Question

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.

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

Short answer

The resultant answer is possible, and the required minimum kinetic energy of the colliding protons is $672\text{\hspace{0.17em}MeV}$.

Step-by-step solution

Given data

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

Concept of rest energy

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

Calculate conservation of charge 

Conservation of charge:

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

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

Conservation of baryons number:

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

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

Therefore, the charge is conserved.

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

Calculate the kinetic energy

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

$\begin{array}{c}{\mathrm{m}}_{\mathrm{p}}{\mathrm{c}}^2+{\mathrm{m}}_{{\mathrm{K}}^{+}}{\mathrm{c}}^2+{\mathrm{m}}_{{\mathrm{\Lambda }}^0}{\mathrm{c}}^2-2{\mathrm{m}}_{\mathrm{p}}{\mathrm{c}}^2={\mathrm{m}}_{{\mathrm{K}}^{+}}{\mathrm{c}}^2+{\mathrm{m}}_{{\mathrm{\Lambda }}^0}{\mathrm{c}}^2-{\mathrm{m}}_{\mathrm{p}}{\mathrm{c}}^2\\ =494\text{\hspace{0.17em}MeV}+1116\text{\hspace{0.17em}MeV}-938\text{\hspace{0.17em}MeV}\\ =672\text{\hspace{0.17em}MeV}\end{array}$

Therefore, the required minimum kinetic energy of the colliding protons is $672\text{\hspace{0.17em}MeV}$.