Physics » Radioactivity and Nuclear Physics » Nuclear Decay and Conservation Laws

# Beta Decay

## Beta Decay

There are actually three types of beta decay. The first discovered was “ordinary” beta decay and is called $${\beta }^{-}$$ decay or electron emission. The symbol $${\beta }^{-}$$ represents an electron emitted in nuclear beta decay. Cobalt-60 is a nuclide that $${\beta }^{-}$$ decays in the following manner:

$${}^{\text{60}}\text{Co}\to {}^{\text{60}}\text{Ni}+{\beta }^{-}+\text{neutrino.}$$

The neutrino is a particle emitted in beta decay that was unanticipated and is of fundamental importance. The neutrino was not even proposed in theory until more than 20 years after beta decay was known to involve electron emissions. Neutrinos are so difficult to detect that the first direct evidence of them was not obtained until 1953. Neutrinos are nearly massless, have no charge, and do not interact with nucleons via the strong nuclear force. Traveling approximately at the speed of light, they have little time to affect any nucleus they encounter. This is, owing to the fact that they have no charge (and they are not EM waves), they do not interact through the EM force. They do interact via the relatively weak and very short range weak nuclear force.

Consequently, neutrinos escape almost any detector and penetrate almost any shielding. However, neutrinos do carry energy, angular momentum (they are fermions with half-integral spin), and linear momentum away from a beta decay. When accurate measurements of beta decay were made, it became apparent that energy, angular momentum, and linear momentum were not accounted for by the daughter nucleus and electron alone. Either a previously unsuspected particle was carrying them away, or three conservation laws were being violated. Wolfgang Pauli made a formal proposal for the existence of neutrinos in 1930. The Italian-born American physicist Enrico Fermi (1901–1954) gave neutrinos their name, meaning little neutral ones, when he developed a sophisticated theory of beta decay (see this figure). Part of Fermi’s theory was the identification of the weak nuclear force as being distinct from the strong nuclear force and in fact responsible for beta decay.

The neutrino also reveals a new conservation law. There are various families of particles, one of which is the electron family. We propose that the number of members of the electron family is constant in any process or any closed system. In our example of beta decay, there are no members of the electron family present before the decay, but after, there is an electron and a neutrino. So electrons are given an electron family number of $$+1$$. The neutrino in $${\beta }^{-}$$ decay is an electron’s antineutrino, given the symbol $${\overline{\nu }}_{e}$$, where $$\nu$$ is the Greek letter nu, and the subscript e means this neutrino is related to the electron.

The bar indicates this is a particle of antimatter. (All particles have antimatter counterparts that are nearly identical except that they have the opposite charge. Antimatter is almost entirely absent on Earth, but it is found in nuclear decay and other nuclear and particle reactions as well as in outer space.) The electron’s antineutrino $${\overline{\nu }}_{e}$$, being antimatter, has an electron family number of $$–1$$. The total is zero, before and after the decay. The new conservation law, obeyed in all circumstances, states that the total electron family number is constant. An electron cannot be created without also creating an antimatter family member. This law is analogous to the conservation of charge in a situation where total charge is originally zero, and equal amounts of positive and negative charge must be created in a reaction to keep the total zero.

If a nuclide $${}_{Z}^{A}{X}_{N}$$ is known to $${\beta }^{-}$$ decay, then its $${\beta }^{-}$$ decay equation is

$${}_{Z}{}^{A}\text{}{\text{X}}_{N}\to {}_{Z+1}{}^{A}\text{}{\text{Y}}_{N-1}+{\beta }^{-}+{\stackrel{-}{\nu }}_{e}\phantom{\rule{0.25em}{0ex}}({\beta }^{-}\phantom{\rule{0.25em}{0ex}}\text{decay}),$$

where Y is the nuclide having one more proton than X (see this figure). So if you know that a certain nuclide $${\beta }^{-}$$ decays, you can find the daughter nucleus by first looking up $$Z$$ for the parent and then determining which element has atomic number $$Z+1$$. In the example of the $${\beta }^{-}$$ decay of $${}^{\text{60}}\text{Co}$$ given earlier, we see that $$Z=\text{27}$$ for Co and $$Z=\text{28}$$ is Ni. It is as if one of the neutrons in the parent nucleus decays into a proton, electron, and neutrino. In fact, neutrons outside of nuclei do just that—they live only an average of a few minutes and $${\beta }^{-}$$ decay in the following manner:

$$\text{n}\to \text{p}+{\beta }^{-}+{\stackrel{-}{\nu }}_{e}.$$

We see that charge is conserved in $${\beta }^{-}$$ decay, since the total charge is $$Z$$ before and after the decay. For example, in $${}^{\text{60}}\text{Co}$$ decay, total charge is 27 before decay, since cobalt has $$Z=\text{27}$$. After decay, the daughter nucleus is Ni, which has $$Z=\text{28}$$, and there is an electron, so that the total charge is also $$28 + (–1)$$ or 27. Angular momentum is conserved, but not obviously (you have to examine the spins and angular momenta of the final products in detail to verify this). Linear momentum is also conserved, again imparting most of the decay energy to the electron and the antineutrino, since they are of low and zero mass, respectively.

Another new conservation law is obeyed here and elsewhere in nature. The total number of nucleons $$A$$ is conserved. In $${}^{\text{60}}\text{Co}$$ decay, for example, there are 60 nucleons before and after the decay. Note that total $$A$$ is also conserved in $$\alpha$$ decay. Also note that the total number of protons changes, as does the total number of neutrons, so that total $$Z$$ and total $$N$$ are not conserved in $${\beta }^{-}$$ decay, as they are in $$\alpha$$ decay. Energy released in $${\beta }^{-}$$ decay can be calculated given the masses of the parent and products.

### Example: $${\beta }^{-}$$ Decay Energy from Masses

Find the energy emitted in the $${\beta }^{-}$$ decay of $${}^{\text{60}}\text{Co}$$.

Strategy and Concept

As in the preceding example, we must first find $$\Delta m$$, the difference in mass between the parent nucleus and the products of the decay, using masses given in this appendix. Then the emitted energy is calculated as before, using $$E=(\Delta m){c}^{2}$$. The initial mass is just that of the parent nucleus, and the final mass is that of the daughter nucleus and the electron created in the decay. The neutrino is massless, or nearly so. However, since the masses given in this appendix are for neutral atoms, the daughter nucleus has one more electron than the parent, and so the extra electron mass that corresponds to the $${\beta }^{–}$$ is included in the atomic mass of Ni. Thus,

$$\Delta m=m({}^{\text{60}}\text{Co})-m({}^{\text{60}}\text{Ni}).$$

Solution

The $${\beta }^{-}$$ decay equation for $${}^{\text{60}}\text{Co}$$ is

$${}_{\text{27}}^{\text{60}}{\text{Co}}_{\text{33}}\to {}_{\text{28}}^{\text{60}}{\text{Ni}}_{\text{32}}+{\beta }^{-}+{\overline{\nu }}_{e}.$$

As noticed,

$$\Delta m=m({}^{\text{60}}\text{Co})-m({}^{\text{60}}\text{Ni}).$$

Entering the masses found in this appendix gives

$$\Delta m=\text{59}\text{.}\text{933820 u}-\text{59.930789 u}=\text{0.003031 u}.$$

Thus,

$$E=(\Delta m){c}^{2}=(\text{0.003031 u}){c}^{2}.$$

Using $$1 u=931.5 MeV/{c}^{2}$$, we obtain

$$E=(0\text{.}\text{003031})(931.5 MeV/{c}^{2})({c}^{2})=2\text{.}\text{82 MeV.}$$

Discussion and Implications

Perhaps the most difficult thing about this example is convincing yourself that the $${\beta }^{-}$$ mass is included in the atomic mass of $${}^{\text{60}}\text{Ni}$$. Beyond that are other implications. Again the decay energy is in the MeV range. This energy is shared by all of the products of the decay. In many $${}^{\text{60}}\text{Co}$$ decays, the daughter nucleus $${}^{\text{60}}\text{Ni}$$ is left in an excited state and emits photons ($$\gamma$$ rays). Most of the remaining energy goes to the electron and neutrino, since the recoil kinetic energy of the daughter nucleus is small. One final note: the electron emitted in $${\beta }^{-}$$ decay is created in the nucleus at the time of decay.

The second type of beta decay is less common than the first. It is $${\beta }^{+}$$decay. Certain nuclides decay by the emission of a positive electron. This is antielectron or positron decay (see this figure).

The antielectron is often represented by the symbol $${e}^{+}$$, but in beta decay it is written as $${\beta }^{+}$$ to indicate the antielectron was emitted in a nuclear decay. Antielectrons are the antimatter counterpart to electrons, being nearly identical, having the same mass, spin, and so on, but having a positive charge and an electron family number of $$–1$$. When a positron encounters an electron, there is a mutual annihilation in which all the mass of the antielectron-electron pair is converted into pure photon energy. (The reaction, $${e}^{+}+{e}^{-}\to \gamma +\gamma$$, conserves electron family number as well as all other conserved quantities.) If a nuclide $${}_{Z}^{A}{X}_{N}$$ is known to $${\beta }^{+}$$ decay, then its $${\beta }^{+}$$decay equation is

$${}_{Z}^{A}{\text{X}}_{N}\to {}_{Z-1}{}^{A}\text{}{\text{Y}}_{N+1}+{\beta }^{+}+{\nu }_{e}\phantom{\rule{0.25em}{0ex}}({\beta }^{+}\phantom{\rule{0.25em}{0ex}}\text{decay}),$$

where Y is the nuclide having one less proton than X (to conserve charge) and $${\nu }_{e}$$ is the symbol for the electron’s neutrino, which has an electron family number of $$+1$$. Since an antimatter member of the electron family (the $${\beta }^{+}$$) is created in the decay, a matter member of the family (here the $${\nu }_{e}$$) must also be created. Given, for example, that $${}^{\text{22}}\text{Na}$$$${\beta }^{+}$$ decays, you can write its full decay equation by first finding that $$Z=\text{11}$$ for $${}^{\text{22}}\text{Na}$$, so that the daughter nuclide will have $$Z=\text{10}$$, the atomic number for neon. Thus the $${\beta }^{+}$$ decay equation for $${}^{\text{22}}\text{Na}$$ is

$${}_{\text{11}}^{\text{22}}{\text{Na}}_{\text{11}}\to {}_{\text{10}}^{\text{22}}{\text{Ne}}_{\text{12}}+{\beta }^{+}+{\nu }_{e}.$$

In $${\beta }^{+}$$ decay, it is as if one of the protons in the parent nucleus decays into a neutron, a positron, and a neutrino. Protons do not do this outside of the nucleus, and so the decay is due to the complexities of the nuclear force. Note again that the total number of nucleons is constant in this and any other reaction. To find the energy emitted in $${\beta }^{+}$$ decay, you must again count the number of electrons in the neutral atoms, since atomic masses are used. The daughter has one less electron than the parent, and one electron mass is created in the decay. Thus, in $${\beta }^{+}$$ decay,

$$\Delta m=m(\text{parent})-[m(\text{daughter})+{2m}_{e}],$$

since we use the masses of neutral atoms.

Electron capture is the third type of beta decay. Here, a nucleus captures an inner-shell electron and undergoes a nuclear reaction that has the same effect as $${\beta }^{+}$$ decay. Electron capture is sometimes denoted by the letters EC. We know that electrons cannot reside in the nucleus, but this is a nuclear reaction that consumes the electron and occurs spontaneously only when the products have less mass than the parent plus the electron. If a nuclide $${}_{Z}^{A}{X}_{N}$$ is known to undergo electron capture, then its electron capture equation is

$${}_{Z}^{A}{\text{X}}_{N}+{e}^{-}\to {}_{Z-1}{}^{A}\text{}{\text{Y}}_{N+1}+{\nu }_{e}(\text{electron capture, or EC})\text{.}$$

Any nuclide that can $${\beta }^{+}$$ decay can also undergo electron capture (and often does both). The same conservation laws are obeyed for EC as for $${\beta }^{+}$$ decay. It is good practice to confirm these for yourself.

All forms of beta decay occur because the parent nuclide is unstable and lies outside the region of stability in the chart of nuclides. Those nuclides that have relatively more neutrons than those in the region of stability will $${\beta }^{-}$$ decay to produce a daughter with fewer neutrons, producing a daughter nearer the region of stability. Similarly, those nuclides having relatively more protons than those in the region of stability will $${\beta }^{-}$$ decay or undergo electron capture to produce a daughter with fewer protons, nearer the region of stability.