Question

Sketch the Feynman diagram if the proposed decay is possible.

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

Short answer

The proposed decay is not possible.

Step-by-step solution

Given data

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

Concept of Feynman diagram

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

Step 3:Find whether the charge is conserved

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.

 Step 4: Find conservation of mass

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$1530\text{MeV}/{\text{c}}^2$ .

 Step 5: Find conservation of baryons number

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.

Find conservation of leptons number

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.