Sketch the Feynman diagram if theproposed decay is possible.

${\tau }^{+}\to e^{+}+v_e+\overrightarrow{v_{\tau }}$

The proposed reaction is${\tau }^{+}\to e^{+}+v_e+\overrightarrow{v_{\tau }}$ .

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 can be checked as:

The charge of ${\tau }^{+}$is $+e$, the charge of $e^{+}$is $+e$, the charge of $v_e$ is 0 , and the charge of $\overrightarrow{v_{\tau }}$ is also 0 .

$$\begin{gathered} {\tau }^{+}\to e^{+}+v_e+\overrightarrow{v_{\tau }} \\ (+e)\to (+e)+0+0 \\ (+e)\to (+e) \end{gathered}$$

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

Therefore, the charge is conserved.

Conservation of energy of mass is given below.

$$\begin{gathered} {\tau }^{+}\to e^{+}+v_e+\overrightarrow{v_{\tau }} \\ \left(1.78\times {10}^3\text{MeV}/c^2\right)\to \left(0.511\text{MeV}/c^2\right) \end{gathered}$$

The mass after the decay is $0.511\text{MeV}/{\text{c}}^2$ less than that of the initial mass $1.78\times {10}^3\text{MeV}/c^2$.

Conservation of lepton numbersis shown below.

$$\begin{gathered} {\tau }^{+}\to e^{+}+v_e+\overrightarrow{v_{\tau }} \\ (0,-1)\to (-1,0)+(+1,0)+(0,-1) \\ (0,-1)\to (0,-1) \end{gathered}$$

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

Therefore, the lepton numbers are conserved.

From the above result, we conclude that the decay reaction is possible.

This is a weak decay because it involves W-bosons.

The Feynman diagram is shown in figure 1.

Figure 1

Therefore, the proposed decay is possible.