 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 $2\to 1$ and $3\to 1$, with counts $N_3+N_4=38500+3770=42270$. Its only downward transition is $1\to 0$, whose count is $N_2=1$. Its direct beta feeding would therefore have to bewhich is impossible.

For scheme A the corresponding direct feeding values, in the same arbitrary count normalization, areAll are non-negative. Thus 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.