nuclear-physicsradiation
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:
You're assuming that nuclei with exactly a magic number of neutrons are more stable than all their non-magic neighbors in the chart of nuclides, but there's no reason to think that.
If a nucleus has a magic number of neutrons, that means one shell is completely full, and the next shell is empty. Therefore the next neutron you add (going to magic+1) will have significantly greater single-particle energy, so that nucleus should be less stable than the magic one. This is born out by both of your examples for the well-known magic number 126:
However, if you remove one neutron, then nothing changes about the other (magic−1) neutrons; they are still in the same shells as in the magic case. No extra stability is caused by having a completely full shell as opposed to an almost full one. (In this respect, the word "magic" is misleading.)
Here's another way to think of it: In alpha decay (which is the main decay branch for all of your examples), two neutrons are removed from the nucleus. It will always be the two most energetic neutrons that are removed (those in the highest shells). In a magic nucleus, the lower shell is completely full, so two neutrons are removed from that lower shell. In a magic+1 nucleus, there is a lone neutron in the upper shell, so when two neutrons are removed, one comes from the upper shell and one from the lower shell. Since one comes from the upper shell, more energy is released in the alpha decay and in general the half-life will be shorter.
However, in a magic−1 nucleus, there are no neutrons in the upper shell, so both neutrons come from the lower shell. This is the same situation as the magic case, so there's no reason to expect the decay energies or half-lives to be drastically different. (Of course they won't be exactly the same, but the differences come from other, more subtle effects.)
BTW, 152 is not a "canonical" or "universal" magic number. It shows up on the plot you have there of some specific elements, but if you look at other elements, the gap between the shells occurs at a different place. 126 is a universal magic number, but at 152 the situation is more complicated because changing the neutron/proton ratio also shifts the shells relative to each other. This is why the thing I said about magic+1 always being less stable than magic doesn't hold for one of those. Nuclear structure is really complicated.
NB: I feel like this is a pretty half-assed job, and I apologize for that but having opened my mouth in the comments I guess I have to write something to back it up.
We start with Fermi's golden rule for all transitions. The probability of the transition is $$ P_{i\to f} = \frac{2\pi}{\hbar} \left|M_{i,f}\right|^2 \rho $$ where $\rho$ is the density of final states which is proportional to $p^2$ for massive particles. To find the rate1 for all possible final states we sum over these probabilities incoherently. When the mass difference between the initial and final states is much less than the $W$ mass the matrix element $M_{i,f}$ depends only weakly (hah!) on the particular state and the sum is well approximated by a sum only over the density of states: $$P_\text{decay} \approx \frac{2\pi}{\hbar} \left|M_{}\right|^2 \int_\text{all outcomes} \rho .$$ This sum is collectively called the phase space available to the decay. In these cases the matrix element is also quite small for the reason that Dr BDO discusses.
The phase space computation can be quite complicated as it must be taken over all unconstrained momenta of the products. For decays to two body states it turns out to be easy, there is no freedom in the final states except the $4\pi$ angular distribution in the decay frame (their are eight degrees of freedom in two 4-vectors, but 2 masses and the conservation of four momentum account for all of them except the azimuthal and polar angles of one of the particles).
The decays that you have asked about are to three body states. That gives us twelve degrees of freedom less three constraints from masses, four from conservation of 4-momentum which leaves five. Three of these are the Euler angles describing the orientation of the decay (and a factor of $8\pi^2$ to $\rho$), so our sum is over two non-trival momenta. The integral looks something like $$ \begin{array}\\ \rho \propto \int p_1^2 \mathrm{d}p_1 \int p_2^2 \mathrm{d}p_2 \int \mathrm{d}(\cos\theta) &\delta(m_0 - E_1 - E_2-E_3 ) \\ &\delta(E_1^2 - m_1^2 - p_1^2) \\ &\delta(E_2^2 - m_2^2 - p_2^2) \\ &\delta(E_2^2 - m_2^2 - p_2^2) \\ &\delta(\vec{p}_1 + \vec{p}_2 + \vec{p}_3) \end{array} $$ which is easier to compute in Monte Carlo than by hand. (BTW--the reason for introducing the seemingly redundant integral over the angle $\theta$ between the momenta of particles 1 and 2 will become evident in a little while).
For beta decays the remnant nucleus is very heavy compared to the released energy, which simplifies the above in one limit.
In the case of muon decay, it is not unreasonable to treat all the products as ultra-relativistic, and the above reduces to $$ \begin{array}\\ \rho \propto \int p_1^2 \mathrm{d}p_1 \int p_2^2 \mathrm{d}p_2 \int \mathrm{d}(\cos\theta) &\delta(m_0 - E_1 - E_2 - E_3 ) \\ &\delta(E_1 - p_1) \\ &\delta(E_2 - p_2) \\ &\delta(E_3 - p_3) \\ &\delta(\vec{p}_1 + \vec{p}_2 + \vec{p}_3) \\ = \int p_1^2 \mathrm{d}p_1 \int p_2^2 \mathrm{d}p_2 \int \mathrm{d}(\cos\theta) &\delta(m_0 - p_1 - p_2 - p_3 ) \\ &\delta(\vec{p}_1 + \vec{p}_2 + \vec{p}_3) \\ = \int p_1^2 \mathrm{d}p_1 \int p_2^2 \mathrm{d}p_2 \int \mathrm{d}(\cos\theta) &\delta(m_0 - p_1 - p_2 - \left|\vec{p}_1 + \vec{p}_2\right| )\\ = \int p_1^2 \mathrm{d}p_1 \int p_2^2 \mathrm{d}p_2 \int \mathrm{d}(\cos\theta) &\delta\left(m_0 - p_1 - p_2 - \sqrt{p_1^2 + p_2^2 - p_1p_2\cos\theta} \right) \end{array} $$ The integral over the angle will evaluate to one in some regions and zero in others and as such is equivalent to correctly assigning the limits of the other two integrals, so writing $\delta m = m_0 - m_1 - m_2 - m_3$ we get $$ \begin{array} \rho & \propto \int_0^{\delta m/2} p_1^2 \mathrm{d}p_1 \int_0^{\delta m-p_1} p_2^2 \mathrm{d}p_2 \\ & \propto \int_0^{\delta m/2} p_1^2 \mathrm{d}p_1 \left[ \frac{p_2^3}{3}\right]_{p_2=0}^{\delta m-p_1} \\ & \propto \int_0^{\delta m/2} p_1^2 \mathrm{d}p_1 \frac{(\delta m - p_1)^3}{3} \end{array} $$ which I am not going to bother finishing but shows that that phase space can vary as a high power of the mass difference (up to the sixth power in this case).
1 The lifetime of the state is inversely proportional to the probability
Best Answer
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.