42.6.1 Alpha Decay

An alpha particle, symbolized as $α$ , is a $^{4} He$ nucleus, a strongly bound system of two protons and two neutrons. An unstable nucleus that ejects an alpha particle will lose two protons and two neutrons, so we can write the decay as

^{A} X_{Z} \to^{A - 4} Y_{Z - 2} + α + energy

(42.14)

FIGURE 42.18 shows the alpha-decay process. The original nucleus X is called the parent nucleus, and the decay-product nucleus Y is the daughter nucleus. This reaction can occur only when the mass of the parent nucleus is greater than the mass of the daughter nucleus plus the mass of an alpha particle. This requirement is met for heavy, high- $Z$ nuclei well above the maximum on the Figure 42.6 curve of binding energy. It is energetically favorable for these nuclei to eject an alpha particle because the daughter nucleus is more tightly bound than the parent nucleus.

The figure consists of two parts that illustrate the before and after of the alpha decay process.

FIGURE 42.18 Full Alternative Text

Although the mass requirement is based on the nuclear masses, we can express it—as we did the binding energy equation—in terms of atomic masses. The energy released in an alpha decay, essentially all of which goes into the alpha particle’s kinetic energy, is

Δ E \approx K_{α} = (m_{X} - m_{Y} - m_{He}) c^{2}

(42.15)

EXAMPLE 42.5 Alpha decay of uranium

The uranium isotope $^{238} U$ undergoes alpha decay to $^{234} Th$ . The atomic masses are 238.0508 u for $^{238} U$ and 234.0436 u for $^{234} Th$ . What is the kinetic energy, in MeV, of the alpha particle?

MODEL Essentially all of the energy release $Δ E$ goes into the alpha particle’s kinetic energy.

SOLVE The atomic mass of helium is 4.0026 u. Thus

\begin{array}{l} K_{α} & = & (238.0508 u - 234.0436 u - 4.0026 u) c^{2} \\ = & (0.0046 u \times \frac{931.5 MeV/ c^{2}}{1 u}) c^{2} = 4.3 MeV \end{array}

REVIEW This is a typical alpha-particle energy. Notice how the $c^{2}$ canceled from the calculation so that we never had to evaluate $c^{2}$ .

Alpha decay is a purely quantum-mechanical effect. FIGURE 42.19 shows the potential energy of an alpha particle, where the $^{4} He$ nucleus of an alpha particle is so tightly bound that we can think of it as existing “prepackaged” inside the parent nucleus. Both the depth of the energy well and the height of the Coulomb barrier are twice those of a proton because the charge of an $α$ particle is 2 $e$ .

The figure illustrates the potential-energy diagram of an alpha particle in the parent nucleus.

FIGURE 42.19 Full Alternative Text

Because of the high Coulomb barrier (alpha decay occurs only in high- $Z$ nuclei), there may be one or more allowed energy levels with $E > 0$ . Energy levels with $E < 0$ are completely bound, but an alpha particle in an energy level with $E > 0$ can tunnel through the Coulomb barrier and escape. That is exactly how alpha decay occurs.

Energy must be conserved, so the kinetic energy of the escaping $α$ particle is the height of the energy level above $E = 0$ . That is, potential energy is transformed into kinetic energy as the particle escapes. Notice that the width of the barrier decreases as $E$ increases. The tunneling probability depends very sensitively on the barrier width, as you learned in conjunction with the scanning tunneling microscope. Thus an alpha particle in a higher energy level should have a shorter half-life and escape with more kinetic energy. The full analysis is beyond the scope of this text, but this prediction is in excellent agreement with measured energies and half-lives.

42.6.1 Alpha Decay

FIGURE 42.18 Alpha decay.

EXAMPLE 42.5 Alpha decay of uranium

FIGURE 42.19 The potential-energy diagram of an alpha particle in the parent nucleus.