1. Chapter 8 Alpha Decay ◎ Introduction and some other properties of α-decay ● The simple theory of Coulomb barrier penetration ◎ The angular momentum barrier

2. § 8-1 Introduction and some other properties of α-decay 1.The theory of α-decay is an application of simple quantum mechanics and its presentation will not, in the simple approach that we shall adopt, add much to our knowledge of nuclear structure. 2.We should answer the following question: why do the mean lives of α -emitting nuclei vary so dramatically from τ= 2.03 × 10 10 years for to τ= 4.3 × 10 -7 seconds for This is a 24 orders of magnitude variation in transition rates !!

3. 3. There are many unstable heavy nuclei in nature. They tend to give away excessive energies and charges by emitting α- particles. Only a few will undergo nucleon emission. 4. We need to find out a reasonable explanation of why the α- particle decay turns out to be the preferable choice for a decay process to occur in unstable heavy nuclei. 5. An α- particle being kicked out from an unstable nucleus is basically the effect of Coulomb repulsion. An α- particle is much less massive than a parent nucleus and has a more stable structure with large binding energy (E B = 28.3 MeV). A heavy nucleus with too many protons can reduce some Coulomb repulsion energy by emitting an α- particle.

4. Emitted Particle Energy Release (MeV) Emitted Particle Energy Release (MeV) n-7.26 4 He+5.41 1H1H-6.12 5 He-2.59 2H2H-10.70 6 He-6.19 3H3H-10.24 6 Li-3.79 3 He-9.92 7 Li-1.94 Energy Release (Q value) for Various Modes of Decay of 232 U

5. The energy-level diagram for two nucleic connected by α-decay The energy-level diagram for the α-decay of 242 Pu α-particle decay The available energy Q α goes into the kinetic energies of the α-particle and of the recoil of the daughter nucleus. If Q α > 0, α-decay is energetically possible; however, it may not occur for other reasons.

6. We now have to apply the energy conditions for α-decay to occur in real nuclei and to find where in the periodic table it is expected to occur. Rewrite the definition of Q α in terms of the nuclear binding energies. Thus α-decay is energetically allowed if (1) (2)

7. which is Above A ≈ 120, d(B/A)/dA is about −7.7×10 −3 MeV. Now B(2,4), the helium nuclear binding energy, is 28.3 MeV, so the critical A must satisfy the following relation: A = 151 (3) Above this A the inequality of equation (3) is satisfied by most nuclei and α-decay becomes, in principle, energetically possible. In fact from A = 144 to A = 206, 7 α- emitters are known amongst the naturally occurring nuclides.

8. From A = 144 to A =206, there are 7 α-emitters of naturally occurring nuclides. When α-emitters are found in this range of A, the energies of the emitted α-particle are normally less than 3 MeV. It is known that the lower the energy release the greater is the lifetime. Their existence implies mean lifetimes comparable to or greater than the age of the earth (about 4 × 10 9 years). Most nuclei in this range on the line stability may be energetically able to decay by α-emission. They do not do so at a detectable level because the transition rate is too small. From A = 144 to A =206 Most of the heavy nuclei to be found on earth were probably produced in one or more supernova explosions of early massive stars. Such explosions can produce very heavy nuclei including trans-uranic elements (Z > 92) and their subsequent decay by α-emission will take them down the periodic table in steps of ΔA = −4. Each α-decay increases the ration N/Z until a β - decay intervenes to restore the nucleus closer to the line of stability. Above Z = 82 many naturally occurring α-emitters are found, many with short lives. Why are they to be found when their lifetime is so short? Above Z = 82 (A > 206)

9. α - emitter Natural AbundanceMean life τ 144 Nd 84 23.8%1.04×10 16 years 147 Sm 125 15.1%2.74×10 11 years 190 Pt 112 0.0127%8.51×10 11 years 192 Pt 114 0.78%≈ 10 15 years 209 Bi 126 100%3×10 17 years 232 Th 142 100%2×10 10 years 238 U 146 99.2739%6.3×10 9 years 7 α-emitters of naturally occurring nuclides. The age of the earth is ~ 4×10 9 years.

10. Very long lifetime comparable to the age of the earth Relatively long lifetime Fast-decaying daughter nuclei are in secular equilibrium.

11. Early observations on α-decay established that, for a unique source, the majority of the emitted α- particles had the same kinetic energy. For each α-emitter, this kinetic energy, T α, is a fraction M D /(M D +M α ) of Q α where M D and M α are the masses of the daughter nucleus and of an α- particle respectively. From the previous transparency we see the values of Q α and the mean life of the principal α-emitter in one of the naturally occurring radioactive series. It is clear that that transition rates (ω) are a strong function of the kinetic energy. The empirical rule connection the two is known as the Geiger-Nuttal rule (1911). R α is the range in air at 15ºC and 1 atmosphere pressure of the α-particles emitted in a decay with transition rate ω. (4)

12. About 6 MeV which is the nucleon separation energy neutronproton Effective potential for an α-particle Z = 90, A ≈ 236

13. For heavy nuclei, the nucleon separation energy is about 6 MeV, so the nucleons fill energy levels up to about 6 MeV below zero total energy. If two protons and two neutrons from the top of the filled levels amalgamate into an α-particle, the binding energy of 28.3 MeV is sufficient to provide the four separation energies and leave the α-particle with positive energy of about 4 MeV. Now we have an α-particle with positive energy leaving it in a potential well. The effective potential is the result of nuclear and Coulomb repulsion potentials. An α-particle is able to tunnel through the “Coulomb barrier” and become free. The tunneling probability can be calculated quantum mechanically. 6 MeV

14. This is the effective mechanical potential for an α-particle as a function of distance between the center of the α-particle and the center of the system which is the parent nucleus less the α-particle. The whole range of potential is separated into three regions: Region I At distances less than R, approximately the nuclear radius, the α-particle is in a potential well of unspecified depth but representing the effect of the nuclear binding force on the α-particle. Region II At a distances R this potential becomes positive and reaches a maximum value of U(R) = zZe 2 /4πε 0 R, where z = 2 and Z is the atomic number of the remaining nucleus. Region III At a distances greater than R the potential is Coulomb, U(r) = zZe 2 /4πε 0 r.

15. If the parent nucleus Z+2, is energetically capable of emitting an α-particle of kinetic energy T α, then there are two possibilities: (1). T α > U(R): the α-particle, if inside the nucleus, is free to leave and will do so almost instantaneously. (That means in a time comparable to the time taken for the α-particle to cross the nucleus, which is less than 10 -21 second.) (2). T α < U(R): classically the α-particle is confined to the nucleus. Quantum mechanically it is free to tunnel through the potential barrier, emerging with zero kinetic energy at radius b (where b = zZe 2 /4 πε 0 T α, z = 2) and to move to large r, where it will have the full kinetic energy T α. We need to find the barrier penetration probability.

16. First we consider a simple square potential barrier with height U and thickness t. The whole area can be separated into three parts. (5) The wave function of the α-particle must satisfy the Schrödinger’s equation. (6) m is the mass of the α-particle u is the wave function of the α-particle [u = u (r)] E is the energy of the α-particle

17. Solutions of the Schrödinger’s equation in three different sections: where T is the kinetic energy of the α-particle and p is its linear momentum. (I) (II) (III) (7) In the section (III) there is no reflection wave therefore D = 0. The probability of transmission is then proportional to |C| 2.

18. Since the wave function u(r) and its first derivative du/dr are continuous on boundaries we are able to summarize the following equations: r r = 0 r = t On the boundary r = t, (1) We may have (8)

19. r r = 0 r = t On the boundary r = 0, (2) We may have (9)

20. Combine four equations from (8) and (9) we have the following relation: (10) In evaluating the quantity K But t ~ several fm The value of e -Kt is extremely small and can be neglected. (11)

22. (13)

26. T α (MeV)lnT α 3.6999 - 4.6791(T α ) -1/2 4.01.386291.36035 4.51.504081.49415 5.01.609481.60734 5.51.704751.70473 6.01.791761.78967 6.51.871801.86461 7.01.945911.93137

27. (19)

28. This is a plot of lnω against the values of Z(M/M α Q α ) 1/2 for the ground state to ground state transitions of many of the naturally occurring and man-made α-active elements. There are deviations from a single straight line but there is a general tendency for the points to cluster near a linear relation between lnω and Z(M/M α Q α ) 1/2 with a slope somewhat less steep than ̶ 3.97 MeV 1/2. The deviation is due to the neglect of the term f’ of equation (19). Thus we have a theory which goes some way towards adequately explaining the range of these mean lives.

29. § 8-3 The angular momentum barrier There is an important effect in all processes involving nuclear and particle reactions and decays. We want to introduce the idea in the context of α-particle decay. Consider the α-particle decay: Suppose the parent and daughter nuclei have spins of quantum number j p and j D. The total angular momentum must be conserved. If j P ≠j D the α-particle must emerge with relative orbital angular momentum (with quantum number l ) with respect to the recoiling daughter nucleus. With the zero spin of the α-particle the conservation of the vector of angular momentum requires that: (20)

30. The quantum number l must be zero or a positive integer. Let us now write down the Schrödinger equation for an α-particle ( z = 2) leaving a recoiling nucleus ( Z, A ). For r > R Here M is the reduced mass of the system. We do the usual separation of variables: let The spherical harmonic Y defines the orbital angular momentum l and its z-component m for the outgoing α-particle. (20) (21) (22)

31. Putting R (r) = U(r)/r and substituting into equation (21) the radial function U(r) should satisfy the following equation: (23) The Coulomb barrier The angular momentum barrier It is clear that the total barrier is harder to penetrate and the transition rate will be lower (and the mean life longer) than with the Coulomb barrier alone.

32. Blatt and Weisskopf (1952) have given some figures for the suppression factors in α-decay transition rates due to the angular momentum barrier. l0123456 ω l /ω 0 1.00.70.370.1370.0370.00710.0011 Values of the suppression factor due to the angular momentum barrier in an α-decay for which Z = 86, T α = 4.88 MeV, R = 9.87 fm Note that we have assumed that the particle is emerging from the nucleus. However, both the Coulomb and angular momentum barrier effects can apply also to particles entering the nucleus. And this is relevant to the rates of nuclear reactions where the first step is the penetration into the nucleus by an incident particle.

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