= Solution
For every daughter <energy level>, the total number leaving it must equal direct beta feeding plus cascades entering it. Direct feeding cannot be negative. This <nuclear level population balance> gives a decisive test without assuming any electromagnetic transition-strength model.
In scheme B, the first excited level is populated by $2\to1$ and $3\to1$, with counts $N_3+N_4=38\,500+3770=42\,270$. Its only downward transition is $1\to0$, whose count is $N_2=1$. Its direct beta feeding would therefore have to be
$$
F_1^{(B)}=N_2-N_3-N_4=1-38\,500-3770=-42\,269,
$$
which is impossible.
For scheme A the corresponding direct feeding values, in the same arbitrary count normalization, are
$$
F_1=N_1-N_3-N_5=1\,230\,470,\qquad
F_2=N_3+N_4-N_2=42\,269,\qquad
F_3=N_2+N_5+N_6=2501.
$$
All are non-negative. Thus \b[scheme A is the compatible scheme]. Under the stipulated complete decay sample and the six-transition model, scheme B is ruled out by population conservation, rather than merely being less probable. This inference assumes that the reported emitted-photon counts account for de-excitation: substantial unreported <internal conversion> or other decay paths would require additional terms.
Back to article page
