Please look at the attached file. The first part of the question: why it doesn't decay to the GS. I think thats simply because the nuclear spin changes by 4 and therefore this decay is very supressed. But how "is it possible for the isomeric state at 140kev which decays by gamma emisson, to have a half life of several hours." (ie. a half life comparable to the original state (Mo) decaying to the allowed excited of Tc. Any tips on what approach I can take to show this? Thanks.
It can decay into its ground state, however it is less probable, however the decay pathways are generally toward the next lower energy spin state until the ground state is reached. The reason why it is 'negligible' is because of the extreme energy difference between the energy levels of 1.21 Mev -> 0.14 Mev, where the 0.14 Mev energy level is extremely close to the ground state energy level. The spin states are also a factor, the difference in energy levels in a decay of one spin state into its ground spin state.
thanks for your reply Orion1. I would have thought the difference in spin states is the main reason - I mean nuclear spin changes by 4 (if there was a transition to the ground state). Its the fourth forbidden transition so it would be very, very supressed. What about the second part of the question: any thoughts on how to show it has a half life of several hours?
In simplest terms, if the decay slope is linear, the half life is proportional to the difference in energy between the two spin states. [tex]\frac{T_{\frac{1}{2} a}}{T_{\frac{1}{2} b}} \propto \frac{Q_{\beta1}}{Q_{\beta2}}[/tex] [tex]T_{\frac{1}{2} b} \propto T_{\frac{1}{2} a} \left( \frac{Q_{\beta2}}{Q_{\beta1}} \right) = 66 \; \text{hrs} \left( \frac{0.14 \; \text{Mev}}{1.21 \; \text{Mev}} \right) = 7.636 \; \text{hrs}[/tex] [tex]\boxed{T_{\frac{1}{2} b} = 7.636 \; \text{hrs}}[/tex] The remaining difference being due to the effects of differential spin states. [tex]\boxed{T_{\frac{1}{2} b} = 7.636 \; \text{hrs} - \Delta S \cdot 0.409 \; \text{hrs}}[/tex]
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