KAHIBARO
Table of Contents
Energy Release in Alpha Decay
In alpha decay, a nucleus emits an alpha particle, which is a helium nucleus containing 2 protons and 2 neutrons. The key question in this chapter is how much energy is released in this process, and how that energy is shared between the emitted alpha particle and the daughter nucleus.
A typical alpha decay can be written as
$$
{}^{A}_{Z}X \rightarrow {}^{A-4}_{Z-2}Y + {}^{4}_{2}\alpha
$$
where $X$ is the parent nucleus and $Y$ is the daughter nucleus.
The Q-value of Alpha Decay
The energy released in alpha decay is called the $Q$-value. It comes from the difference in mass between the initial nucleus and the final products. If the parent nucleus has a greater mass than the combined mass of the daughter nucleus and alpha particle, the missing mass appears as released energy.
The $Q$-value is
$$
Q = \left[M_X - M_Y - M_\alpha\right]c^2
$$
where $M_X$ is the mass of the parent nucleus, $M_Y$ is the mass of the daughter nucleus, and $M_\alpha$ is the mass of the alpha particle.
If $Q > 0$, the decay can release energy spontaneously. If $Q < 0$, the decay is not energetically allowed.
For alpha decay to occur spontaneously, the released energy must satisfy
$$
Q = \left[M_X - M_Y - M_\alpha\right]c^2 > 0
$$
A positive mass defect means energy is released.
Using Atomic Masses
In practice, atomic masses are often used instead of bare nuclear masses. This works conveniently for alpha decay because the electron numbers balance on both sides.
The parent atom has $Z$ electrons. The daughter atom has $Z-2$ electrons, and the helium atom associated with the alpha particle has 2 electrons. The total number of electrons is the same before and after decay, so electron masses cancel.
Therefore, using atomic masses,
$$
Q = \left[M_{\text{atom}}(X) - M_{\text{atom}}(Y) - M_{\text{atom}}(\text{He})\right]c^2
$$
This is very useful in nuclear data tables.
Conversion to Energy Units
Since nuclear masses are very small, it is common to express the released energy in mega electron volts, MeV. The standard conversion is
$$
1\ \text{u} \, c^2 = 931.5\ \text{MeV}
$$
If the mass difference is given in atomic mass units, then
$$
Q(\text{MeV}) = \Delta m(\text{u}) \times 931.5
$$
where
$$
\Delta m = M_X - M_Y - M_\alpha
$$
Important conversion:
$$
Q(\text{MeV}) = \Delta m(\text{u}) \times 931.5
$$
This formula is used constantly in nuclear energy calculations.
How the Released Energy Is Shared
The parent nucleus is usually at rest before decay. After decay, both the alpha particle and the daughter nucleus move away from each other. Because momentum must be conserved, they have equal magnitudes of momentum in opposite directions.
If $p$ is the momentum of each product, then their kinetic energies are
$$
K_\alpha = \frac{p^2}{2M_\alpha}, \qquad K_Y = \frac{p^2}{2M_Y}
$$
The total released energy becomes
$$
Q = K_\alpha + K_Y
$$
Since the alpha particle is much lighter than the daughter nucleus, it receives most of the kinetic energy.
From the ratio of kinetic energies,
$$
\frac{K_\alpha}{K_Y} = \frac{M_Y}{M_\alpha}
$$
So the lighter alpha particle gets the larger share.
Solving for each kinetic energy gives
$$
K_\alpha = Q \frac{M_Y}{M_Y + M_\alpha}
$$
and
$$
K_Y = Q \frac{M_\alpha}{M_Y + M_\alpha}
$$
Because $M_Y \gg M_\alpha$, usually
$$
K_\alpha \approx Q
$$
and the daughter recoil energy is much smaller.
In alpha decay,
$$
Q = K_\alpha + K_Y
$$
and
$$
K_\alpha = Q \frac{M_Y}{M_Y + M_\alpha}, \qquad
K_Y = Q \frac{M_\alpha}{M_Y + M_\alpha}
$$
The alpha particle carries most of the released energy.
Recoil of the Daughter Nucleus
The daughter nucleus cannot remain at rest after the alpha particle is emitted. If the parent nucleus started at rest, then conservation of momentum requires
$$
\vec p_\alpha + \vec p_Y = 0
$$
So the daughter nucleus recoils in the opposite direction.
This recoil is important because it means the alpha particle energy is slightly less than the full $Q$-value. The missing small part goes into the daughter nucleus kinetic energy.
Example Calculation
Suppose the mass difference in an alpha decay is
$$
\Delta m = 0.00500\ \text{u}
$$
Then the released energy is
$$
Q = 0.00500 \times 931.5\ \text{MeV} = 4.66\ \text{MeV}
$$
If the daughter nucleus is much heavier than the alpha particle, then the alpha particle gets almost all of this energy, roughly a little less than $4.66\ \text{MeV}$, while the daughter gets a small recoil energy.
Typical Alpha Decay Energies
Alpha decay energies are usually in the range of a few MeV. This is large on the nuclear scale, even though it is tiny on an everyday scale.
| Quantity | Typical value |
|---|---|
| Alpha decay $Q$-value | $4$ to $9\ \text{MeV}$ |
| Alpha particle kinetic energy | Nearly the full $Q$-value |
| Daughter recoil energy | Small fraction of $Q$ |
These characteristic energies are one reason alpha particles from a given radioactive nucleus often appear with well-defined energies.
Excited Daughter Nuclei
Sometimes the daughter nucleus is produced not in its lowest energy state, but in an excited state. In that case, part of the available decay energy goes into nuclear excitation instead of kinetic energy.
Then the effective kinetic energy release is smaller:
$$
Q_{\text{kin}} = Q - E^*
$$
where $E^*$ is the excitation energy of the daughter nucleus.
So the emitted alpha particle may have less kinetic energy than expected from the ground-state mass difference alone.
If the daughter nucleus is left excited, then
$$
Q_{\text{kin}} = Q - E^*
$$
Only the remaining energy appears as kinetic energy of the decay products.
Why Alpha Energies Are Discrete
Unlike beta decay, alpha decay usually produces alpha particles with specific, discrete energies. This happens because nuclei have discrete energy levels. If decay occurs to one particular daughter state, the released energy is fixed, so the alpha particle energy is also fixed apart from the small recoil correction.
If several daughter energy levels are possible, several distinct alpha energies may be observed.
Summary Relations
The central idea of this chapter is that alpha decay energy comes from mass difference and appears mainly as kinetic energy of the emitted alpha particle and, to a smaller extent, as recoil of the daughter nucleus.
| Relation | Meaning |
|---|---|
| $Q = \left[M_X - M_Y - M_\alpha\right]c^2$ | Energy released |
| $Q = K_\alpha + K_Y$ | Energy goes into kinetic energy |
| $\frac{K_\alpha}{K_Y} = \frac{M_Y}{M_\alpha}$ | Lighter particle gets more kinetic energy |
| $Q(\text{MeV}) = \Delta m(\text{u}) \times 931.5$ | Mass to energy conversion |
Key result of alpha decay energetics:
The released energy is determined by the mass defect, and because of momentum conservation, most of that energy appears as kinetic energy of the alpha particle, with a small part going to recoil of the daughter nucleus.