ArmillarySphere
Muse
- Joined
- Jul 28, 2006
- Messages
- 829
Well, I don't remember much of my subatomic physics lessons (they're about 15 years in the past now), but thanks to you guys, now I know a little more. So your time wasn't entirely wasted 
Seems to me that to apply the principle of least action to a QM system, you'd need to look at the total energy post-decay (i.e. the sum of the bound-state energy for the daughter nucleus and the emitted particle), and pick the one with the lowest value. More energy difference compared to the bound-state energy would give us more KE to plug into half-life calculations. I don't know if what conservation laws are at play in this - for us poor lurkers, could someone explain why deuteron decay is disadvantaged?
Once we've picked a resulting state, you would get the resulting half-life from something like Geiger-Nuttall's law:
[latex]\lambda=\frac{\ln 2}{t_{1/2}} = a_1e^{-a_2\frac{Z}{\sqrt{E}}}[/latex]
That's as far as wiki can take me - I take it modern nuclear theories will give some predictions about a1 and a2.
Seems to me that to apply the principle of least action to a QM system, you'd need to look at the total energy post-decay (i.e. the sum of the bound-state energy for the daughter nucleus and the emitted particle), and pick the one with the lowest value. More energy difference compared to the bound-state energy would give us more KE to plug into half-life calculations. I don't know if what conservation laws are at play in this - for us poor lurkers, could someone explain why deuteron decay is disadvantaged?
Once we've picked a resulting state, you would get the resulting half-life from something like Geiger-Nuttall's law:
[latex]\lambda=\frac{\ln 2}{t_{1/2}} = a_1e^{-a_2\frac{Z}{\sqrt{E}}}[/latex]
That's as far as wiki can take me - I take it modern nuclear theories will give some predictions about a1 and a2.