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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 [link] .

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 [link] . 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 [link] . Note that the masses given in [link] 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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Beta decay

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

60 Co 60 Ni + β + neutrino. size 12{"" lSup { size 8{"60"} } "Co" rightarrow "" lSup { size 8{"60"} } "Ni"+β rSup { size 8{-{}} } +" 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 [link] ). 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.

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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