I know atoms undergoing alpha decay emit alpha particles, but I was wondering why specifically helium nuclei? If these atoms wanted to emit the smallest unit of matter (I'm talking about hadrons here, not subatomic) they could just emit hydrogen nuclei, or if they wanted to emit the smallest unit of protons/neutrons attached together they could just emit deuterium. Yet for some reason, they choose to emit helium nuclei. Any thoughts?
Asked
Active
Viewed 1,240 times
12
-
3An alpha particle is tightly bound (per nucleon) and low mass such that tunneling is easier. Wikipedia covers the general basics in its article on 'Alpha decay', so perhaps you could read that over and come back with a more specific question? (Note that I don't think the nucleus has a choice in the matter - I doubt it has free will). – Jon Custer Jul 31 '18 at 21:49
-
1@John Custer- I read Wikipedia, but all I found is information on how the alpha particles are emitted rather than why alpha particles are always helium nuclei. Also, I agree that helium nuclei are tightly bound, but according to the answers to the question on Quora, they are not tighter than deuterium ( see https://www.quora.com/How-is-it-that-an-atom-of-helium-weighs-less-than-2-hydrogen-atoms for details). Moreover, one deuterium atom is much lighter than a helium atom. – Anthony Ducharme Jul 31 '18 at 22:00
-
3https://en.wikipedia.org/wiki/Alpha_decay is the correct article, and has a section on the mechanism and various aspects of 'why alphas'. It points out the need to preserve wave function symmetry which helps rule out deuterium. Note that D has 1112keV per nucleon binding energy, while 4He has 7073keV per nucleon, so 4He is much more tightly bound than D. – Jon Custer Jul 31 '18 at 22:13
-
Ok, so I read the article you sent but from what I could tell it does not rule out deuterium. The Bose-Einstein/Fermi-Dirac statistics prohibit alphas being an odd number nuclei. Also, it seems we both misunderstood how binding energy plays a role here. What matters is the binding energy between the alpha and the parent nucleus rather than the binding energy of the alpha itself. The equation they used seemed to exhibit a pattern where the lower the mass of the alpha the more favorable the value of the binding energy is. I could be wrong though. – Anthony Ducharme Jul 31 '18 at 23:06
2 Answers
0
The 'Q' value is positive for only the emission of alpha particle as compared to other light nuclei including the neutron. To understand the 'Q' value, refer some standard text book of Nuclear Physics like one authored by Krane.
-
it’s not terribly useful to answer by suggesting someone read a book. Maybe you can expand your answer and make it self-contained? – ZeroTheHero Apr 27 '20 at 13:30
-
Q value is the energy equivalent of net change in the mass of entities during a nuclear reaction. For Ex, if we take decay of U 234 through deuterium nucleus the equation would be : (234,92) U = (2,1) H + (232, 91) Pa. Now, Q factor is defined as (Atomic Mass of U - Atomic mass of Pa - Atomic mass of Deuterium) multiplied by square of speed of light. The value you get is negative, therefore, the decay is not possible. If you include alpha particle i.e (4,2) He in place of (2,1) H , the Q factor becomes positive and hence the decay is possible. – vicky gautam Apr 28 '20 at 09:01
-4
The atoms are lazy and do what costs them the least effort.
On a more serious note, yes, that captures it. The He nucleus has a high binding energy of 7 MeV per nucleon. Other light nuclei have much smaller values. Nuclear fission occurs when the daughter nuclei have a higher binding energy per nucleon then the parent. This means that reactions emitting He is an energetically very attractive light candidate daughter nucleus.
my2cts
- 24,097
-
7Please consider giving answers that actually teach the reader something instead of pithy one-liners that are only intellegible to those already having an idea of what the answer might be. – ACuriousMind Jul 31 '18 at 21:40
-
But if that's true then why do they choose helium? If it is in terms of mass, hydrogen should theoretically take less effort to emit. But if it were in terms of neutron/proton ratio, deuterium should take the least effort to emit. – Anthony Ducharme Jul 31 '18 at 21:40
-
-
5It seems that some users are lazy too and answer what costs them the least effort. – user171780 Jul 31 '18 at 22:06
-
@user171780 After midnight I sometimes get lazy. As long as my answers are correct, so what ? See https://physics.stackexchange.com/questions/403460/why-is-moon-light-cold/403470#403470 – my2cts Jul 31 '18 at 23:00
-
4@my2cts It is worth noting that the tooltips for the up- and down-vote controls read "This answer is useful" and "This answer is not useful". A correct but unhelpful answer is a bad answer by that standard. – dmckee --- ex-moderator kitten Aug 01 '18 at 00:04
-
@my2cts Such posts frequently get flagged as VLQ and NAA, and I've noticed them in the review queues. Although they do not qualify for deletion, the fact that they're being flagged indicates how the community thinks of them. This post, yours post here and here, among several others, while correct, could use a some more description. – Aug 01 '18 at 04:13
-
-
@my2cts But the whole point of SE is to reduce the signal-to-noise ratio on the internet (I think Jeff Atwood said that somewhere). Answering in a humorous way counts as noise, and it's not the same as slipping a little pun into answers every now and then. – Aug 01 '18 at 07:14
-
@Chair Perhaps my formulation distracted from the fact the answer is accurate ? – my2cts Aug 01 '18 at 07:46
-
1@my2cts Exactly. The answers are usually correct, but as ACM said, it's only apparent if you know the more elaborate justification. – Aug 01 '18 at 07:47
-
There are a lot of critical comments here, but very few answers. It would have been far better for the "signal to noise ratio" if the commenters had spent their effort to write better answers than mine. Commenting others is lazy. – my2cts Apr 27 '20 at 20:33