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7.4 Nuclear decay and conservation laws  (Page 4/25)

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If a nuclide Z A X N is known to β − decay, then its β − decay equation is

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} } ``` \( β rSup { size 8{ - {}} } `"decay" \) ,} {}

where Y is the nuclide having one more proton than X (see [link] ). So if you know that a certain nuclide β − 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 β − decay of 60 Co size 12{"" lSup { size 8{"60"} } "Co"} {} given earlier, we see that Z = 27 for Co and Z = 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 β − decay in the following manner:

n → p + β − + ν - e . size 12{n rightarrow p+β rSup { size 8{ - {}} } + { bar {ν}} rSub { size 8{e} } } {}
Image shows parent nucleus before beta decay and daughter nucleus after beta decay.
In β − 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.

We see that charge is conserved in β − decay, since the total charge is Z size 12{Z} {} before and after the decay. For example, in 60 Co decay, total charge is 27 before decay, since cobalt has Z = 27 . After decay, the daughter nucleus is Ni, which has Z = 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 60 Co decay, for example, there are 60 nucleons before and after the decay. Note that total A is also conserved in α decay. Also note that the total number of protons changes, as does the total number of neutrons, so that total Z size 12{Z} {} and total N size 12{N} {} are not conserved in β − size 12{β rSup { size 8{ - {}} } } {} decay, as they are in α size 12{α} {} decay. Energy released in β − size 12{β rSup { size 8{ - {}} } } {} decay can be calculated given the masses of the parent and products.

β − size 12{β rSup { size 8{ - {}} } } {} Decay energy from masses

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

Strategy and Concept

As in the preceding example, we must first find Δ m , 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 = ( Δ 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 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 β – is included in the atomic mass of Ni. Thus,

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

Solution

The β − decay equation for 60 Co size 12{"" lSup { size 8{"60"} } "Co"} {} is

27 60 Co 33 → 28 60 Ni 32 + β − + ν ¯ e .

As noticed,

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

Entering the masses found in Appendix A gives

Δ m = 59 . 933820 u − 59.930789 u = 0.003031 u .

Thus,

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

Using 1 u = 931.5 MeV / c 2 , we obtain

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

Discussion and Implications

Perhaps the most difficult thing about this example is convincing yourself that the β − size 12{β rSup { size 8{ - {}} } } {} mass is included in the atomic mass of 60 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 60 Co decays, the daughter nucleus 60 Ni is left in an excited state and emits photons ( γ 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 β − decay is created in the nucleus at the time of decay.

Questions & Answers

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Source:  OpenStax, Basic physics for medical imaging. OpenStax CNX. Feb 17, 2014 Download for free at http://legacy.cnx.org/content/col11630/1.1
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