What Is the Minimum Beam Energy for Proton-Proton Collisions?

[J_M_R]

Homework Statement

Calculate the minimum beam energy in a proton-proton collider to initiate the p+p→p+p+n0 reaction. The rest energy of the no is 547.3MeV.

Homework Equations

For a head on collision between particles a and b, from conservation of total energy: Ex = Ep + Ep' = 2Ep

so that mx^2c^4 = Ex^2 - px^2c^2 = (2Ep)^2

The Attempt at a Solution

[/B]
Rearranging the above, gives mx = 2Ep/c^2

So using the values I have been provided: Ep = (547.3MeV)/2 = 0.2737Gev

I am not sure if my method is correct?

 

[mfb]

I guess "beam energy" means the total energy, including the rest energy of the protons.
For the additional energy: yes it is correct. The lab frame is also the center of mass frame, so both protons simply contribute half of the n0 energy plus their rest energy.

 

[J_M_R]

So to get Ep, I am also required to add the rest energy of the two protons?:

Ep = (547.3MeV + 2(938.3MeV))/2 = 1.212GeV

If rest energy of a proton is 938.3Mev.

 

[mfb]

Right.