0
$\begingroup$

The International Atomic Energy Authority (IAEA)'s Nuclear Data Services list tables of nuclear data, including a table of atomic masses and beta decay energies, data taken from Huang et al., Chin Phys C, volume 45, 2021.

The decay energy for most $\beta$- decays is positive, as one would expect from the event which must be exothermic as the ejected electron has kinetic energy. However, some of the decay energies in the table linked above (and the original paper from where the data was taken) is negative.

For example, we can see that the beta-decay energy of $^{40}_{17}\textrm{Cl}$ is 7480 keV, whereas the beta-decay energy of $^{40}_{18}\textrm{Ar}$ is -1504 keV. Does this mean that the $\beta$- of $^{40}_{18}\textrm{Ar}$ is endothermic? It seems unbelievable to me that a mass of $^{40}_{18}\textrm{Ar}$ would cool its environment due to decay.

$\endgroup$
2
  • 1
    $\begingroup$ Better question: would it decay through that channel? $\endgroup$
    – Jon Custer
    May 30 at 12:38
  • 2
    $\begingroup$ Ar-40 is a stable isotope of Argon. $\endgroup$
    – Farcher
    May 30 at 15:42

2 Answers 2

1
$\begingroup$

From comments on the question it appears the answer is: it wouldn't decay.

$\endgroup$
-1
$\begingroup$

When a negative beta electron leaves the nucleus, its potential energy is negative until it leaves the pull of the nucleus. If it escapes with some velocity, then its total energy was positive. If it gets pulled back (possibly into an orbital) then its total energy was negative.

$\endgroup$
3
  • $\begingroup$ The binding energy of the most tightly bound electron in argon is 3.6 keV, far too small to balance the 1504 keV in the question. $\endgroup$
    – John Doty
    May 30 at 17:38
  • $\begingroup$ So, the negative energy would have to close to zero for the electron to achieve orbit status? $\endgroup$
    – R.W. Bird
    May 31 at 14:00
  • $\begingroup$ Yes. Very close to zero for a neutral atom, since all the tightly bound orbitals are already occupied. $\endgroup$
    – John Doty
    May 31 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.