hi,
we have a test on 'the standard model' coming up, and i was wandering if you could please help me with this question?
when a muon and an anti-muon collide they can annihiate each other and release their mass-energy as 2 photons. assuming that these two photons are identical,
a) what will each of their energies be
b) what wavelength will they have
c) why does there need to be 2 photons produced and not just one?
d) in what directions would they have to travel relative to each other and why?
e) in what part of the electromagnetic spectrum are they located?
i don't have any answers to these so i even the ones that i have attempted, i don't know if the answer is right or wrong
thankyou soooo much in advance
Hey mq123! I think we have another case of the WA course not overlapping with ours, so I can't give full solutions. I don't have all the info. However, the help I can give:
For Part A, you would need the total energy of the muon/anti-muon system. I wager this would be some combination of their masses, and perhaps their speed. Once you have this, the conservation of energy dictates that this energy must now be contained in the two photons. So,
For Part B, use the formula linking a photons energy to its frequency, and then convert this to a wavelength. In one step, it looks like this:
For Part C/D, I think your answer lies in the
Conservation of Momentum. When two particles collide, they are moving (presumably) in opposite directions. There is momentum in one direction, and then an identical momentum in the opposite direction, and so the total momentum of the system is likely to be zero (same mass, velocities in opposite directions, and so the momentum of one cancels out the other when we add them as vectors). Photons, being a particle, must obey this principle, and so the total momentum of the photons must also be 0. Thus, there must be two photons, as they move in opposite directions to obey the Conservation of Momentum. We can't have just 1, because there would then be momentum in one direction, and the momentum in the opposite direction would have been "lost."
Let me know if this is a little confusing and I will elaborate.For Part E, take your wavelength and map it to the EM Spectrum, it will probably be in the visible spectrum.
I hope this helps!