Does antineutrino capture preferentially form neutrons?

[snorkack]

Besides the energetic preference (lower threshold, and more phase space above)?
Antineutrino capture is a weak process, so it can and does change quark flavour.
p+ν=n+e⁺
is actually
uud+ν=udd+e⁺
that is
u+ν=d+e⁺
But given enough energy (like cosmic ray neutrinos), do antineutrinos also get captured:
p+ν=Λ+e⁺?
Because this is just
udu+ν=uds+e⁺
that is
u+ν=s+e⁺
As you see, even though baryon charges match, a process
p+ν=Ξ+e⁺
would be obstructed, because Ξ has 2 s quarks. But process
p+ν=Λ_b+e⁺
should be just
u+ν=b+e⁺

Obviously these processes are impossible below energy threshold, and above they have a phase space factor. But at high energies, does proton conversion into n vs conversion into any specified flavour of Λ approach ratio of unity, or will any difference remain?

 

[Dr.AbeNikIanEdL]

You would have different CKM matrix elements in there. And $|V_{ud}| > |V_{us}| > |V_{ub}|$. So I would say yes, $p+\bar{\nu}\to n+e^+$ is preferred over $p+\bar{\nu}\to \Lambda+e^+$ even setting aside phase space constraints.

Edit: changed $\nu\to\bar{\nu}$, see below

 

[vanhees71]

But it should be $p+\bar{\nu} \rightarrow n+ e^+$.

 

[Dr.AbeNikIanEdL]

Indeed, I just copied the reactions as noted in the OP (all $\nu$ there should also be $\bar{\nu}$ but the text correctly says antineutrino).