When an electron of charge –e and positron of charge +e are brought close to each other, they annihilate each other to produce charge-less $\gamma $ photons.
A. True
B. False
Answer
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Hint: We see that, in the question we are given two oppositely charged particles which when brought in contact with each other has the possibility of annihilating each other. By the law of conservation of charge, this annihilation should result in a neutral particle. Find out whether this neutral particle is photon and hence answer the question.
Complete answer:
We are given a statement that ‘When an electron of charge –e and positron of charge +e are brought close to each other, they annihilate each other to produce charge-less $\gamma $ photons’ and are asked whether this given statement is true or false.
The given statement is true. Electron-positron annihilation does occur when an electron and a positron collide. In case you don’t know, the positron is the antiparticle of an electron. But when they collide there are two possibilities: one is a lower energy case and the other is the high energy case.
${{e}^{-}}+{{e}^{+}}\to \gamma +\gamma $
Let us have a closer look at both these cases.
High energy case:
This is the case when either the electron or the positron or both have significant Kinetic energies. Since there is enough energy to provide the rest energies for heavier particles like D mesons or B mesons, they can also be produced. There also exists a possibility of production of photons and other light particles with high kinetic energies in this case.
Low energy case:
In this case the most probable final state is the production of two or more protons. Each of them will have energy equal to the rest energy of the electron/ positron.
Therefore, we see that $\gamma $ photons are produced when an electron and a positron collide.
So, the correct answer is “Option A”.
Note:
You may wonder why the collision does not produce a single photon instead of two or more. The reason is that conservation of energy and linear momentum forbid them to do so. However, tightly bound atomic electrons are an exception. The electron-positron creation is the reverse reaction to that discussed here.
Complete answer:
We are given a statement that ‘When an electron of charge –e and positron of charge +e are brought close to each other, they annihilate each other to produce charge-less $\gamma $ photons’ and are asked whether this given statement is true or false.
The given statement is true. Electron-positron annihilation does occur when an electron and a positron collide. In case you don’t know, the positron is the antiparticle of an electron. But when they collide there are two possibilities: one is a lower energy case and the other is the high energy case.
${{e}^{-}}+{{e}^{+}}\to \gamma +\gamma $
Let us have a closer look at both these cases.
High energy case:
This is the case when either the electron or the positron or both have significant Kinetic energies. Since there is enough energy to provide the rest energies for heavier particles like D mesons or B mesons, they can also be produced. There also exists a possibility of production of photons and other light particles with high kinetic energies in this case.
Low energy case:
In this case the most probable final state is the production of two or more protons. Each of them will have energy equal to the rest energy of the electron/ positron.
Therefore, we see that $\gamma $ photons are produced when an electron and a positron collide.
So, the correct answer is “Option A”.
Note:
You may wonder why the collision does not produce a single photon instead of two or more. The reason is that conservation of energy and linear momentum forbid them to do so. However, tightly bound atomic electrons are an exception. The electron-positron creation is the reverse reaction to that discussed here.
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