Why some isotopes with positive alpha-decay energy are stable? For example, alpha-decay energy of stable 194Pt is about 1.5 MeV.

But there is no stable isotopes with positive beta-decay energy. For example, half life of 197Pt with beta-decay energy 0.7 MeV is 19 hours

  • $\begingroup$ LBL's online Table of the Isotopes doesn't support the basic claim that $^{197}\mathrm{Au}$ has a alpha line (neither in the ground state nor the meta-stable excited state $^{197\mathrm{m}}\mathrm{Au}$). Can you give a reference? $\endgroup$ – dmckee Feb 16 '13 at 21:25
  • $\begingroup$ @dmckee, It's easy to calculate that reaction Au(197,79)+He(4,2)=Tl(201,81) requires 1.5 MeV to take place. Masses of the atoms (atom.kaeri.re.kr/ton/main.shtml): Au(197,79) - 196.9665516; He(4,2)- 4.0026032; Tl(201,81)-200.9708038 $\endgroup$ – voix Feb 18 '13 at 19:04
  • $\begingroup$ One or the other of us has misunderstood the other. I claim that Gold-197 does not emit alpha particle on it's own. That's what is means to be stable, and it is so because additional energy is required to generate the end-state. For that reason I don't understand what the question is here. $\endgroup$ – dmckee Feb 18 '13 at 19:08
  • $\begingroup$ Yes, Gold-197 is stable. But Au(197,79)=Tl(201,81)+ He(4,2) should be exothermic reaction. Why it does not take place? $\endgroup$ – voix Feb 18 '13 at 19:20
  • $\begingroup$ Count the nucleons in the reaction you propose. It's not balanced. The alpha decay of Tl-201: $^{201}\mathrm{Tl} \to ^{197}\mathrm{Au} + ^4\mathrm{He}$ is balanced, but taking your numbers it does not appear to be energetically favored. $\endgroup$ – dmckee Feb 18 '13 at 19:23

There are two issues at work here. Firstly, there is relatively little phase space for this decay and little choice in the matter of which nucleons go into making the alpha as the Osmium nucleus must be left in either the ground state or a only very modestly excited state. Both effects would drive a long halflife and if the resultant nucleus is left in the ground state there will be only a low energy alpha line which will straggle out in all but the thinnest layers of source material making the detection of the decay difficult.

Earlier comments

It occurs to me that the question might be

"Why doesn't Tallium-201 exhibit alpha decay when there is sufficent energy there?".

I believe the answer would be that angular momentum and parity conservation can not both be satisfied at once in that decay.

Tl-201 has $\mathrm{J}^\mathrm{P}$ of $\left.\frac{1}{2}\right.^+$, Au-197 has $\left.\frac{3}{2}\right.^+$, and the alpha particle has $0^+$. To make the angular momentum work out the alpha would need to be emitted in a relative p state, but that imposes a factor of -1 on parity and we're stuck.

  • $\begingroup$ There is no sufficent energy for Tallium-201 alpha decay. There is sufficent energy for Gold-197 alpha decay. $\endgroup$ – voix Feb 18 '13 at 19:27
  • $\begingroup$ The alpha decay of Gold-197, would be $^{197}\mathrm{Au} \to ^{193}\mathrm{Ir} + ^4\mathrm{He}$ (please note the use of $\to$ (\to) rather than $=$ here to indicate the directionality of the reaction, it matters in this case). The reaction you propose violates baryon conservation number by 8: you've made four each of protons and neutrons out of nowhere. $\endgroup$ – dmckee Feb 18 '13 at 19:35
  • $\begingroup$ I'm so sorry, you are right of course :) $\endgroup$ – voix Feb 18 '13 at 19:40
  • $\begingroup$ Tallium-201 is not stable, so I corrected the question. $\endgroup$ – voix Feb 18 '13 at 19:58

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