Neutrino was discovered from the seemingly violation of conservation laws.

Supposedly, it was suprising to scientists, when they found electrons were emmited at various energies during beta decay Hence, the continuous range of energies was accounted for using this new particle, the neutrino.

However, it is unclear to me why other forms of decay must be discrete. Why was it suprising that electrons were emitted at various energies? Consider alpha decay. A parent nucleus decays into a daughter nucleus and an alpha particle. Couldn't the decay energy be variably shared across the daughter nucleus and the alpha particle, such that the alpha particle is emitted with a range of energies as well?

Watching a video, I found the following explanation.

When a nuclear reaction takes place, some amount of energy is released. This energy released can be predicted using the mass defect. From this energy, we can then predict the kinetic energy of the emitted particle, as the daughter nuclei is so massive compared to the emitted particle. Hence, the vast amount of decay energy is carried off by the emitted particle. So we can theoretically predict the kinetic energy of this emitted particle.

What is not clear to me, is why the emitted particle carries nearly all the kinetic energy. Kinetic energy is $\frac{1}{2}mv^2$. As the daughter nuclei has much more mass, wouldn't the kinetic energy be roughly the same between the daughter nuclei and the emitted particle?

It is unclear to me why the decay energy cannot be shared in variable proportions with the daughter nuclei and the alpha particles, therefore giving a continuous range of energies for the alpha particle. What is wrong with my intuition?

  • 1
    $\begingroup$ I think this question is a duplicate of the following question: Why is the spectrum of the β-decay continuous? $\endgroup$
    – Cleonis
    Commented Oct 7, 2023 at 4:11
  • 1
    $\begingroup$ In the answers to the above question the following things are pointed out: in a two-body decay the resulting velocities of the decay products are on a single line, with momentum conservation constraining how the energy distribution comes out. In a three-body decay the velocities of the decay products are in a plane, but for the angles there is a degree of freedom, allowing a continuous spectrum. $\endgroup$
    – Cleonis
    Commented Oct 7, 2023 at 4:23
  • $\begingroup$ What is conserved is the energy of the system, not just the kinetic energy $\endgroup$
    – anna v
    Commented Oct 7, 2023 at 5:01

2 Answers 2


It is unclear to me why the decay energy cannot be shared in variable proportions with the daughter nuclei and the alpha particles.

The decay must conserve both energy and momentum. For a two-body decay, there's only one way to accomplish this.

Let's consider the decay of a heavy nucleus into an alpha particle with mass $m=4\rm\,u$ and a heavy daughter with mass $M\gg m$. In the rest frame of the parent nucleus, the decay products have momenta obeying $$ \vec p_\text{nucleus} + \vec p_\alpha = 0 $$ and kinetic energies obeying $$ \frac{p_\text{nucleus}^2}{2M} + \frac{p_\alpha^2}{2m} = Q, $$ where the value $Q$ (sometimes called the "$Q$-value") is the total energy liberated in the decay. The momentum equation tells us that we need consider only the magnitude of the momentum $p_\text{nucleus} = p_\alpha = p$, and the energy equation tells us that $p = 2Q \cdot \left( M^{-1} + m^{-1} \right)^{-1}$ is uniquely determined.

You can see from the energy relation $p^2/2m$ that, for a given momentum, the lower-mass particle will carry more of the kinetic energy.

To conserve energy and momentum in a three-body decay, you have two equations but three unknowns to conserve momentum and energy. The third degree of freedom is convenient in different places, depending on exactly what you're trying to compute, but a nice way to think of it is the angle between the electron and neutrino momenta.

The comments about the ratio of masses are a red herring in a question about why the energies are discrete instead of continuous. For another example, consider the disintegration of lithium as induced by thermal neutrons, $$ \rm n + {^6Li} \to {^3H} + {^4He} $$ This reaction is quite efficient with thermal neutrons, who have energies best measured in milli-eV, while its $Q$-value is better measured in mega-eV. The relatively tiny initial momentum in thermal neutron capture makes it reasonable to think of the neutron-lithium system as being initially at rest. But when you neglect the incident neutron's energy, the two-body final state is discrete by the same arguments as above.

  • $\begingroup$ Great thanks for the explanation! $\endgroup$ Commented Oct 7, 2023 at 6:35

Don't use velocity, use:

$$ E = \frac{p^2}{2M}$$

Then, in the rest frame of the nucleus:

$$ \vec p_N + \vec p_{\alpha} = 0 $$


$$ p_N = p_{\alpha} \equiv p$$

so the energies are:

$$E_N = \frac{p^2}{2M_A}$$ $$ E_{\alpha} = \frac{p^2}{2m_{\alpha}}$$

and the ratio of alpha energy to nuclei energy is:

$$ R = \frac{E_{\alpha}}{E_A} = \frac{M_A}{m_{\alpha}} \approx \frac A 4$$

and heavy nuclei alpha decay, say $A\approx 200$, so the ratio is around 50.

For a beta, the maximum energy is around $1836A$...and that it was variable was a problem, requiring the postulation of a light mass (that's another story) neutral particle that interacts weakly.


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