4.2.3 Alpha Decay

Alpha decay was discovered by Rutherford, who placed an isotope-emitting alpha radiation into a thin glass foil and put the foil into a glass vessel closed at the bot­tom with mercury. The alpha particles have great energy, so they can penetrate the foil into the glass vessel and transform to helium by reacting with two electrons. The gas, of course, shows helium spectrum under excitation. Helium gas, however, cannot penetrate the foil because of the low energy of the atoms, so helium cannot be detected outside the foil.

Apha decay is characteristic for nuclei with great atomic and mass numbers. Thermodynamically, alpha decay can take place at A > 150, but it is only common at A > 210, except for samarium and neodymium, which have isotopes emitting alpha radiation. Alpha decay is possible when the mass decreases in Eq. (4.75).

Alpha particles consist of two protons and two neutrons. By emitting an alpha particle, the ratio of protons to neutrons changes and the atomic and mass number decreases by 2 or 4, respectively.

AM! A 2 4m + a (4.74)

The alpha particle, consisting of two protons and two neutrons, is very stable because of the filled energy levels for protons and neutrons.

The energy of the alpha radiation is in the range of 4—9 MeV. The energy can be calculated from the difference of the rest masses between the parent nuclide and the daughter nuclide, the alpha particle, and the emitting electrons:

Дт = MA — MA-4 2 ma 2 2mt

where MA, MA-4, ma, and me are the rest masses of the parent nuclide, daughter nuclide, alpha particle, and electron, respectively. Since 1 a. m.u. is equivalent to 931 MeV energy, the energy of the alpha particle can be expressed as:

AE = 931 MeV X Am (4.76)

The energy of the alpha particle, however, is smaller than the value calculated by Eq. (4.76) because a portion of the energy recoil the daughter nuclide. The energy of the recoiling can be calculated on the basis of the law of conservation of linear momentum:

mava + Mv = 0

where ma and M are the masses of the alpha particle and the daughter nuclide, and va and v are the rates of the alpha particle and the daughter nuclide, respectively. The rate of the daughter nuclide is expressed from Eq. (4.77):

The total energy emitted in alpha decay is the sum of the energies of the daugh­ter nuclide and the alpha particle:

E = — Mv2 + — mav2n

By substituting Eq. (4.79) into Eq. (4.78), you get the following:

The energy of alpha radiation can be measured in a calorimeter, so the kinetics of the alpha decay can be studied by calorimetry.

In the case of alpha decay, the decay constants of alpha emitters (A) in a decay series correlate to the radiation energy (E) and the range R of the alpha particles in air. This relation can be expressed by the Geiger—Nuttall rule:

log A = a + b X log R (4.81)

log A = a! + b X log E (4.82)

where a, b, a0, and b are constants in a decay series.

The log A—log E function for the alpha-emitting members of the 238U decay series is shown in Figure 4.7.

Figure 4.7 Log A—log E function for the alpha-emitting members of the 238U decay series, illustrating the Geiger—Nuttall rule.

The alpha particles have well-determined, discrete energies. An alpha emitter, however, can produce alpha particles with different energies. This phenomenon can be interpreted by the shell model of nuclei: after the decay, the nucleus is in an excited energy state. The energy of the alpha particles is lower than the value cal­culated from the differences of the rest masses (Eq. (4.75)), and the difference cor­responds to the excitation energy of the nucleus. The excited nucleus may return to a lower excited state or ground state, emitting photons with a characteristic energy. These photons are called gamma photons (described in Section 4.4.6). Since the nucleus may return to the ground state via excited states, the emission of an alpha particle can be followed by more than one gamma photon. In the case of intermedi­ate members of decay series, the energy of a small number of alpha particles may be greater than the value calculated from the differences of the rest masses if the parent nuclide has been in an excited state at the moment of the alpha emission.

In Figure 4.8, decay schemes of two alpha emitter nuclides (230Th and 241Am) are shown. Similar schemes are constructed for all radioactive nuclides. All impor­tant information on the nuclides (parent and daughter nuclides), the mechanism of the decay, and the half-life of the parent nuclide can be found. In addition, the ratio of the lines with different energy is given, and the spin and parity (+ or —) are also included.

Models describing the alpha decay postulate two stages of alpha emission: (1) the separation of the parent nuclide into the alpha particle and the daughter nuclide; (2) the penetration of the alpha particle through a potential barrier that is formed by the joint action of nuclear forces and a Coulomb (electrostatic) interaction of the alpha particle with the remaining portion of the nucleus (daughter nucleus; see Figure 4.9). The range of nuclear forces is very short (see Section 2.2), and at greater distances, the Coulomb interaction is determining. As seen previously, alpha particles have two positive charges. Since the daughter nucleus is also positive, the alpha particle and the daughter nucleus repulse each other. The energy

Figure 4.8 Decay scheme й * 10 1 years of alpha-emitting nuclides (230Th and 241Am).

Figure 4.9 Potential barrier against alpha emission. The center of the nucleus is in the origin, Fs are the wave function at the different spaces, ECb is the Coulomb repulsion energy, r0 is the radius of the nucleus, rx is the outer wall of the Coulomb barrier, and the emitted alpha particle is outside this barrier.

of the repulsion between the alpha particle and the daughter nuclide (the height of the potential barrier) is:

2Ze2

ECb =—— (4.83)

Г0

where ECb is the height of the potential barrier, Z is the atomic number, and r0 is the Coulomb radius. ECb can be determined experimentally in nuclear reactions with charged particles. It is about 20—25 MeV. The energy of the alpha radiation, however, is about 4—9 MeV. According to classical physics, the kinetic energy of the alpha particles is too low to penetrate the potential barrier. Therefore, alpha decay cannot be interpreted by classical physics. The problem of alpha decay can be solved by quantum physics, assuming the wave- particle dual nature. This means that each particle can be described by a wave function, an energy, a linear moment, and a direction of which is the same as those of the particle.

The total energy of the wave or the particle is:

(4.84)

where v is the frequency.

The moment of the wave function (g) is:

g = hk = h = — (4.85)

A c

where A is the wavelength of the alpha particle, its reciprocal (k) is the wave num­ber, and c is the velocity of light in a vacuum.

The intensity of the wave is:

/ = (Ф)2 (4.86)

where Ф is the probability amplitude of the wave function.

The kinetic energy of the particle at a place with U potential can be expressed by the difference between the total energy and the potential energy:

1 2 g2

Ekin = E — U = — mv2 = ^ (4.87)

2 2m

From here,

g = J 2m(E — U) = hk (4.88)

The probability amplitude of the alpha particle is found as follows:

In the nucleus:

2ni(k-r—v-1)

At the place with U potential of the barrier:

ф2 = B2 e2ni(k2r v21)

and over the barrier (outside the nucleus):

Ф3 = B3 e2ni(k3r-v3t)

As a consequence of Eq. (4.88), k is an imaginary number if the potential U is greater than the total energy of the particle (E). When the imaginary k is multiplied by 2ni as in the power, the power of Eq. (4.91) will be a real number. Therefore, the alpha particles can be present outside the nucleus.

The wave functions defined in Eqs. (4.89)—(4.91) can be summarized in the Schrodinger equation. An approximate solution of the Schrhdinger equation for the alpha radiation is discussed here.

When the rate of the alpha particle in the nucleus is v, the number of collisions on the potential barrier in 1 s is nc:

where r0 is the radius inside the nucleus where the nuclear field is homogeneous. The number of hits can be given by means of the de Broglie wavelength (A) of the particle:

When substituting the rate (v) into Eq. (4.92), we obtain:

The ratio of the probability amplitude of the alpha particle existing outside and inside the nucleus is:

ІФ3І2

ІФ1І2

The decay constant is given as the product of the number of collisions and the ratio of the probability amplitude as follows:

A =

4mrg |Ф! І2

When substituting Eqs. (4.89) and (4.91) into Eq. (4.96) and integrating the SchrOdinger equation from r0 to rx, we obtain:

The approximate solution of Eq. (4.97) for heavy nuclei is:

log A = 20.47 — 1.191 X 109 —-—= 14.084 X 106-Z — 2 X —r0 (4.98)

v E

In this way, the decay constant (A) is obtained in seconds. Equation (4.98) is formally similar to the Geiger—Nuttall rule.

The radius of the nucleus (r0 in Eq. (4.98)) calculated from the decay constant is always smaller than the radius calculated from the alpha backscattering. For exam­ple, the radius of 238U is 9.5 X 10-15m calculated from the decay constant and 4 X 10-14m from alpha backscattering. The differences originate in a different place than the location of the collision of the alpha particles: in the case of alpha decay, the alpha particles collide at the inner side of the potential barrier, while in the case of backscattering, alpha particles collide at the outer side of the potential barrier.