Question

Sketch the Feynman diagram if the proposed decay is possible.

${\Delta }^{+}\to {\Sigma }^{+}+{\pi }^0+\gamma$

Short answer

The proposed decay is not possible

Step-by-step solution

Given data

The proposed reaction is ${\Delta }^{+}\to {\Sigma }^{+}+{\pi }^0+\gamma$

Concept of Feynman diagram 

A Feynman diagram is drawn to express the particle interaction in nature.

Step 3:Find whether the charge is conserved

Conservation of charge is shown below:

$$\begin{gathered} {\Delta }^{+}\to {\Sigma }^{+}+{\pi }^0+\gamma \\ (+e)\to (+e)+\left(0\right)+\left(0\right) \\ (+e)\to (+e) \end{gathered}$$

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

Therefore, the charge is conserved.

Find conservation of baryons number

Conservation of baryons number is given below:

$$\begin{gathered} {\Delta }^{+}\to {\Sigma }^{+}+{\pi }^0+\gamma \\ (+1)\to (+1)+\left(0\right)+\left(0\right) \\ (+1)\to (+1) \end{gathered}$$

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

Therefore, the baryons' number is conserved.

Find conservation of energy

Conservation of energy:

$$\begin{gathered} {\Delta }^{+}\to {\Sigma }^{+}+{\pi }^0+\gamma \\ (1232{\text{MeV/c}}^2)\to (1232{\text{MeV/c}}^2)+(135{\text{MeV/c}}^2) \\ 1232{\text{MeV/c}}^2\to 1367{\text{MeV/c}}^2 \end{gathered}$$

The energy $1232\text{MeV}$ before the decay is less than the energy $1367\text{MeV}$ that alters the decay.

Thus, energy conservation is violated.From the above results, we conclude that the decay reaction is not possible.

The proposed decay is not possible.