Official Nuclid Data

• Wallet Card at Brookhaven, USA
• KAERI Table of Nuclides
• Nuclid Data at Berkely, USA
• Swedish Data
• Japanese Nuclear data
• old korean nuclides chart mirrored

Other data sources

• Table of Isotopes to buy
• Karlsruher Nuklidkarte to buy
• Good introduction
• single html chart

Quality of nuclide data/discovery

1. dubious, contradictive data [118-293]
2. only Z and A of the nuclid published [Rf-264] list of only known-to-exist nuclides
3. 2) and (one) mode of decay given [Hs-263] list of shallow known nuclides
4. 3) and (estimation of) T_1/2 given [Lr-262]
5. 4) and spin given [Fm-255]
6. 5) and mass given [Sg-265]

Distribution of "stable" nuclides

• There are 277 nuclides with Z <= 83 and which are stable to simple beta-decay.
• Of these 277 nuclides, 68 may decay by double-beta-decay.
• Also, of these 277 nuclides 84 may decay by alpha-decay.
• Therefore 147 nuclides are stable to any kind of alpha or beta decay.
• One of these 147, namely He-5, decays by n-emittion in about 10^-21 sec.
• and 54 are instable to cluster-decay (C-12, O-16, Ne-20, e.t.c.).
• But, as of March 2002, there are 237 nuclides for which a decay (still) has not been observed.
• There are 275 nuclides which have half life > 10¹² years, none of these has Z > 83.
• Only 6 of these 275 nuclides are beta-instable (In-115, Te-123, Cd-113, V-50, Ca-48, Zr-96), but Ca-48 and Zr-96 decay by double beta-decay.
• There are 9 further nuclides having a half life > 10⁹ years, namely Pt-190, Sm-147, La-138, Rb-87, Re-187, Lu-176, Th-232, U-238, K-40.
• There are no stable isobares of A= 5, 8, 147, 149, 182, 183, 184, 186 and >= 210.
  Moreover there are no stable isotopes for Z=43, 61, 74(Wolfram) and >= 84 --- further to discover.

Distribution of half life of (groundstate) nuclides

Observed Modi of Decay

alpha-decay

emittion of an He-4 nuclid: ^a_zX --> ^a-4_z-2Y^2- + ⁴₂He²⁺

beta-decay

the number of nucleons stay constant. Transformation from neutrons into protons or reverse.

beta- -decay

^a_zX --> ^a_z+1Y⁺ + e^-
An electron will be emitted.

beta+ -decay

^a_zX --> ^a_z-1Y^- + e⁺
A positron will be emitted.

Electron capture

^a_zX --> ^a_z-1Y
An inner electron of the atom will be captured by its parent nucleus and transform so a proton-electron pair into a neutron.

double-beta-decay

either ^a_zX --> ^a_z+2Y²⁺ + 2e^- or ^a_zX --> ^a_z-2Y^2- + 2e⁺

p emission

^a_zX --> ^a-1_z-1Y^- + p

n emission

^a_zX --> ^a-1_zY + n

cluster decay

emittion of a nuclide heavier than ⁴₂He, but far lighter than the parent nuclide. These cluster nuclides are strong bound nuclides. Typical examples are ¹⁴₆C, ²⁰₈O. Maybe considered as a generalization of the alpha-decay.

spontanious fission

Breaking of the nuclide into two parts and some single neutrons. The exact values of p and z of the two child nuclides will variate.

gamma decay

The nucleons stay the same. Also denoted as internal transition. A so-called isomere will emitte a photon going into a lower internal nuclide state.

"Binding Energy" of an Atom

For the physicaly irrelevance look only at the beta- decay of ³H into ³He. Tritium (H-3) has a higher binding energy than He-3 but it decays exotherm! Ridiculous. Or ⁶²Ni has a higher "binding energy" than ⁵⁶Fe, which is the strongest bound nuclide at all, and the correct statement is of important astrophysical relevance.

The error comes from the assumption that in a nucleous are distinguishable particles called neutrons and protons, but this is false. There is quantum-mechanically only one kind of particle which maybe interpreted like a superposition of proton and neutron --- not to question whether this is really a particle. :) Another (more intuitive) interpretation is the continous exchange of pions between the nucleons which blurs the difference between proton and neutron. Thus the correct formula to measure the binding energy must be of the form:
(Z+N)m_nucleon - M.
But what is "the" nucleon --- respectively its mass? It doesn't matter much, you may select m_p or m_n or any value between these two. This only shifts the zero level of your binding energy. You could even accept the mass defect, defined to be M - (Z+N)m_C-12/12, as a kind of (inverse) binding energy, if you like the zero level to be obtained by the Carbon-12 isotope. I personally prefer to choose m_n, because then all nuclides get positive binding energy. Have a look at the first atoms and the strongest bound atoms:
.