# What are continuous energy spectra?

Apr 11, 2018

"Continuous Energy spectra" in nuclear chemistry typically refers to the fact that kinetic energy of electrons (or positrons) released in beta decays can take any value from a specific range of energies.

#### Explanation:

The sum of all energy released in a nuclear reaction can be calculated from mass defect, the difference in the mass of the products and the reactants, by the equation $E = m \cdot {c}^{2}$.

The amount of mass lost in a particular process of beta is definite, meaning that the sum of kinetic energy of all product particles shall have discrete values.

It is possible to set up a system of equations to solve for the final kinetic energy of the electron:

1. The sum of final kinetic energy of the nucleus and the electron equals to the energy released in the decay;
2. Momentum conserves

This system will yield a finite number of solution (one or two) if the decay produces only two particles: the nucleus and an electron. Hence one might expect to detect electrons that travel at some particular velocity near a collection of nucleus undergoing beta-minus decay.

However, experimental results disagree with the prediction; instead of giving discrete points, plotting kinetic energy against the number of particles possessing that amount of energy will produce a distribution similar to that of a Maxwell-Boltzmann distribution curve.

Both energy and momentum still have to conserve; the release of the antineutrino in beta-plus decays (or neutrino for beta-minus decays) as a third product of beta decays accounts for the continuous energy spectra. For a general beta-minus decay:
${\textcolor{w h i t e}{X}}_{Z}^{A} X \to$

${\textcolor{w h i t e}{X}}_{Z + 1}^{A} X ' + {e}^{-} + \overline{v}$

The sum of kinetic energy of all three products- the daughter nuclei, the electron, and the antineutrino- is a definite value.
$K {E}_{\text{X'" +KE_(e^-)+KE_(bar v)="Mass Defect}} \cdot {c}^{2}$

Hence
$K {E}_{{e}^{-}} = \Delta m \cdot {c}^{2} - K {E}_{\text{X'}} - K {E}_{\overline{v}}$

The mass of the nucleus is much larger than that of the electron and the antineutrino such that its share of kinetic energy is negligible; the kinetic energy of the antineutrino, however, can vary significantly from $0$ all the way to $\text{Mass Defect} \cdot {c}^{2}$- where it takes all the energy from the decay. The presence of a third particle makes it impossible to find unique solutions to the kinetic energy of the electron; similarly, it can take more than one possible value ranging from $0$ all the way to $\text{Mass Defect} \cdot {c}^{2}$ and therefore gives a continuous energy spectrum.