Problem on tritons and deuterons

[krateesh]

Suppose we have a beam balance capable of measuring even smallest differences in weights.
Now on one side we have a Triton and Deuteron.
And on the other side we have He-nucleus and a neutron

I want to know which will weigh heavier and why?

I saw on a book cover with balance tilting towards the former one thus i concluded that the deuterons and tritons will weigh more but my explanation on that conclusion don't seem to be logical.

Is that book cover right or I have been just fooled and my fumndamentals are just checked?

 

[Rade]

See table 3.1 in this report: http://book.nc.chalmers.se/KAPITEL/CH03NY3.PDF
The mass of the triton + deuteron ~ 5.030 u, that of helium-4 + neutron ~ 5.011 u. The book cover is correct.

 

[krateesh]

This means helium nucleus is more stable as it do have such even arrangement of both protons and neutrons.

 

[gjshep]

This means helium nucleus is more stable as it do have such even arrangement of both protons and neutrons.

Why would even=stable? Yes, He4 is stable AND seems to be a fundamental building block for heavier elements as He4 is often a decay component. I believe there is a fundamental "structure" to He4 that gives it its stability, though I will admit that the "even" P/N count may provide for this stable structure.

 

[Rade]

Having an equal number of protons and neutrons does not always result in stability--for example see this isotope chart for Beryllium:
http://environmentalchemistry.com/yogi/periodic/Be-pg2.html#Nuclides
Note that Beryllium-8, which has 4 P and 4 N [NNNNPPPP] is not stable.

 

[Meir Achuz]

2p + 2n is the smallest number of nucleons that can fit into the same ground S-state. At the same time, each of these nucleons feels the attractive force of the nearby other three. That is why He^4 is the stablest nucleus.
Be^8 is not stable because it decays ilnto a pair of these most stable He^4.

 

[gjshep]

2p + 2n is the smallest number of nucleons that can fit into the same ground S-state. At the same time, each of these nucleons feels the attractive force of the nearby other three. That is why He^4 is the stablest nucleus.

I wouldn't categorize that explanations as "why", but it's the current leading theory. "feels attractive force" is a concept with no structure, but I don't disagree. True, He^4 is the most stable known nuclei, but there may still be baryonic nuclear structures "larger" but more stable. One possible candidate would be a baryonic neutrolino, but the current theory gives a huge range of energy values and no composition or structure. Another could be n^4 - no charge, no magnetic moment, very stable and very dense. The later is predicted by W.F. Hagen's second paper on Energiewirbel.

Be^8 is not stable because it decays ilnto a pair of these most stable He^4.

Hmm.. How about Be^8 is not stable AND it decays into a pair of He^4. Why it is unstable remains unexplained. This decay example does however reinforce one of my first points that He^4 is a building block of larger nuclei.

 

[Rade]

One hypothesis as to why Be-8 is unstable derives from the Close-Packed Spheron Model of Linus Pauling. Here is quote from Pauling: "I assert that nuclei beyond A = 4 are stable only if two or more resonating structures involving hydrogen-3, helium-3, and helium-4 can be written". Dr. Linus Pauling, Research Notebook, #26 . 13 February 1966 :http://osulibrary.oregonstate.edu/specialcollections/rnb/26/rnb26.html

According to this Pauling model there "must" be present a three nucleon cluster, either triton [NPN] or [PNP], rotating against either (1) another three nucleon cluster or (2) a helium-4 cluster (the alpha)for any isotope with A > 4 to be "stable". Hence, Be-8, being composed of two helium-4 rotating clusters [PNPN]~[PNPN] violates this basic "rule of assembly" dynamic and is thus "unstable". Note the same explanation holds for He-5, in fact no mass 5 or 8 isotopes are stable. Now, before anyone starts throwing darts, Linus Pauling was a genius, holds two Nobel Prizes-- and is it not interesting that his model that explains why no stable mass 5 or 8 isotopes exist, published in Science and Proc. National. Acad. Sci. in 1965, has never been falsified.

 

[gjshep]

One hypothesis as to why Be-8 is unstable derives from the Close-Packed Spheron Model of Linus Pauling. Here is quote from Pauling: "I assert that nuclei beyond A = 4 are stable only if two or more resonating structures involving hydrogen-3, helium-3, and helium-4 can be written". Dr. Linus Pauling, Research Notebook, #26 . 13 February 1966 :http://osulibrary.oregonstate.edu/specialcollections/rnb/26/rnb26.html

According to this Pauling model there "must" be present a three nucleon cluster, either triton [NPN] or [PNP], rotating against either (1) another three nucleon cluster or (2) a helium-4 cluster (the alpha)for any isotope with A > 4 to be "stable". Hence, Be-8, being composed of two helium-4 rotating clusters [PNPN]~[PNPN] violates this basic "rule of assembly" dynamic and is thus "unstable". Note the same explanation holds for He-5, in fact no mass 5 or 8 isotopes are stable. Now, before anyone starts throwing darts, Linus Pauling was a genius, holds two Nobel Prizes-- and is it not interesting that his model that explains why no stable mass 5 or 8 isotopes exist, published in Science and Proc. National. Acad. Sci. in 1965, has never been falsified.

Great link, thanks!