When reading about the history of the discovery of the neutrino I came across the following paragraphs in my lecturer's notes:

(...) physicists were puzzled by the continuous spectrum of $\beta$-decays. In this process, a nucleus transforms itself into another one with the emission of an electron:

$${}^A_ZX \to {}^A_{Z+1}X' + e^−(+...) . \tag{1}$$

Only the parent and daughter nuclei and the electron could be seen. Based on these observations, according to energy-momentum conservation, the electron should carry away an energy corresponding to the difference in mass between parent and daughter nuclei. Therefore there should be a monochromatic line in the $\beta$-spectrum, in disagreement with observations.

I don't understand how would a monochromatic line in the $\beta$-spectrum indicates that the electron carries the energy corresponding to the mass difference? How would this neglect the concept of a neutrino, if it was true of course.


1 Answer 1


If a particle (or nucleus) decays to 2 daughter particles then their energies are fixed. This is easy to see. If the parent is at rest and one of them has momentum $\vec p$ then the other must have $-\vec p$, by conservation of momentum. If the energy available is $\Delta E$ then conservation of energy says $\Delta E = p^2/2m_1 + p^2 /2m_2$, or the relativistic equivalent if that's applicable. So in a given decay ($\Delta E, m_1,$ and $ m_2$ are specified) you always have the same $p$ and a daughter has the same energy. This is what happens in $\alpha$ decay and $\gamma$ decay.

If a particle (or nucleus) decays to 3 daughter particles then more options are open. $\Delta E$ can be shared between them in different ways, still conserving momentum. Interesting limiting cases occur if one particle has zero momentum, and the other two are then back to back - so the electron can have a kinetic energy of 0, all the way up to its maximum when the daughter nucleus is at rest and the energy is shared between the electron and the neutrino. Which is what happens in $\beta$ decay.

Another way of looking at it is that for 2 body decays there are 6 parameters (the momentum components of the two daughters) and 4 energy-momentum constraint equations, leaving only two free parameters, which correspond to the angular orientation of the decay. For 3 body decays there are 9-4=5 free parameters, which can be written as the three Euler angles of the decay plane and two relative energy fractions.

  • $\begingroup$ Unless I’m misunderstanding, doesn’t a monochromatic line represent a fixed energy? But the momentum $\vec p$ is free to vary unconstrained. This wouldn’t lead to a monochromatic line. $\endgroup$ Aug 21, 2020 at 12:37
  • $\begingroup$ @SuperfastJellyfish - in the two body situation, energy and momentum conservation are directly linked, as indicated in the answer. There are no free parameters. In a 3 body problem that is no longer the case. $\endgroup$
    – Jon Custer
    Aug 21, 2020 at 15:42
  • $\begingroup$ @JonCuster But can’t $\Delta E$ be anything greater than binding energy? Then the excess will be kinetic energy of the constituents. This way the kinetic energies can be anything above that threshold just like in photoelectric effect. $\endgroup$ Aug 21, 2020 at 17:45
  • $\begingroup$ @SuperfastJellyfish - the constraints to satisfy both energy and momentum conservation mean there are no free parameters. The energy and momentum of one object are directly related as are the energy and momentum of the other. $\endgroup$
    – Jon Custer
    Aug 21, 2020 at 17:48
  • $\begingroup$ $\Delta E$ is the binding energy. That is, $c^2$ times the excess of the mass of the parent over the sum of the masses of the two (or three, or whatever) constituents. $\endgroup$ Aug 22, 2020 at 13:09

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