Why is the excited state of 116 Indium more stable than ground state? Both undergo beta decay, but the ground state has a half-life of 14 seconds, while the excited state has a half-life of 54 minutes. Is there something special about the ground state?

I found a diagram of the typical decay of the excited state of 116 indium to 116 Tin:

level diagram

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    $\begingroup$ Why do you think the excited state is more stable? $\endgroup$
    – Jon Custer
    Jun 6, 2015 at 13:55
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    $\begingroup$ @JonCuster The ground state has a half-life of 14 seconds, but the $5^+$ isomer has a half-life of 54 minutes. $\endgroup$
    – rob
    Jun 6, 2015 at 14:03
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    $\begingroup$ OK, good point. So we are looking at a difference of going spin 1 to spin 0 vs spin 5 to one of three spin 4 states. The multiple spin 4 states could easily explain the increased transition probability. $\endgroup$
    – Jon Custer
    Jun 6, 2015 at 14:08
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    $\begingroup$ I'm going to plead lack of coffee! $\endgroup$
    – Jon Custer
    Jun 6, 2015 at 14:43
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    $\begingroup$ For a much clearer example, tantalum-180, the ground state, decays in 8.152 hours, while the excited state tantalum-180m1 has a half-life that exceeds 1.2 quadrillion years. See Isotopes of tantalum. $\endgroup$ Jun 6, 2015 at 16:58

2 Answers 2


In general, decays which release a lot of energy are faster than than decays which release only a little energy.

As Carl Witthoft points out, the first excited state is unlikely to decay to the ground state partially because the energy difference is small, and partially because the spin difference is large. (The photon for the $5^+\to1^+$ transition within $\rm ^{116}In$ would have to carry angular momentum $4\hbar$: a so-called electric hexadecapole (E4) transition.)

We can make the same combination angular momentum / energy argument for the beta decays. Beta decays are parity-violating: the electron tends to come out left-handed, and the antineutrino comes out right-handed. The two leptons are most likely to leave the nucleus without orbital angular momentum (that is, in an $s$-wave state), and if you detect them in roughly opposite directions then together they carry one unit $\hbar$ of spin. We therefore expect the beta transition to strongly prefer $\Delta J=1$ transitions. If the beta decay were to have $\Delta J > 1$, the leptons would have to be emitted with some $p$-wave or higher orbital angular momentum; those wavefunctions have much less overlap with the nucleus than the $s$-wave.

So the energetic decay $\rm^{116}In(5^+) \not\to {}^{116}Sn(0^+)$ is strongly suppressed by angular momentum considerations, and the preferred decays for the isomer $\rm^{116}In(5^+) \to {}^{116}Sn(4^+)$ only have an energy of about $\rm1.5\,MeV$. That low-energy decay proceeds more slowly than the $\rm3.9\,MeV$ decay $\rm^{116}In(1^+) \to {}^{116}Sn(0^+)$.

For internal consistency in this seat-of-my-pants, hand-waving argument, notice that the decay to the lowest-energy $4^+$ state is about five times more likely than the decay to the highest-energy $4^+$ state.

My golden-rule skills are too weak to make this argument quantitative; I'd love to see a definitive answer.


I'm not an expert on the notation, but I think the answer you're after is this info, from Trent U. Physics Lab Notes. Sorry about the loss of formatting.

In this experiment foils of the stable element 49 115 In are bombarded by neutrons produced in a radioactive
source (described below). The reaction is
0 1 49 115 49 116,m
n + In In
The m indicates that the 116 In nuclei are formed in metastable states. These states could in principle decay by
prompt emission of gamma rays leading to lower excited states of 116 In . Normally such gamma emission
would very quickly leave the nucleus in its ground state which could then decay by −
emission to states of
50 116 Sn . However the first excited state of 116 In has a very low probability of gamma-emission. This is
because it is only slightly more energetic than the ground state (0.127 MeV) and the difference in spins is
rather large (5 +
). Therefore this state decays instead by −
emission to excited states of 50 116 Sn . The
half-life for this decay is about 54 minutes.

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