= Solution
The observed dominant two-body transition is $\boxed{\Sigma^0\to\Lambda^0\gamma}$. Both <baryons> have <valence quarks> $uds$ and the same <strangeness>; an <electromagnetic interaction> transition is therefore possible. Their <mass> difference is only about $77\,\mathrm{MeV}/c^2$, below the lightest <pion> <mass>, so no additional on-shell strongly interacting particle fits. A <photon> is neutral and massless, and can carry the required <angular momentum> through a <magnetic dipole transition> between the two positive-<parity> spin-$1/2$ states. A single <neutrino> would violate <lepton number> and would not give the required fermion parity.
At rest, <four-momentum conservation> gives $p_\Lambda=p_\Sigma-k$ and $k^2=0$. Thus
$$
M_\Lambda^2c^4=M_\Sigma^2c^4-2M_\Sigma c^2E_\gamma,
\qquad
\boxed{E_\gamma=\frac{M_\Sigma^2-M_\Lambda^2}{2M_\Sigma}c^2.}
$$
The result is slightly below $(M_\Sigma-M_\Lambda)c^2$ because the <Lambda baryon> recoils; rounded <masses> give about $74.5\,\mathrm{MeV}$.
The channels $n\pi^0$ and $p\pi^-$ are actually kinematically open: their total <rest energies> are about $1075$ and $1078\,\mathrm{MeV}$, below $M_\Sigma c^2\simeq1193\,\mathrm{MeV}$. Their suppression is therefore \b[not] a <decay threshold> prohibition. Each changes <strangeness> from $-1$ to zero and so requires a <weak interaction>. The much faster allowed electromagnetic transition makes such weak <branching fractions> extremely small. “Not observed” should be understood as overwhelming electromagnetic competition, not an exact ban in the <Standard Model>. The photon identification likewise uses the known low-<energy> spectrum and dominant transition mechanism, not neutrality alone.
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