What Is the Minimum Photon Energy Needed to Produce a Proton-Antiproton Pair?

  • Thread starter Thread starter PhyzicsOfHockey
  • Start date Start date
AI Thread Summary
To produce a proton-antiproton pair, the minimum photon energy required is determined by the rest mass energy of the particles involved. The relevant equation is E^2 = p^2c^2 + m^2c^4, where the rest mass of the proton and antiproton is crucial. Since the photon has zero rest mass, its energy is entirely based on momentum. The total energy needed for the pair production is twice the rest mass energy of a proton, which is approximately 1.88 GeV. Understanding the conservation of momentum and energy is essential in solving this problem.
PhyzicsOfHockey
Messages
41
Reaction score
0

Homework Statement



How much photon energy would be required to produce a proton-antiproton pair? Give the answer in SI units.

Homework Equations



E^2=p^2c^2+m^2c^4
E=pc+pc
p=mc
E=2*m*c^2

The Attempt at a Solution



I seem to be confused I know the KE would be zero I also know this would not exist in real life but I need to find the min energy so KE of the photon pairs is 0. I am not sure if the equation has to do with the mass of a photon or the momentum of the photons. In either case I don't know what the mass of a photon would be. I read it was zero but can't figure out how to use this formula then.
 
Physics news on Phys.org
ok... mass of a photon is zero... I think we all know that. But its momentum is not... we can agree on that too I think. Now, your key equation is
E^2=p^2 c^2+m^2 c^4[\tex]<br /> where m here is the rest mass of your particle, p is its momentum and E is its total energy. To do your question: remember 4-momentum is conserved...etc
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Thread 'Struggle to understand the concept of the question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »

Similar threads

Can a 2MeV photon produce a proton-antiproton pair?
Replies
3
Views
5K
Minimum energy of a photon to produce ##\pi^+##
Replies
8
Views
3K
Invariant mass and energy balance
Replies
4
Views
2K
Pair Production Velocity Calculation
Replies
5
Views
2K
Photon-electron collision with pair production
Replies
4
Views
2K
Kinetic Energy of Colliding Protons
2
Replies
54
Views
10K
Energy of the photon emitted in a gamma ray decay
Replies
19
Views
3K
Pair Production: Which Conservation Law?
Replies
2
Views
2K
Muon production and decay from relativistic annihilation
Replies
4
Views
2K
Change in the energy of photon due to recoil of the nucleus
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top