4}\text{He}\) nucleus), which moves away at high speed. The energy carried away by the recoil of the \({}^{\text{235}}\text{U}\) nucleus is much smaller in order to conserve momentum. The \({}^{\text{235}}\text{U}\) nucleus can be left in an excited state to later emit photons (\(\gamma \) rays).

This decay is spontaneous and releases energy, because the products have less mass than the parent nucleus. The question of why the products have less mass will be discussed in Binding Energy. Note that the masses given in this appendix are atomic masses of neutral atoms, including their electrons. The mass of the electrons is the same before and after \(\alpha \) decay, and so their masses subtract out when finding \(\Delta m\). In this case, there are 94 electrons before and after the decay.

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