Which of the following reactions obey the conservation of baryon number?

(a) $\mathit{p}\mathbf{+}\mathit{p}\mathbf{\to }\mathit{p}\mathbf{+}{\mathit{e}}^{\mathbf{+}}$

(b) $\mathit{p}\mathbf{+}\mathit{n}\mathbf{\to }{\mathit{e}}^{\mathbf{-}}\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{+}}\mathbf{;}$

(c)$\mathit{p}\mathbf{\to }\mathit{n}\mathbf{+}{\mathit{e}}^{\mathbf{-}}\mathbf{+}{\overline{\mathbf{\nu }}}_{\mathbf{e}}\mathbf{;}$

(d)$\mathit{p}\mathbf{+}\overline{\mathbf{p}}\mathbf{\to }\mathbf{2}\mathit{\gamma }\mathbf{.}$

Baryonsinclude nucleons and several particles called hyperons. These resemble nucleons but are more massive. Baryons have half-integer spin, and therefore all are fermions. The only stable baryon is the proton; a free neutron decays to a proton, and hyperons decay to other hyperons or nucleons by various processes.

Baryons obey the law of the conservation of baryon numbers, analogous to the conservation of lepton numbers, again with no counterpart in classical physics. We assign a baryon number B= 1 to each baryon (p, n) and B= -1 to each antibaryon ($\overline{p}$).

Table representing baryon number of some particles:

(a)

The reaction is $p+p\to p+e^{+}$.

The baryon number on the left-hand side of the reaction is:

$+1+1=+2$

The baryon number on the right-hand side of the reaction is:

$+1+0=+1$

Thus, the baryon number is not conserved

(b)

The reaction is $p+n\to e^{-}+2e^{+}$

The baryon number on the left-hand side of the reaction is:

$+1+1=+2$

The baryon number on the right-hand side of the reaction is:

$0+2\left(0\right)=0$

Thus, the baryon number is not conserved.

(c)

The reaction is $p\to n+e^{-}+{\overline{\nu }}_e$

The baryon number on the left-hand side of the reaction is:

$+1$

The baryon number on the right-hand side of the reaction is:

$+1.$

Thus, the baryon number is conserved.

(d)

The reaction is $p+\overline{p}\to 2\gamma$

The baryon number on the left-hand side of the reaction is:

$+1-1=0$

The baryon number on the right-hand side of the reaction is:

$2\left(0\right)=0$

Thus, the baryon number is conserved.

Therefore, The reactions that obey the conservation of baryon number are (c)$p\to n+e^{-}+{\overline{v}}_e$ and (d)$p+\overline{p}\to 2\gamma$