Sketch the Feynman diagram if the proposed decay is possible.

$\overline{n}\to p+e^{-}+{\overline{v}}_e$

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

A Feynman diagram is used to show the interaction of particles in quantum field theory. The particles are represented by the lines in the diagram.

Conservation of charge is shown below:

$$\begin{gathered} \overline{n}\to p+e^{-}+{\overline{v}}_e \\ 0\to (+e)+(-e)+\left(0\right) \\ 0\to 0 \end{gathered}$$

Thus, the charge before the decay and after the decay is equal.

Therefore, the charge is conserved.

Conservation of mass or energy is given below:

$$\begin{gathered} \overline{n}\to p+e^{-}+{\overline{v}}_e \\ 940\text{MeV}/{\text{c}}^2\to \left(938\text{MeV}/{\text{c}}^2\right)+\left(0.511\text{MeV}/{\text{c}}^2\right)+\left(0\right) \\ 940\text{MeV}/{\text{c}}^2\to 938.511\text{MeV}/{\text{c}}^2 \end{gathered}$$

The mass after the decay is $938.511\text{MeV}/{\text{c}}^2$ less than that of the initial mass $940\text{MeV}/{\text{c}}^2$.

Conservation of lepton number:

$$\begin{gathered} \overline{n}\to p+e^{-}+{\overline{v}}_e \\ \left(0\right)\to \left(0\right)+(+1)+(-1) \\ \left(0\right)\to \left(0\right) \end{gathered}$$

Thus, the lepton numbers before the decay are equal to the lepton numbers after the decay.

Therefore, the lepton numbers are conserved.

The conservation of baryon numbers is shown below:

$$\begin{gathered} \overline{n}\to p+e^{-}+{\overline{v}}_e \\ (-1)\to (+1)+\left(0\right)+\left(0\right) \\ (-1)\to (+1) \end{gathered}$$

The baryon number is not conserved.

Hence, the proposed decay is not possible.