Sketch the Feynman diagram if the proposed decay is possible.

${\Xi }^{*-}\to {\Sigma }^0+e^{-}+v_e$

The proposed reaction is ${\Xi }^{*-}\to {\Sigma }^0+e^{-}+v_e$

Particle interactions can be represented as Feynman diagrams. Different types of lines are drawn in the diagrams.

Conservation of charge is shown below:

$$\begin{gathered} {\Xi }^{*}\to {\Sigma }^0+e^{-}+v_e \\ (-e)\to \left(0\right)+(-e)+\left(0\right) \\ (-e)\to (-e) \end{gathered}$$

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 as:

$$\begin{gathered} {\Xi }^{*-}\to {\Sigma }^0+e^{-}+v_e \\ 1530\text{MeV}/c^2\to \left(1193\text{MeV}/c^2\right)+\left(0.511\text{MeV}/c^2\right) \\ 1530\text{MeV}/c^2\to 1193.511\text{MeV}/c^2 \end{gathered}$$

The mass after the decay is $1193.511\text{MeV}/{\text{c}}^2$ less than that of the initial mass$1530\text{MeV}/{\text{c}}^2$ .

Conservation of baryons number:

$$\begin{gathered} {\Xi }^{*-}\to {\Sigma }^0+e^{-}+v_e \\ (+1)\to (+1)+\left(0\right)+\left(0\right) \\ (+1)\to (+1) \end{gathered}$$

Thus, the baryons number before the decay is equal to the baryons number after the decay.

Therefore, the baryon's number is conserved.

Conservation of leptons number is shown below:

$$\begin{gathered} {\Xi }^{*}-\to {\Sigma }^0+e^{-}+v_e \\ \left(0\right)\to \left(0\right)+(+1)+(+1) \\ \left(0\right)\to (+2) \end{gathered}$$

The lepton number is not conserved.

Therefore, the proposed decay is not possible.