91影视

The proposed reaction is $\stackrel{炉}{n}鈫�p+e^{-}+{\stackrel{炉}{v}}_e$

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:

$$\begin{gathered} \stackrel{炉}{n}鈫�p+e^{-}+{\stackrel{炉}{v}}_e \\ 0鈫�(+e)+(-e)+\left(0\right) \\ 0鈫�0 \end{gathered}$$

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:

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

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:

$$\begin{gathered} \stackrel{炉}{n}鈫�p+e^{-}+{\stackrel{炉}{v}}_e \\ \left(0\right)鈫�\left(0\right)+(+1)+(-1) \\ \left(0\right)鈫�\left(0\right) \end{gathered}$$

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:

$$\begin{gathered} \stackrel{炉}{n}鈫�p+e^{-}+{\stackrel{炉}{v}}_e \\ (-1)鈫�(+1)+\left(0\right)+\left(0\right) \\ (-1)鈫�(+1) \end{gathered}$$

The baryon number is not conserved.

Hence, the proposed decay is not possible.