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

Z A X N Z 2 A 4 Y N 2 + 2 4 He 2 ( α decay ) size 12{α} {}

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 239 Pu α 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 α size 12{α} {} decay. You can see from the equation Z A X N Z 2 A 4 Y N 2 + 2 4 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 α 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 = ( Δ m ) c 2 .

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

Alpha decay energy found from nuclear masses

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

Strategy

Nuclear reaction energy, such as released in α decay, can be found using the equation E = ( Δ m ) c 2 size 12{E= \( Δm \) c"" lSup { size 8{2} } } {} . We must first find Δ m 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 239 Pu size 12{"" lSup { size 8{"239"} } "Pu"} {} ; it is

239 Pu 235 U + 4 He .

Thus the pertinent masses are those of 239 Pu , 235 U , and the α particle or 4 He , all of which are listed in Appendix A . The initial mass was m ( 239 Pu ) = 239 . 052157 u . The final mass is the sum m ( 235 U ) + m ( 4 He ) = 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.

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

E = ( Δ m ) c 2 = ( 0 .005631 u ) c 2 .

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

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

Discussion

The energy released in this α size 12{α} {} decay is in the MeV size 12{"MeV"} {} range, about 10 6 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 α size 12{α} {} particle (or 4 He size 12{"" lSup { size 8{4} } "He"} {} nucleus), which moves away at high speed. The energy carried away by the recoil of the 235 U size 12{"" lSup { size 8{"235"} } U} {} nucleus is much smaller in order to conserve momentum. The 235 U size 12{"" lSup { size 8{"235"} } U} {} nucleus can be left in an excited state to later emit photons ( γ 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 α decay, and so their masses subtract out when finding Δ m . In this case, there are 94 electrons before and after the decay.

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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