91Ó°ÊÓ

The proposed decay reaction is $\mathrm{p}+\mathrm{p}→{\mathrm{Î\pounds }}^{+}+\mathrm{p}+{\mathrm{K}}^0$.

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

Conservation of charge:

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

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

Conservation of baryons number:

$\begin{array}{l}\mathrm{p}+\mathrm{p}→{\mathrm{Î\pounds }}^{+}+\mathrm{p}+{\mathrm{K}}^0\\ (+1)+(+1)→(+1)+(+1)+(+0)\\ (+2)→(+2)\end{array}$

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.

$\begin{array}{l}={\mathrm{m}}_{{\mathrm{K}}^0}{\mathrm{c}}^2+{\mathrm{m}}_{{\mathrm{Î\pounds }}^{+}}{\mathrm{c}}^2−{\mathrm{m}}_{\mathrm{p}}{\mathrm{c}}^2\\ =498\text{ M±ð³Õ}+1189\text{ M±ð³Õ}−938\text{ M±ð³Õ}\\ =749\text{ M±ð³Õ}\end{array}$

Therefore, the required minimum kinetic energy of the colliding protons is $749\text{ M±ð³Õ}$.