1. From : 1.Introduction to Nuclear and Particle Physics
Nuclear radiation 1
From : 1.Introduction to Nuclear and Particle Physics
A.Das and T.Ferbel 2.

2. Alpha Decay
we have seen before, a-decay represents the disintegration of a parent to a daughter through the emission of the nucleus of a helium atom, the transition can be characterized as

3. a-decay can be regarded as the spontaneous fission of the parent nucleus into daughter nuclei with highly asymmetric masses. If we assume that the parent nucleus is initially at rest, then the conservation of energy requires
Where MP, MD and Ma are the masses of the parent, daughter and the a-particle respectively . Similarly TD and Ta represent the kinetic energies of the daughter and of the a-particle. Equation (4.2) can also be rewritten
(4-3)

4. Although the right hand side of Eq. (4
Although the right hand side of Eq. (4.3) involves nuclear masses, we can in fact, use atomic masses in the expression since the masses of the electrons cancel. Thus, we can write
(4-4)
where we have defined the disintegration energy or Q-value as the difference in the rest masses of the initial and final states. It is clear that Q also equals the sum of the kinetic energies of the final state particles. For nonrelativistic particles, the kinetic energies can be written as
(4-5)

5. with vD and va representing the magnitude of the velocities of the daughter and of the a-particle.
Since the parent nucleus decays from rest, the daughter nucleus and the a-particle must necessarily move in opposite directions to conserve momentum, satisfying
(4-6)

7. When the mass of the daughter nucleus is much greater than that of a-particle , then vD<<va and consequently the kinetic energy of the daughter nucleus is far smaller than that of the a-particle. Let us eliminate vD and write expressions for TD and Ta in terms of the Q-value

8. (4-7)
(4-8)

9. The kinetic energy of the emitted a-particle cannot be negative, that is, Ta ≥0. Consequently, for a-decay to occur, we must have an exothermic process
∆𝑴≥𝟎, 𝑸≥𝟎 (4-9)
For massive nuclei, which is our main interest, most of the energy is carried off by the a-particle. The kinetic energy of the daughter nucleus is obtained from Eqs. (4.4) and (4.8)
𝑻 𝑫 =Q - 𝑻 𝜶 = 𝑴 𝜶 𝑴 𝜶 + 𝑴 𝑫 Q = 𝑴 𝜶 𝑴 𝑫 𝑻 𝜶 ≪ 𝑻 𝜶 (4-10)
If we use the approximation 𝑴 𝜶 𝑴 𝑫 ≅ 𝟒 𝑨−𝟒 , we can then write
(4-11)
which can be used to estimate the energy released in the decay.
Important

10. We note from Eq. (4.8) that the kinetic energy (and therefore the magnitude of the velocity) of the a-particle in the decay is unique,. This is a direct consequence of the fact that the process is a two body decay of a parent initially at rest .
Careful measurements, however, have revealed a fine splitting in the energies of a-particles emitted from any radioactive material, corresponding to possibly different Q values. The most energetic a-particles are observed to be produced alone, but less energetic a-decays are always accompanied by the emission of photons.

11. This suggests the presence of energy levels and of underlying quantum structure of discrete states in nuclei. If this is correct then a parent nucleus can transform to the ground state of the daughter nucleus by emitting an a-particle with energy corresponding to the entire Q value, or it can decay to an excited state of the daughter nucleus, . In which case the effective Q value is lower.

12. And, as in the case of atomic transitions, the daughter nucleus can subsequently de-excite to its ground state by emitting a photon. Hence, the decay chain would involve
(4-12)

13. The difference in the two Q values would then correspond to the energy of the emitted photon. For example, the spectrum of observed a-particle: energies in the decay of 228Th90 to 224Ra88 can be associated schematically with the level structure shown in Fig. 4.1.

15. The underlying level structure shown in Fig. 4
The underlying level structure shown in Fig. 4.1 can be determined by the kinetic energies of the different a-particles observed in these decays, which in turn yield the Q values for the transitions through Eq- (4.8). Based on the assumption of discrete nuclear levels, the difference in the Q values will then yield the expected energies of the emitted photons . The measured energies of such accompanying (coincident)photons have, in fact ,confirmed the overall picture and therefore the existence of discrete nuclear levels.

16. Example 1 Consider the a-decay of 240Pu94 240Pu94 —236U92 + 4He2
The emitted a-particles are observed to have energies of 5.17 MeV and 5.12 MeV. Substituting these two values into the first relation in Eq. (4.11)
𝑸≈ 𝑨 𝑨−𝟒 𝑻 𝜶 ,
We obtain the two Q values
𝑸 𝟏 ≈ 𝟐𝟒𝟎 𝟐𝟑𝟔 ×𝟓.𝟏𝟕 𝑴𝒆𝑽≈𝟏.𝟎𝟏𝟕×𝟓.𝟏𝟕 𝑴𝒆𝑽≈𝟓.𝟐𝟔 𝑴𝒆𝑽
𝑸 𝟐 ≈ 𝟐𝟒𝟎 𝟐𝟑𝟔 ×𝟓.𝟏𝟐 𝑴𝒆𝑽≈𝟏.𝟎𝟏𝟕×𝟓.𝟏𝟐 𝑴𝒆𝑽≈𝟓.𝟐𝟏 𝑴𝒆𝑽

17. Thus, when 240pu decays with disintegration energy Q2 ≈5
Thus, when 240pu decays with disintegration energy Q2 ≈5.21 MeV, the daughter nucleus 236U92 is left in an excited state and transforms to the ground state by emitting a photon of energy
Q1 —Q2 =5.26MeV MeV= 0.05 MeV.
This is, indeed, consistent with the observed energy of MeV for the photon. Thus, we can conclude from such studies of a-decays that there are discrete energy levels in nuclei, very much like those found in atoms, and that the spacing between nuclear levels is about 100 keV, whereas the corresponding spacing in atomic levels is of the order of 1 eV.

18. Example:

20. 4.3 Barrier Penetration
The a-particles emitted in nuclear decay have typical energies of about 5 MeV. When such low-energy particles are scattered from a heavy nucleus they cannot penetrate the Coulomb barrier and get sufficiently close to nucleus to interact through the strong force. The height of the Coulomb barrier for A ≈ 200 is about 20—25 MeV, and a 5 MeV a-particle therefore cannot overcome this barrier to get absorbed into the center. On the other hand, a low-energy a-particle that is bound in a nuclear potential sees that same barrier, and yet is able to escape. How this could happen constituted a great puzzle, until it was recognized that the emission of a-particles was a quantum-mechanical phenomenon.

21. The first quantitative understanding of a-decay came in 1929 from the work of George Gamow and of Ronald Gurney and Edward Condon . assuming that the a-particle and the daughter nucleus exist within the parent nucleus prior to its dissociation, we can treat the problem as an a-particle moving in the potential of the daughter nucleus, with the Coulomb potential of preventing their separation (see Fig. 4.2). For concreteness, consider the decay
The kinetic energy of the emitted a-particle is observed to be E=4.05 MeV, and the lifetime of 232 Th is t= 1.39 x 1010 years. The radius of the thorium nucleus obtained from the formula R= 1.2× 𝟏𝟎 −𝟏𝟑 A1/3 cm is ≈7.4 x cm.

22. The a-particle must penetrate the coulomb barrier in order for the decay to take place. The calculation of barrier penetration for a three- differential Coulomb potential is rather complicated. However, since we will ignore the angular momentum of the Schrödinger equation and consider the potential as effectively one-dimensional. Furthermore, we will replace the Coulomb potential by a square barrier of equal area, which approximates the effect of the coulomb repulsion, and is calculationally much simpler (see Fig. 4B). As long as V0 is chosen so that it is larger than E, then the transmission through the barrier is sensitive primarily to the product of 𝑽 𝟎 −𝑬 and a, and not to the precise value of V0. For Z ≈90, we can choose

23. A straightforward quantum mechanical treatment of the transmission through the square barrier shown in Fig. 4.3, yields the following transmission ,coefficient

25. Where Ma is the rest mass and E is the kinetic energy of the emitted a particle (outside of the barrier). For Mac2 ≈4000 MeV, E= 4.05 MeV, V0=14 MeV and Uo ≈ 40 MeV (the calculation is not very sensitive to the nuclear potential), we have

26. We see that the transmission coefficient T is determined essentially by this exponent, and is not very sensitive to the choice of k1 and k. Because we are interested only in estimating T, we can therefore simplify Eq. (4.15) by taking the limit of large k1 (i.e 𝒌 𝟏 𝟐 » 𝜿2 and 𝒌 𝟏 𝟐 » k2). In this limit, the transmission coefficient of Eq. (4.15) becomes

27. Thus, the a-particle has an exceedingly small probability for penetrating the barrier. This explains why low energy a-particles cannot be absorbed by heavy nuclei. However, for an a-particle bound in a nucleus, the situation is quite different. The kinetic energy of the a-particle within the well is

28. Being confined to a small region of ≈ cm ,the a-particle will bounce against the barrier with a frequency given approximately by
Every time the 𝜶-particle hits the barrier, the probability of escape is given in (4.19). We conclude therefore, that the probability for the a-particle to escape per second is simply

31. This result provides a quantitative relationship between the decay constant and the energy of the decaying particle , and is known as the Geiger-Nuttal rule . This relation was discovered in data prior to the development of theoretical formulation.