In the following exercises, two protons are smashed together in an attempt to convert kinetic energy into mass and new particles. Indicate whether the proposed reaction is possible. If not, indicate which rules are violated. Consider only those for charge, angular momentum, and baryon number If the reaction is possible, calculate the minimum kinetic energy required of the colliding protons.

$\mathbf{p}\mathbf{+}\mathbf{p}\mathbf{\to }{\mathbf{\Sigma }}^{\mathbf{+}}\mathbf{+}\mathbf{p}\mathbf{+}{\mathbf{K}}^{\mathbf{0}}$

The proposed decay reaction is $\mathrm{p}+\mathrm{p}\to {\mathrm{\Sigma }}^{+}+\mathrm{p}+{\mathrm{K}}^0$.

A nucleus has mass $\mathbf{m}$ . The rest energy can be calculated as ${\mathbf{mc}}^{\mathbf{2}}$.

Conservation of charge:

$\begin{array}{l}\mathrm{p}+\mathrm{p}\to {\mathrm{\Sigma }}^{+}+\mathrm{p}+{\mathrm{K}}^0\\ (+\mathrm{e})+(+\mathrm{e})\to (+\mathrm{e})+(+\mathrm{e})+\left(0\right)\\ (+2\mathrm{e})\to (+2\mathrm{e})\end{array}$

Thus, the charge before the decay and after the decay is equal, which implies that the charge is conserved.

Conservation of baryons number:

$\begin{array}{l}\mathrm{p}+\mathrm{p}\to {\mathrm{\Sigma }}^{+}+\mathrm{p}+{\mathrm{K}}^0\\ (+1)+(+1)\to (+1)+(+1)+(+0)\\ (+2)\to (+2)\end{array}$

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

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

The required minimum kinetic energy of the colliding protons can be calculated as shown below.

$\begin{array}{l}={\mathrm{m}}_{{\mathrm{K}}^0}{\mathrm{c}}^2+{\mathrm{m}}_{{\mathrm{\Sigma }}^{+}}{\mathrm{c}}^2-{\mathrm{m}}_{\mathrm{p}}{\mathrm{c}}^2\\ =498\text{\hspace{0.17em}MeV}+1189\text{\hspace{0.17em}MeV}-938\text{\hspace{0.17em}MeV}\\ =749\text{\hspace{0.17em}MeV}\end{array}$

Therefore, the required minimum kinetic energy of the colliding protons is $749\text{\hspace{0.17em}MeV}$.