Why are exited states of an Isotope metastable?

[kuecken]

Why are exited states of an Isotope metastable? Is it because they have a large spin and therefore the final states they decay to have to be excited as well? And therefore they have less energy gain?
I am thinking about 116 In (the 1+ and 5+ state) both can undergo a β- decay.
Thank you!

 

[kjhskj75]

I'm not an expert but I think the reason is that if a decay involves a large spin change, then to conserve angular momentum the nucleon has to be at some distance from the centre, in fact outside the nucleus.
And it doesn't spend much time out there, leading to a low probability of decay. This applies to both beta and gamma.

Unless anyone has a better theory.

 

[mfb]

Most of the excited states are not metastable. Large spin differences are a common reason for long lifetimes, they make transitions very unlikely as they need higher orders of the electromagnetic interaction.

I'm not an expert but I think the reason is that if a decay involves a large spin change, then to conserve angular momentum the nucleon has to be at some distance from the centre, in fact outside the nucleus.

That does not make sense.

 

[kjhskj75]

That does not make sense.

Sorry, what I was trying to say was this:

In the O.P.'s example of the first excited state of ¹¹⁶In, which normally emits a $\gamma$ with energy 127.267 keV or 1.039$\times$10^-14 Joules

Such a photon has a momentum of E/c or 6.8$\times$10^-23 kg m² s^-2

Since this state has J=5 and decays to one of J=1, it has to lose 4 units of angular momentum, and the photon can only carry off 1, and as (orbital angular momentum) $L = r \times p$, then the emission must be at a point at least L/p = 4.65$\times$10^-12 m from the centre of the nucleus.

Given that the nucleus has radius roughly 1.5$\times$10^-15 m times the cube root of the mass number, say 7$\times$10^-15 m. This means that the nucleon that decays must be considerably outside the normal boundary of the nucleus, and because the probability of it being that far out is extremely low, the half-life is quite long. Longer than the beta decay half-life of the unexcited nucleus, which has to lose only 1 unit of spin to become ¹¹⁶Sn (J=0).

The same applies to a beta decay for the excited state, (I don't have the figures for that), but again the large delta-J would tend to suppress it, since an electron and antineutrino together can only carry off one unit of J.

Have I got this right ?

 

[mfb]

The decay is an effect of the whole nucleus (as the nucleons all interact with each other), but I agree with the basic concept.