Question

Two protons, with equal kinetic energy, collide head-on. What is the minimum kinetic energy Kp of one of these protons necessary to make a pion-antipion pair? The rest energy of a pion is 139.6MeV.

Answer 1

`K_p=139.6\ MeV`

Step-by-step explanation:

It is given that,

The rest energy of a pion is 139.6 MeV. Here, two protons having equal kinetic energy collides elastically. We need to find the minimum kinetic energy of one of these protons necessary to make a pion-antipion pair.

It can be calculated using conservation of energy and momentum, the total energy of the particles gets converted into rest mass energy of new particles. So,

`2K_p=2* E_(\pi^+)`

`K_p=* E_(\pi^+)`

`K_p=139.6\ MeV`

So, the minimum kinetic energy of one of these protons necessary to make a pion-antipion pair is 139.6 MeV. Hence, this is the required solution.