Suppose it is possible to produce a particle of rest-mass energy Mc2 = 91.2 GeV
ID: 1407549 • Letter: S
Question
Suppose it is possible to produce a particle of rest-mass energy Mc2 = 91.2 GeV by colliding together a proton and an anti-proton. Let’s call this particle D; then the reaction is given by
p + p D .
(a) Suppose we set up a stationary proton and accelerate an anti-proton up to a total energy E and slam it into the proton. What is minimum value of E that will result in the production of the particle D?
(b) Now suppose we accelerate both the proton and the anti-proton up to the same energy E such that they have equal momenta and are directed towards each other for a head-on collision. What is the minimum value of E that can result in the production of the particle D? You should find that E is considerably smaller than E.
Explanation / Answer
We need , proton rest mass energy(E0)=938 MeV/c2 = anti proton rest mass
Total energy before collision= E + E0=(0.938+E)GeV/c2 ;
Minimum Total energy after collision= Rest mass energy of D=91.2 GeV
Apply Conservation of energy-mass
0.938+E = 91.2
minimum E= 90.262 GeV/c2
(b)as before
Total energy before collision= 2E'
Total energy after collision= 91.2
apply Conservation of enrgy-mometum law
minimum E' =45.6 GeV/c2