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olsyboy
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#1
In annihilation, do the two photons produced have the same energy?
This question has the answer C, however I thought it would be D.
Two gamma photons are produced when a muon and an antimuon annihilate each other.
What is the minimum frequency of the gamma radiation that could be produced?
A 2.55 × 10^16 Hz
B 5.10 × 10^16 Hz
C 2.55 × 10^22 Hz
D 5.10 × 10^22 Hz
This question has the answer C, however I thought it would be D.
Two gamma photons are produced when a muon and an antimuon annihilate each other.
What is the minimum frequency of the gamma radiation that could be produced?
A 2.55 × 10^16 Hz
B 5.10 × 10^16 Hz
C 2.55 × 10^22 Hz
D 5.10 × 10^22 Hz
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honestlyidk
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From what I understood is that the rest energy of a Muon is about 106MeV (given in the data booklet) and then you convert MeV into Joules so, 106 X (1.6x10^-13) = Ans. And because E=hf, (we have the energy in joules now cos we worked it out above), you divide the answer you got above by Planck's constant to get your frequency. Hope this helped! 
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Joealex24
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(Original post by honestlyidk)
From what I understood is that the rest energy of a Muon is about 106MeV (given in the data booklet) and then you convert MeV into Joules so, 106 X (1.6x10^-13) = Ans. And because E=hf, (we have the energy in joules now cos we worked it out above), you divide the answer you got above by Planck's constant to get your frequency. Hope this helped!
From what I understood is that the rest energy of a Muon is about 106MeV (given in the data booklet) and then you convert MeV into Joules so, 106 X (1.6x10^-13) = Ans. And because E=hf, (we have the energy in joules now cos we worked it out above), you divide the answer you got above by Planck's constant to get your frequency. Hope this helped!
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Thanquinho
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(Original post by Joealex24)
you only mentioned the muon. Does the antimuon not have energy?
you only mentioned the muon. Does the antimuon not have energy?
Consider the conservation of momentum, when a particle and its antiparticle collide, there cannot be just 1 photon created, as it would break the conversation of momentum, therefore 2 photons are created that travel in opposite directions to converse momentum.
Now, checking the data booklet, you use the rest energy of a muon and convert it into joules. this energy will be the same for the antimuon, so you can double it. Then, because E = hf, you do E/h to find out the frequency. But as mentioned earlier, there are 2 photons created, not just one, so you have to divide the frequency by 2 as well. But because you multiply the energy by 2 and divide the frequency by 2, you might as well not include that 2 in ur multiplication/division as you end up with the same answer. But if it helps you understand the concept better, there is nothing wrong with multiplying E by 2 and dividing f by 2.
As I already said, you will end up with the same answer.
Hope this helps a little!
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