Troubleshooting nuclear decay, electron binding energies, internal contributions

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The discussion focuses on the appropriate binding energy shell to use in nuclear decay calculations, specifically questioning the choice of L2 over L1 or L3. It raises confusion about the inclusion of L2, which has a binding energy lower than 20 keV, while also questioning the relevance of omitting electrons with energies below this threshold. Participants suggest that there may be typographical errors in the referenced equations, indicating that L1 should be K and L2 should be simplified to L. Additionally, it is noted that the binding energy values for mercury suggest that the 20 keV restriction may not be necessary, as other relevant energies are significantly higher. Clarification on these points is needed for accurate calculations.
Graham87
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Homework Statement
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Relevant Equations
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How do you know which binding energy shell to use? In the solution it uses K and L2. Why specifically L2 and not L3 or L1 for example?

And what should I do with the information to omit electrons lower than 20kev? I initially thought that meant to omit the electron binding energies lower than 20kev. But L2 which is lower than 20kev is included, so which expression represents electron energy? If it is ΔE - B(L) then shouldn’t L3 be included since it also has a higher energy than 20kev?

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My guess is that there are some typos as indicated below:
1683822652238.png

The ##L_1## should be ##K## and the ##L_2##'s should just be ##L##. This corresponds with the given binding energies:

1683822987537.png


Perhaps the value ##B(L)_{Hg} = 14.2087## keV is a weighted average over the ##L_1##, ##L_2##, and ##L_3## levels.

I'm not sure about the 20 keV restriction. Since ##\overline E_\beta##, ##B(K)_{Hg}## and ##B(L)_{Hg}## are all in the hundreds of keV, there doesn't appear to be any need to worry about requiring the electrons to have an energy greater than 20 keV.
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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