Talk:Isotopes of promethium

OO61Pm 142

This isotope is noted to have an uncharacteristically low half-life (40.5 seconds) with reference to the half-lives of its nearest OO neighbors, namely OO61Pm144 (360days), and OO61Pm140 (5.87 minutes), thus causing an irregularity in the OO stability profile line.WFPM (talk) 19:49, 29 September 2011 (UTC)[reply]

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How does the decay of Promethium-147 "work"?

This list and all other sources I can find says that Promethium-147 decays to Samarium-147 by emitting a beta particle. The mass of Promethium-147 is 146.9151385 amu the mass of Samarium-147 is 146.9148979 amu. The difference between the two masses is 0.0002406 amu. A beta particle is nothing other than an electron, which has a rest mass of 5.48579909070×10−4 amu or 0.000548579909070 amu. That means the mass of the electron is greater than the mass difference between the two nuclei. This is impossible, isn't it? If the difference in masses were greater than the mass of the electron, that could be explained by it being carried away in the kinetic energy of the electron or in gamma rays. But you cannot gain mass without consuming energy. And where is this energy coming from? There is about a half electron mass missing (notice that I cut off the "uncertainty" in the brackets in each case and the order of magnitude of the problem is well outside the claimed measurement error). So... How does Promethium-147 emit an electron, create Samarium-147 and in the end there's mass to spare? Where is the mistake? An electron capture reaction can of course produce a mass defect that is smaller than an electron mass, but it cannot ever produce a net mass gain. To gain mass, there must be energy coming from somewhere. Where? Hobbitschuster (talk) 14:59, 19 February 2022 (UTC)[reply]

The actual decay process is a bit more complicated. The mass of ¹⁴⁷
Pm
(which you quote) is

61~m_{H}~(=61~m_{p}+61~m_{e})+86~m_{n}-E_{binding}

. For ¹⁴⁷
Sm
, we have

62~m_{H}~(=62~m_{p}+62~m_{e})+85~m_{n}-E_{binding}

. Beta decay thus sees a neutron converted into a proton and electron. the electron then orbits the newly formed Sm nucleus as the 62nd electron, otherwise every beta decay would produce an ion. The remaining energy (about half an electron mass) is carried away by the photon, so all the conservation laws hold. In the end, there is no missing mass, as the mass of ¹⁴⁷
Sm
already includes the additional electron from beta decay, and a photon does the rest. ComplexRational (talk) 15:53, 19 February 2022 (UTC)[reply]

This isn't quite accurate. Bound state beta decay, where the emitted electron is captured into an orbit around the nucleus, is actually fairly rare, as the highest possible binding energy for an electron (for Hydrogen) is 13.6 keV. Most nuclei have substantially lower electron binding energy, and most emitted electrons have substantially higher kinetic energy. And that's the answer to the question - the energy goes into the kinetic energy of the emitted electron (and the neutrino, a tiny bit).

In terms of conservation of charge, neutral Promethium-147 has 61 protons, 86 neutrons, and 61 electrons. Neutral Samarium-147 has 62 protons, 85 neutrons, and 60 electrons. To keep things neutral (and not ionized) the electron MUST escape, to balance the conversion of a neutron to a proton in the nucleus. PianoDan (talk) 16:44, 19 February 2022 (UTC)[reply]

So the solution to the problem is basically to ignore the fact that beta decay involves a beta particle because atomic weights are always given for the neutral atom and the beta particle basically is automatically included on whichever side of the reaction one looks at, because the weights are given for the atoms including all electrons... That's an answer I didn't expect, but come to think of it, I could've thought of. Thanks for clearing that up! Hobbitschuster (talk) 16:46, 19 February 2022 (UTC)[reply]

wait, so now I'm more confused than I was before... We can all agree that (ignoring the beta particle) the Promethium-147 has a higher atomic weight than the Sanarium-147. That difference is lower than the mass of an electron. So if the mass of the electron must be ignored, it works out. But if you have to account for the mass of the electron, the Q-value is negative... Hobbitschuster (talk) 16:50, 19 February 2022 (UTC)[reply]

I was unclear above - by "not ionized" I meant the whole system. Beta decay DOES produce ions. PianoDan (talk) 17:48, 19 February 2022 (UTC)[reply]

Sorry about the confusion. I should rather say that the "missing" mass is already factored into the calculation when taking the masses of promethium-147 and samarium-147. The electron may subsequently escape and leave behind an ion, and charge conservation within the system still holds. ComplexRational (talk) 17:58, 19 February 2022 (UTC)[reply]

wouldn't it be easier to do Q-value calculations with only the masses of the nuclei and ignoring all other particles unless they come from or "disappear in" the core? Hobbitschuster (talk) 20:12, 19 February 2022 (UTC)[reply]

Positron emission of ¹⁴⁴Pm

According to page 100 of NUBASE2020, the possibility of positron emission among beta plus decay of ¹⁴⁴Pm is <8×10^-5%. which is extremely low considering the very large energy difference between ¹⁴⁴Pm and ¹⁴⁴Nd (2.3321 MeV). 129.104.241.214 (talk) 00:45, 24 December 2023 (UTC)[reply]

Possible alpha decay of ¹⁴⁶Pm and ¹⁴⁷Pm

According to [1], ¹⁴⁶Pm and ¹⁴⁷Pm should have a partial alpha decay half-life at the order of 10¹⁸ years and 10²² years. 129.104.241.214 (talk) 11:28, 27 January 2024 (UTC)[reply]

The section Promethium-147

"The isotopes ¹⁴²Nd, ¹⁴³Nd, ¹⁴⁴Nd, ¹⁴⁵Nd, ¹⁴⁶Nd, ¹⁴⁸Nd, and ¹⁵⁰Nd are either stable or nearly so": It would be more clear to say that all these isotopes are beta-stable, including the non-stable ¹⁴⁴Nd and ¹⁵⁰Nd. 14.52.231.91 (talk) 00:47, 26 August 2024 (UTC)[reply]

@Double sharp: I saw that the wording had been changed. That's much better, thanks a lot! 129.104.241.231 (talk) 11:53, 19 September 2024 (UTC)[reply]