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Nuclear Decay and Conservation Laws

Module by: OpenStax College. E-mail the author

Summary:

  • Define and discuss nuclear decay.
  • State the conservation laws.
  • Explain parent and daughter nucleus.
  • Calculate the energy emitted during nuclear decay.

Nuclear decay has provided an amazing window into the realm of the very small. Nuclear decay gave the first indication of the connection between mass and energy, and it revealed the existence of two of the four basic forces in nature. In this section, we explore the major modes of nuclear decay; and, like those who first explored them, we will discover evidence of previously unknown particles and conservation laws.

Some nuclides are stable, apparently living forever. Unstable nuclides decay (that is, they are radioactive), eventually producing a stable nuclide after many decays. We call the original nuclide the parent and its decay products the daughters. Some radioactive nuclides decay in a single step to a stable nucleus. For example, 60Co60Co size 12{"" lSup { size 8{"60"} } "Co"} {} is unstable and decays directly to 60Ni60Ni size 12{"" lSup { size 8{"60"} } "Ni"} {}, which is stable. Others, such as 238U238U size 12{"" lSup { size 8{"238"} } U} {}, decay to another unstable nuclide, resulting in a decay series in which each subsequent nuclide decays until a stable nuclide is finally produced. The decay series that starts from 238U238U size 12{"" lSup { size 8{"238"} } U} {} is of particular interest, since it produces the radioactive isotopes 226Ra226Ra size 12{"" lSup { size 8{"226"} } "Ra"} {} and 210Po210Po size 12{"" lSup { size 8{"210"} } "Po"} {}, which the Curies first discovered (see Figure 1). Radon gas is also produced (222Rn222Rn size 12{"" lSup { size 8{"222"} } "Rn"} {} in the series), an increasingly recognized naturally occurring hazard. Since radon is a noble gas, it emanates from materials, such as soil, containing even trace amounts of 238U238U size 12{"" lSup { size 8{"238"} } U} {} and can be inhaled. The decay of radon and its daughters produces internal damage. The 238U238U size 12{"" lSup { size 8{"238"} } U} {} decay series ends with 206Pb206Pb size 12{"" lSup { size 8{"206"} } "Pb"} {}, a stable isotope of lead.

Figure 1: The decay series produced by 238U238U size 12{"" lSup { size 8{"238"} } U} {}, the most common uranium isotope. Nuclides are graphed in the same manner as in the chart of nuclides. The type of decay for each member of the series is shown, as well as the half-lives. Note that some nuclides decay by more than one mode. You can see why radium and polonium are found in uranium ore. A stable isotope of lead is the end product of the series.
A graph is shown in which decay of alpha and beta is shown. Also half lives of each isotope are shown. Uranium decays in one mode but some isotopes decay by more than one mode. Finally a stable isotope of lead results.

Note that the daughters of αα size 12{α} {} decay shown in Figure 1 always have two fewer protons and two fewer neutrons than the parent. This seems reasonable, since we know that αα size 12{α} {} decay is the emission of a 4He4He size 12{"" lSup { size 8{4} } "He"} {} nucleus, which has two protons and two neutrons. The daughters of ββ size 12{β} {} decay have one less neutron and one more proton than their parent. Beta decay is a little more subtle, as we shall see. No γγ size 12{γ} {} decays are shown in the figure, because they do not produce a daughter that differs from the parent.

Alpha Decay

In alpha decay, a 4He4He size 12{"" lSup { size 8{4} } "He"} {} nucleus simply breaks away from the parent nucleus, leaving a daughter with two fewer protons and two fewer neutrons than the parent (see Figure 2). One example of αα size 12{α} {} decay is shown in Figure 1 for 238U238U size 12{"" lSup { size 8{"238"} } U} {}. Another nuclide that undergoes αα size 12{α} {} decay is 239Pu239Pu size 12{"" lSup { size 8{"239"} } "Pu"} {}. The decay equations for these two nuclides are

238 U 234 Th 92 234 + 4 He 238 U 234 Th 92 234 + 4 He
(1)

and

239Pu235U+4He.239Pu235U+4He. size 12{"" lSup { size 8{"239"} } "Pu" rightarrow "" lSup { size 8{"235"} } U+"" lSup { size 8{4} } "He"} {}
(2)
Figure 2: Alpha decay is the separation of a 4He4He size 12{"" lSup { size 8{4} } "He"} {} nucleus from the parent. The daughter nucleus has two fewer protons and two fewer neutrons than the parent. Alpha decay occurs spontaneously only if the daughter and 4He4He size 12{"" lSup { size 8{4} } "He"} {} nucleus have less total mass than the parent.
The image shows conditions before and after alpha decay. Before alpha decay the nucleus is labeled parent and after decay the nucleus is labeled daughter.

If you examine the periodic table of the elements, you will find that Th has Z=90Z=90 size 12{Z="90"} {}, two fewer than U, which has Z=92Z=92 size 12{Z="92"} {}. Similarly, in the second decay equation, we see that U has two fewer protons than Pu, which has Z=94Z=94 size 12{Z="94"} {}. The general rule for αα size 12{α} {} decay is best written in the format ZAXNZAXN. If a certain nuclide is known to αα size 12{α} {} decay (generally this information must be looked up in a table of isotopes, such as in Appendix B), its αα size 12{α} {} decay equation is

ZA XN Z2A4 YN2 + 24 He2 (αdecay) ZA XN Z2A4 YN2 + 24 He2 (αdecay) size 12{α} {}
(3)

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 239Pu239Pu αα 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 αα size 12{α} {} decay. You can see from the equation ZA XN Z2A4 YN2 + 24 He2 ZA XN Z2A4 YN2 + 24 He2 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 αα size 12{α} {} 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 = ( Δm ) c 2 . E = ( Δm ) c 2 .
(4)

Here, EE size 12{E} {} is the nuclear reaction energy (the reaction can be nuclear decay or any other reaction), and ΔmΔm size 12{Δm} {} is the difference in mass between initial and final products. When the final products have less total mass, ΔmΔm size 12{Δm} {} is positive, and the reaction releases energy (is exothermic). When the products have greater total mass, the reaction is endothermic (ΔmΔm size 12{Δm} {} is negative) and must be induced with an energy input. For αα size 12{α} {} decay to be spontaneous, the decay products must have smaller mass than the parent.

Example 1: Alpha Decay Energy Found from Nuclear Masses

Find the energy emitted in the αα size 12{α} {} decay of 239Pu239Pu size 12{"" lSup { size 8{"239"} } "Pu"} {}.

Strategy

Nuclear reaction energy, such as released in α decay, can be found using the equation E=(Δm)c2E=(Δm)c2 size 12{E= \( Δm \) c"" lSup { size 8{2} } } {}. We must first find ΔmΔm size 12{Δm} {}, the difference in mass between the parent nucleus and the products of the decay. This is easily done using masses given in Appendix A.

Solution

The decay equation was given earlier for 239Pu239Pu size 12{"" lSup { size 8{"239"} } "Pu"} {} ; it is

239Pu235U+4He.239Pu235U+4He.
(5)

Thus the pertinent masses are those of 239Pu239Pu, 235U235U, and the αα particle or 4He4He, all of which are listed in Appendix A. The initial mass was m(239Pu)=239.052157 um(239Pu)=239.052157 u. The final mass is the sum m(235U) + m(4He) = 235.043924 u + 4.002602 u = 239.046526 um(235U) + m(4He) = 235.043924 u + 4.002602 u = 239.046526 u. Thus,

Δ m = m ( 239 Pu ) [ m ( 235 U ) + m ( 4 He ) ] = 239.052157 u 239.046526 u = 0.0005631 u. Δ m = m ( 239 Pu ) [ m ( 235 U ) + m ( 4 He ) ] = 239.052157 u 239.046526 u = 0.0005631 u.
(6)

Now we can find EE size 12{E} {} by entering ΔmΔm size 12{Δm} {} into the equation:

E=(Δm)c2=(0.005631 u)c2.E=(Δm)c2=(0.005631 u)c2.
(7)

We know 1 u=931.5 MeV/c21 u=931.5 MeV/c2 size 12{1" u =931" "." "5 MeV/"c rSup { size 8{2} } } {}, and so

E=(0.005631)(931.5 MeV/c2)(c2)=5.25 MeV.E=(0.005631)(931.5 MeV/c2)(c2)=5.25 MeV. size 12{E= \( 0 "." "005631" \) \( "931" "." 5" MeV"/c rSup { size 8{2} } \) \( c rSup { size 8{2} } \) =5 "." "25"" MeV"} {}
(8)

Discussion

The energy released in this αα size 12{α} {} decay is in the MeVMeV size 12{"MeV"} {} range, about 106106 size 12{"10" rSup { size 8{6} } } {} times as great as typical chemical reaction energies, consistent with many previous discussions. Most of this energy becomes kinetic energy of the αα size 12{α} {} particle (or 4He4He size 12{"" lSup { size 8{4} } "He"} {} nucleus), which moves away at high speed. The energy carried away by the recoil of the 235U235U size 12{"" lSup { size 8{"235"} } U} {} nucleus is much smaller in order to conserve momentum. The 235U235U size 12{"" lSup { size 8{"235"} } U} {} nucleus can be left in an excited state to later emit photons (γγ size 12{γ} {} 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 Appendix A are atomic masses of neutral atoms, including their electrons. The mass of the electrons is the same before and after αα decay, and so their masses subtract out when finding ΔmΔm. In this case, there are 94 electrons before and after the decay.

Beta Decay

There are actually three types of beta decay. The first discovered was “ordinary” beta decay and is called ββ size 12{β rSup { size 8{ - {}} } } {} decay or electron emission. The symbol ββ size 12{β rSup { size 8{ - {}} } } {} represents an electron emitted in nuclear beta decay. Cobalt-60 is a nuclide that ββ size 12{β rSup { size 8{ - {}} } } {} decays in the following manner:

60Co60Ni+β+ neutrino.60Co60Ni+β+ neutrino. size 12{"" lSup { size 8{"60"} } "Co" rightarrow "" lSup { size 8{"60"} } "Ni"+β rSup { size 8{-{}} } +" neutrino"} {}
(9)

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 Figure 3). 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.

Figure 3: Enrico Fermi was nearly unique among 20th-century physicists—he made significant contributions both as an experimentalist and a theorist. His many contributions to theoretical physics included the identification of the weak nuclear force. The fermi (fm) is named after him, as are an entire class of subatomic particles (fermions), an element (Fermium), and a major research laboratory (Fermilab). His experimental work included studies of radioactivity, for which he won the 1938 Nobel Prize in physics, and creation of the first nuclear chain reaction. (credit: United States Department of Energy, Office of Public Affairs)
Photo of physicist Enrico Fermi.

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+1. The neutrino in ββ size 12{β rSup { size 8{ - {}} } } {} decay is an electron’s antineutrino, given the symbol ν¯eν¯e, where νν 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 ν¯eν¯e, being antimatter, has an electron family number of –1–1. 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 ZAXNZAXN is known to ββ decay, then its ββ decay equation is

Z A X N Z + 1 A Y N 1 + β + ν - e ( β decay ) , Z A X N Z + 1 A Y N 1 + β + ν - e ( β decay ) , size 12{"" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } rightarrow "" lSub { size 8{Z+1} } lSup { size 8{A} } Y rSub { size 8{N - 1} } +β rSup { size 8{ - {}} } + { bar {ν}} rSub { size 8{e} } ``` \( β rSup { size 8{ - {}} } `"decay" \) ,} {}
(10)

where Y is the nuclide having one more proton than X (see Figure 4). So if you know that a certain nuclide ββ decays, you can find the daughter nucleus by first looking up ZZ for the parent and then determining which element has atomic number Z+1Z+1. In the example of the ββ decay of 60Co60Co size 12{"" lSup { size 8{"60"} } "Co"} {} given earlier, we see that Z=27Z=27 for Co and Z=28Z=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 ββ decay in the following manner:

n p + β + ν - e . n p + β + ν - e . size 12{n rightarrow p+β rSup { size 8{ - {}} } + { bar {ν}} rSub { size 8{e} } } {}
(11)
Figure 4: In ββ size 12{β rSup { size 8{ - {}} } } {} decay, the parent nucleus emits an electron and an antineutrino. The daughter nucleus has one more proton and one less neutron than its parent. Neutrinos interact so weakly that they are almost never directly observed, but they play a fundamental role in particle physics.
Image shows parent nucleus before beta decay and daughter nucleus after beta decay.

We see that charge is conserved in ββ decay, since the total charge is ZZ size 12{Z} {} before and after the decay. For example, in 60Co60Co decay, total charge is 27 before decay, since cobalt has Z=27Z=27. After decay, the daughter nucleus is Ni, which has Z=28Z=28, and there is an electron, so that the total charge is also 28 + (–1)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 AA is conserved. In 60Co60Co decay, for example, there are 60 nucleons before and after the decay. Note that total AA is also conserved in αα decay. Also note that the total number of protons changes, as does the total number of neutrons, so that total ZZ size 12{Z} {} and total NN size 12{N} {} are not conserved in ββ size 12{β rSup { size 8{ - {}} } } {} decay, as they are in αα size 12{α} {} decay. Energy released in ββ size 12{β rSup { size 8{ - {}} } } {} decay can be calculated given the masses of the parent and products.

Example 2: ββ size 12{β rSup { size 8{ - {}} } } {} Decay Energy from Masses

Find the energy emitted in the ββ size 12{β rSup { size 8{ - {}} } } {} decay of 60Co60Co size 12{"" lSup { size 8{"60"} } "Co"} {}.

Strategy and Concept

As in the preceding example, we must first find ΔmΔm, the difference in mass between the parent nucleus and the products of the decay, using masses given in Appendix A. Then the emitted energy is calculated as before, using E=(Δm)c2E=(Δm)c2. 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 Appendix A are for neutral atoms, the daughter nucleus has one more electron than the parent, and so the extra electron mass that corresponds to the ββ is included in the atomic mass of Ni. Thus,

Δm=m(60Co )m(60Ni).Δm=m(60Co )m(60Ni). size 12{Δm=m \( "" lSup { size 8{"60"} } "Co" \) -m \( "" lSup { size 8{"60"} } "Ni" \) } {}
(12)

Solution

The ββ decay equation for 60Co60Co size 12{"" lSup { size 8{"60"} } "Co"} {} is

2760Co33 2860Ni32 + β +ν¯e.2760Co33 2860Ni32 + β +ν¯e.
(13)

As noticed,

Δm=m(60Co )m(60Ni).Δm=m(60Co )m(60Ni). size 12{Δm=m \( "" lSup { size 8{"60"} } "Co" \) -m \( "" lSup { size 8{"60"} } "Ni" \) } {}
(14)

Entering the masses found in Appendix A gives

Δm=59.933820 u59.930789 u=0.003031 u.Δm=59.933820 u59.930789 u=0.003031 u.
(15)

Thus,

E=(Δm)c2=(0.003031 u)c2.E=(Δm)c2=(0.003031 u)c2. size 12{E= \( Δm \) c rSup { size 8{2} } = \( 0 "." "003031" \) \( uc rSup { size 8{2} } \) } {}
(16)

Using 1 u=931.5 MeV/c21 u=931.5 MeV/c2, we obtain

E=(0.003031)(931.5 MeV/c2)(c2)=2.82 MeV.E=(0.003031)(931.5 MeV/c2)(c2)=2.82 MeV. size 12{E= \( 0 "." "003031" \) \( "931" "." 5" MeV"/c rSup { size 8{2} } \) \( c rSup { size 8{2} } \) =2 "." "82"" MeV"} {}
(17)

Discussion and Implications

Perhaps the most difficult thing about this example is convincing yourself that the ββ size 12{β rSup { size 8{ - {}} } } {} mass is included in the atomic mass of 60Ni60Ni. 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 60Co60Co decays, the daughter nucleus 60Ni60Ni is left in an excited state and emits photons ( γγ size 12{g} {} 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 ββ 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 β+β+ size 12{β rSup { size 8{+{}} } } {}decay. Certain nuclides decay by the emission of a positive electron. This is antielectron or positron decay (see Figure 5).

Figure 5: β+β+ size 12{β rSup { size 8{+{}} } } {} decay is the emission of a positron that eventually finds an electron to annihilate, characteristically producing gammas in opposite directions.
Image shows parent nucleus before beta plus decay and daughter nucleus after beta plus decay, which results in a positively charged electron called a positron.

The antielectron is often represented by the symbol e+e+ size 12{e rSup { size 8{+{}} } } {}, but in beta decay it is written as β+β+ size 12{β rSup { size 8{+{}} } } {} 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 –1–1. 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++eγ+γe++eγ+γ size 12{e rSup { size 8{+{}} } +e rSup { size 8{-{}} } rightarrow g+g} {}, conserves electron family number as well as all other conserved quantities.) If a nuclide ZAXNZAXN is known to β+β+ decay, then its β+β+ size 12{β rSup { size 8{+{}} } } {}decay equation is

ZA X N Z 1 A Y N + 1 + β + + ν e ( β + decay ) , ZA X N Z 1 A Y N + 1 + β + + ν e ( β + decay ) , size 12{"" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } rightarrow "" lSub { size 8{Z - 1} } lSup { size 8{A} } Y rSub { size 8{N+1} } +β rSup { size 8{+{}} } +ν rSub { size 8{e} } ```` \( β rSup { size 8{+{}} } `"decay" \) ,} {}
(18)

where Y is the nuclide having one less proton than X (to conserve charge) and νeνe is the symbol for the electron’s neutrino, which has an electron family number of +1+1. Since an antimatter member of the electron family (the β+β+) is created in the decay, a matter member of the family (here the νeνe) must also be created. Given, for example, that 22Na22Na β+β+ size 12{β rSup { size 8{+{}} } } {} decays, you can write its full decay equation by first finding that Z=11Z=11 for 22Na22Na, so that the daughter nuclide will have Z=10Z=10 size 12{Z="10"} {}, the atomic number for neon. Thus the β+β+ size 12{β rSup { size 8{+{}} } } {} decay equation for 22Na22Na is

1122Na111022Ne12+β++νe.1122Na111022Ne12+β++νe.
(19)

In β+β+ size 12{β rSup { size 8{+{}} } } {} 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 β+β+ size 12{β rSup { size 8{+{}} } } {} 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 β+β+ size 12{β rSup { size 8{+{}} } } {} decay,

Δm=m(parent)[m(daughter)+2me],Δm=m(parent)[m(daughter)+2me],
(20)

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 β+β+ size 12{β rSup { size 8{+{}} } } {} 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 ZAXNZAXN is known to undergo electron capture, then its electron capture equation is

ZA X N + e Z 1 A Y N + 1 + ν e ( electron capture, or EC ) . ZA X N + e Z 1 A Y N + 1 + ν e ( electron capture, or EC ) . size 12{"" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } +e rSup { size 8{ - {}} } rightarrow "" lSub { size 8{Z - 1} } lSup { size 8{A} } Y rSub { size 8{N+1} } +ν rSub { size 8{e} } ``` \( "electron capture, or EC" \) "." } {}
(21)

Any nuclide that can β+β+ size 12{β rSup { size 8{+{}} } } {} decay can also undergo electron capture (and often does both). The same conservation laws are obeyed for EC as for β+β+ size 12{β rSup { size 8{+{}} } } {} 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 ββ size 12{β rSup { size 8{ - {}} } } {} 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 ββ size 12{β rSup { size 8{ - {}} } } {} decay or undergo electron capture to produce a daughter with fewer protons, nearer the region of stability.

Gamma Decay

Gamma decay is the simplest form of nuclear decay—it is the emission of energetic photons by nuclei left in an excited state by some earlier process. Protons and neutrons in an excited nucleus are in higher orbitals, and they fall to lower levels by photon emission (analogous to electrons in excited atoms). Nuclear excited states have lifetimes typically of only about 10141014 size 12{"10" rSup { size 8{ - "14"} } } {} s, an indication of the great strength of the forces pulling the nucleons to lower states. The γγ size 12{γ} {} decay equation is simply

ZA X N * Z A X N + γ 1 + γ 2 + ( γ decay ) ZA X N * Z A X N + γ 1 + γ 2 + ( γ decay ) size 12{"" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } rSup { size 8{*} } rightarrow "" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } +γ rSub { size 8{1} } +γ rSub { size 8{2} } + dotsaxis ``` \( γ`"decay" \) } {}
(22)

where the asterisk indicates the nucleus is in an excited state. There may be one or more γγ s emitted, depending on how the nuclide de-excites. In radioactive decay, γγ emission is common and is preceded by γγ or ββ size 12{β} {} decay. For example, when 60Co60Co ββ size 12{β rSup { size 8{ - {}} } } {} decays, it most often leaves the daughter nucleus in an excited state, written 60Ni*60Ni*. Then the nickel nucleus quickly γγ size 12{γ} {} decays by the emission of two penetrating γγ size 12{γ} {} s:

60Ni*60Ni+γ1+γ2.60Ni*60Ni+γ1+γ2. size 12{"" lSup { size 8{"60"} } "Ni" rSup { size 8{*} } rightarrow "" lSup { size 8{"60"} } "Ni"+γ rSub { size 8{1} } +γ rSub { size 8{2} } } {}
(23)

These are called cobalt γγ size 12{γ} {} rays, although they come from nickel—they are used for cancer therapy, for example. It is again constructive to verify the conservation laws for gamma decay. Finally, since γγ size 12{γ} {} decay does not change the nuclide to another species, it is not prominently featured in charts of decay series, such as that in Figure 1.

There are other types of nuclear decay, but they occur less commonly than αα, ββ, and γγ size 12{γ} {} decay. Spontaneous fission is the most important of the other forms of nuclear decay because of its applications in nuclear power and weapons. It is covered in the next chapter.

Section Summary

  • When a parent nucleus decays, it produces a daughter nucleus following rules and conservation laws. There are three major types of nuclear decay, called alpha α,α, size 12{ left (α right ),} {} beta β,β, size 12{ left (β right ),} {} and gamma γγ size 12{ left (γ right )} {}. The αα size 12{α} {} decay equation is
    ZAXNZ2A4YN2+24He2.ZAXNZ2A4YN2+24He2. size 12{"" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } rightarrow "" lSub { size 8{Z - 2} } lSup { size 8{A - 4} } Y rSub { size 8{N - 2} } +"" lSub { size 8{2} } lSup { size 8{4} } "He" rSub { size 8{2} } } {}
    (24)
  • Nuclear decay releases an amount of energy EE size 12{E} {} related to the mass destroyed ΔmΔm by
    E=(Δm)c2.E=(Δm)c2. size 12{E= \( Δm \) c rSup { size 8{2} } } {}
    (25)
  • There are three forms of beta decay. The ββ size 12{β rSup { size 8{ - {}} } } {}decay equation is
    ZAXNZ+1AYN1+β+ν¯e.ZAXNZ+1AYN1+β+ν¯e.
    (26)
  • The β+β+ decay equation is
    ZAXNZ1AYN+1+β++νe.ZAXNZ1AYN+1+β++νe.
    (27)
  • The electron capture equation is
    ZAXN+eZ1AYN+1+νe.ZAXN+eZ1AYN+1+νe.
    (28)
  • ββ is an electron, β+β+ size 12{β rSup { size 8{+{}} } } {} is an antielectron or positron, νeνe size 12{v rSub { size 8{e} } } {} represents an electron’s neutrino, and ν¯eν¯e size 12{ {overline {ν rSub { size 8{e} } }} } {} is an electron’s antineutrino. In addition to all previously known conservation laws, two new ones arise— conservation of electron family number and conservation of the total number of nucleons. The γγ decay equation is
    Z A X N * Z A X N + γ 1 + γ 2 + Z A X N * Z A X N + γ 1 + γ 2 + size 12{"" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } rSup { size 8{*} } rightarrow "" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } +γ rSub { size 8{1} } +γ rSub { size 8{2} } + dotsaxis } {}
    (29)
    γγ size 12{γ} {} is a high-energy photon originating in a nucleus.

Conceptual Questions

Exercise 1

Star Trek fans have often heard the term “antimatter drive.” Describe how you could use a magnetic field to trap antimatter, such as produced by nuclear decay, and later combine it with matter to produce energy. Be specific about the type of antimatter, the need for vacuum storage, and the fraction of matter converted into energy.

Exercise 2

What conservation law requires an electron’s neutrino to be produced in electron capture? Note that the electron no longer exists after it is captured by the nucleus.

Exercise 3

Neutrinos are experimentally determined to have an extremely small mass. Huge numbers of neutrinos are created in a supernova at the same time as massive amounts of light are first produced. When the 1987A supernova occurred in the Large Magellanic Cloud, visible primarily in the Southern Hemisphere and some 100,000 light-years away from Earth, neutrinos from the explosion were observed at about the same time as the light from the blast. How could the relative arrival times of neutrinos and light be used to place limits on the mass of neutrinos?

Exercise 4

What do the three types of beta decay have in common that is distinctly different from alpha decay?

Problems & Exercises

In the following eight problems, write the complete decay equation for the given nuclide in the complete ZAXNZAXN size 12{"" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } } {} notation. Refer to the periodic table for values of ZZ size 12{Z} {}.

Exercise 1

ββ size 12{β rSup { size 8{ - {}} } } {} decay of 3H3H size 12{"" lSup { size 8{3} } H} {} (tritium), a manufactured isotope of hydrogen used in some digital watch displays, and manufactured primarily for use in hydrogen bombs.

Solution

1 3 H 2 2 3 He 1 + β + ν ¯ e 1 3 H 2 2 3 He 1 + β + ν ¯ e
(30)

Exercise 2

ββ size 12{β rSup { size 8{ - {}} } } {} decay of 40K40K size 12{"" lSup { size 8{"40"} } K} {}, a naturally occurring rare isotope of potassium responsible for some of our exposure to background radiation.

Exercise 3

β+β+ size 12{β rSup { size 8{+{}} } } {} decay of 50Mn50Mn size 12{"" lSup { size 8{"50"} } "Mn"} {}.

Solution

25 50 M 25 24 50 Cr 26 + β + + ν e 25 50 M 25 24 50 Cr 26 + β + + ν e size 12{"" lSub { size 8{"25"} } lSup { size 8{"50"} } M rSub { size 8{"25"} } rightarrow "" lSub { size 8{"24"} } lSup { size 8{"50"} } "Cr" rSub { size 8{"20"} } +β rSup { size 8{+{}} } +v rSub { size 8{e} } } {}
(31)

Exercise 4

β+β+ size 12{β rSup { size 8{+{}} } } {} decay of 52Fe52Fe size 12{"" lSup { size 8{"52"} } "Fe"} {}.

Exercise 5

Electron capture by 7Be7Be size 12{"" lSup { size 8{7} } "Be"} {}.

Solution

4 7 Be 3 + e 3 7 Li 4 + ν e 4 7 Be 3 + e 3 7 Li 4 + ν e size 12{"" lSub { size 8{4} } lSup { size 8{7} } "Be" rSub { size 8{3} } +e rSup { size 8{ - {}} } rightarrow "" lSub { size 8{3} } lSup { size 8{7} } "Li" rSub { size 8{4} } +v rSub { size 8{e} } } {}
(32)

Exercise 6

Electron capture by 106In106In size 12{"" lSup { size 8{"106"} } "In"} {}.

Exercise 7

αα size 12{α} {} decay of 210Po210Po size 12{"" lSup { size 8{"210"} } "Po"} {}, the isotope of polonium in the decay series of 238U238U size 12{"" lSup { size 8{"238"} } U} {} that was discovered by the Curies. A favorite isotope in physics labs, since it has a short half-life and decays to a stable nuclide.

Solution

84 210 Po 126 82 206 Pb 124 + 2 4 He 2 84 210 Po 126 82 206 Pb 124 + 2 4 He 2 size 12{"" lSub { size 8{"84"} } lSup { size 8{"210"} } "Pb" rSub { size 8{"126"} } rightarrow "" lSub { size 8{"82"} } lSup { size 8{"206"} } "Pb" rSub { size 8{"124"} } +"" lSub { size 8{2} } lSup { size 8{4} } "He" rSub { size 8{2} } } {}
(33)

Exercise 8

αα size 12{α} {} decay of 226Ra226Ra size 12{"" lSup { size 8{"226"} } "Ra"} {}, another isotope in the decay series of 238U238U size 12{"" lSup { size 8{"238"} } U} {}, first recognized as a new element by the Curies. Poses special problems because its daughter is a radioactive noble gas.

In the following four problems, identify the parent nuclide and write the complete decay equation in the ZAXNZAXN size 12{"" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } } {} notation. Refer to the periodic table for values of ZZ size 12{Z} {}.

Exercise 9

ββ size 12{β rSup { size 8{ - {}} } } {} decay producing 137Ba137Ba size 12{"" lSup { size 8{"137"} } "Ba"} {}. The parent nuclide is a major waste product of reactors and has chemistry similar to potassium and sodium, resulting in its concentration in your cells if ingested.

Solution

55 137 Cs 82 56 137 Ba 81 + β + ν ¯ e 55 137 Cs 82 56 137 Ba 81 + β + ν ¯ e size 12{"" lSub { size 8{"55"} } lSup { size 8{"137"} } "Cs" rSub { size 8{"82"} } rightarrow "" lSub { size 8{"56"} } lSup { size 8{"137"} } "Ba" rSub { size 8{"81"} } +β rSup { size 8{ - {}} } + {overline {v rSub { size 8{e} } }} } {}
(34)

Exercise 10

ββ size 12{β rSup { size 8{ - {}} } } {} decay producing 90Y90Y size 12{"" lSup { size 8{"90"} } Y} {}. The parent nuclide is a major waste product of reactors and has chemistry similar to calcium, so that it is concentrated in bones if ingested (90Y90Y size 12{"" lSup { size 8{"90"} } Y} {} is also radioactive.)

Exercise 11

αα size 12{α} {} decay producing 228Ra228Ra size 12{"" lSup { size 8{"228"} } "Ra"} {}. The parent nuclide is nearly 100% of the natural element and is found in gas lantern mantles and in metal alloys used in jets (228Ra228Ra size 12{"" lSup { size 8{"228"} } "Ra"} {} is also radioactive).

Solution

90 232 Th 142 88 228 Ra 140 + 2 4 He 2 90 232 Th 142 88 228 Ra 140 + 2 4 He 2 size 12{"" lSub { size 8{"90"} } lSup { size 8{"232"} } "Th" rSub { size 8{"142"} } rightarrow "" lSub { size 8{"88"} } lSup { size 8{"228"} } "Ra" rSub { size 8{"140"} } +"" lSub { size 8{2} } lSup { size 8{4} } "He" rSub { size 8{2} } } {}
(35)

Exercise 12

αα size 12{α} {} decay producing 208Pb208Pb size 12{"" lSup { size 8{"208"} } "Pb"} {}. The parent nuclide is in the decay series produced by 232Th232Th size 12{"" lSup { size 8{"232"} } "Th"} {}, the only naturally occurring isotope of thorium.

Exercise 13

When an electron and positron annihilate, both their masses are destroyed, creating two equal energy photons to preserve momentum. (a) Confirm that the annihilation equation e++eγ+γe++eγ+γ size 12{e rSup { size 8{+{}} } +e rSup { size 8{ - {}} } rightarrow γ+γ} {} conserves charge, electron family number, and total number of nucleons. To do this, identify the values of each before and after the annihilation. (b) Find the energy of each γγ size 12{γ} {} ray, assuming the electron and positron are initially nearly at rest. (c) Explain why the two γγ size 12{γ} {} rays travel in exactly opposite directions if the center of mass of the electron-positron system is initially at rest.

Solution

(a) charge: +1+1=0; electron family number:+1+1=0; A: 0+0=0charge: +1+1=0; electron family number:+1+1=0; A: 0+0=0

(b) 0.511 MeV

(c) The two γγ size 12{γ} {} rays must travel in exactly opposite directions in order to conserve momentum, since initially there is zero momentum if the center of mass is initially at rest.

Exercise 14

Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for αα decay given in the equation ZAXN Z2A4 YN2 + 24 He2 ZAXN Z2A4 YN2 + 24 He2 . To do this, identify the values of each before and after the decay.

Exercise 15

Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for ββ decay given in the equation ZA XN Z+1A Y N1 + β + ν¯eZA XN Z+1A Y N1 + β + ν¯e size 12{"" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } rightarrow "" lSub { size 8{Z−1} } lSup { size 8{A} } Y rSub { size 8{N - 1} } +β rSup { size 8{ - {}} } + {overline {v rSub { size 8{e} } }} } {}. To do this, identify the values of each before and after the decay.

Solution

Z = Z + 1 1; A = A ; efn : 0 = + 1 + 1 Z = Z + 1 1; A = A ; efn : 0 = + 1 + 1
(36)

Exercise 16

Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for ββ size 12{β rSup { size 8{ - {}} } } {} decay given in the equation ZA XN Z1A Y N1 + β + νeZA XN Z1A Y N1 + β + νe. To do this, identify the values of each before and after the decay.

Exercise 17

Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for electron capture given in the equation ZAXN+eZ1AYN+1+νeZAXN+eZ1AYN+1+νe size 12{"" lSub { size 8{Z} } lSup { size 8{A} } X rSub { size 8{N} } +e rSup { size 8{ - {}} } rightarrow "" lSub { size 8{Z - 1} } lSup { size 8{A} } Y rSub { size 8{N+1} } +v rSub { size 8{e} } } {}. To do this, identify the values of each before and after the capture.

Solution

Z - 1 = Z 1; A = A; efn : + 1 = + 1 Z - 1 = Z 1; A = A; efn : + 1 = + 1 alignl { stack { size 12{Z+1=Z - 1" before/after; captured "e rSup { size 8{ - 1} } " is last term rhs;"} {} # A=A" ; efn : " left (+1 right )= left (+1 right ) {} } } {}
(37)

Exercise 18

A rare decay mode has been observed in which 222Ra222Ra emits a 14C14C nucleus. (a) The decay equation is 222RaAX+14C222RaAX+14C size 12{ {} rSup { size 8{"222"} } "Ra" rightarrow rSup { size 8{A} } "X+" rSup { size 8{"14"} } C} {}. Identify the nuclide AXAX. (b) Find the energy emitted in the decay. The mass of 222Ra222Ra size 12{"" lSup { size 8{"222"} } "Ra"} {} is 222.015353 u.

Exercise 19

(a) Write the complete αα size 12{α} {} decay equation for 226Ra226Ra size 12{"" lSup { size 8{"226"} } "Ra"} {}.

(b) Find the energy released in the decay.

Solution

(a) 88226Ra138 86222 Rn136+ 24 He2 88226Ra138 86222 Rn136+ 24 He2

(b) 4.87 MeV

Exercise 20

(a) Write the complete αα size 12{α} {} decay equation for 249Cf249Cf size 12{"" lSup { size 8{"249"} } "Cf"} {}.

(b) Find the energy released in the decay.

Exercise 21

(a) Write the complete ββ size 12{β rSup { size 8{ - {}} } } {} decay equation for the neutron. (b) Find the energy released in the decay.

Solution

(a) np+β+ν¯enp+β+ν¯e

(b) ) 0.783 MeV

Exercise 22

(a) Write the complete ββ size 12{β rSup { size 8{ - {}} } } {} decay equation for 90Sr90Sr size 12{"" lSup { size 8{"90"} } "Sr"} {}, a major waste product of nuclear reactors. (b) Find the energy released in the decay.

Exercise 23

Calculate the energy released in the β+β+ size 12{β rSup { size 8{+{}} } } {} decay of 22Na22Na, the equation for which is given in the text. The masses of 22Na22Na and 22Ne22Ne size 12{"" lSup { size 8{"22"} } "Ne"} {} are 21.994434 and 21.991383 u, respectively.

Solution

1.82 MeV

Exercise 24

(a) Write the complete β+β+ size 12{β rSup { size 8{+{}} } } {} decay equation for 11C11C size 12{"" lSup { size 8{"11"} } C} {}.

(b) Calculate the energy released in the decay. The masses of 11C11C size 12{"" lSup { size 8{"11"} } C} {} and 11B11B size 12{"" lSup { size 8{"11"} } B} {} are 11.011433 and 11.009305 u, respectively.

Exercise 25

(a) Calculate the energy released in the αα size 12{α} {} decay of 238U238U size 12{"" lSup { size 8{"238"} } U} {}.

(b) What fraction of the mass of a single 238U238U size 12{"" lSup { size 8{"238"} } U} {} is destroyed in the decay? The mass of 234Th234Th size 12{"" lSup { size 8{"234"} } "Th"} {} is 234.043593 u.

(c) Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?

Solution

(a) 4.274 MeV

(b) 1.927×1051.927×105 size 12{1 "." "927" times "10" rSup { size 8{ - 5} } u} {}

(c) Since U-238 is a slowly decaying substance, only a very small number of nuclei decay on human timescales; therefore, although those nuclei that decay lose a noticeable fraction of their mass, the change in the total mass of the sample is not detectable for a macroscopic sample.

Exercise 26

(a) Write the complete reaction equation for electron capture by 7Be.7Be. size 12{"" lSup { size 8{7} } "Be"} {}

(b) Calculate the energy released.

Exercise 27

(a) Write the complete reaction equation for electron capture by 15O15O size 12{"" lSup { size 8{"15"} } O} {}.

(b) Calculate the energy released.

Solution

(a) 815O7+ e 715N8+νe815O7+ e 715N8+νe size 12{"" lSub { size 8{8} } lSup { size 8{"15"} } O rSub { size 8{7} } +e rSup { size 8{ - {}} } rightarrow "" lSub { size 8{7} } lSup { size 8{"15"} } N rSub { size 8{8} } +v rSub { size 8{e} } } {}

(b) 2.754 MeV

Glossary

parent:
the original state of nucleus before decay
daughter:
the nucleus obtained when parent nucleus decays and produces another nucleus following the rules and the conservation laws
positron:
the particle that results from positive beta decay; also known as an antielectron
decay:
the process by which an atomic nucleus of an unstable atom loses mass and energy by emitting ionizing particles
alpha decay:
type of radioactive decay in which an atomic nucleus emits an alpha particle
beta decay:
type of radioactive decay in which an atomic nucleus emits a beta particle
gamma decay:
type of radioactive decay in which an atomic nucleus emits a gamma particle
decay equation:
the equation to find out how much of a radioactive material is left after a given period of time
nuclear reaction energy:
the energy created in a nuclear reaction
neutrino:
an electrically neutral, weakly interacting elementary subatomic particle
electron’s antineutrino:
antiparticle of electron’s neutrino
positron decay:
type of beta decay in which a proton is converted to a neutron, releasing a positron and a neutrino
antielectron:
another term for positron
decay series:
process whereby subsequent nuclides decay until a stable nuclide is produced
electron’s neutrino:
a subatomic elementary particle which has no net electric charge
antimatter:
composed of antiparticles
electron capture:
the process in which a proton-rich nuclide absorbs an inner atomic electron and simultaneously emits a neutrino
electron capture equation:
equation representing the electron capture

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