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Theory1:e+ +e-
η + γ
Assuming Theory 1 to be correct, the energy of a single photon could be determined from:
A. ∆msystemc2
B. ∆msystemc2 +½(∆mve+2 + ∆mve-2 - ∆mvη2)
C. ∆msystemc2 + ½(mve+2 + mve-2 - mvη2)C is the best answer. There are three types of energy to consider in this nuclear reaction: the kinetic energies of the particles, the energy of any photons, and the energy associated with mass change. If Theory 1 were correct, then the energy of a single photon could be found by taking the initial energy of the system and subtracting the final energy of the system (energy of the neutrino and any energy gained by mass loss), not including the photon. The electron and positron both became energy, so all of their kinetic energy must be a part of the final system. In other words, we do not have to consider ∆mass for the electron, proton, or neutron, given that they are present on only one side of the reaction. The relationship is as follows:
KEe++ KEe- = Ehƒ + KEη - ∆msystemc2
½mve+2 + ½mve-2 = hνγ + ½mvη2 - ∆msystemc2
½mve+2 + ½mve-2 - ½mvη2 + ∆msystemc2 = hνγ
½(mve+2 + mve-2 - mvη2) + ∆msystemc2 = hνγ
which becomes:
hνγ = ∆msystemc2 + ½(mve+2 + mve-2 - mvη2)
The best answer is C.
D. ∆msystemc2 + ½(∆pe+2 + ∆pe-2 + ∆pη2)
What I don't understand is why do we subtract mc2 in the equation and not add??
Assuming Theory 1 to be correct, the energy of a single photon could be determined from:
A. ∆msystemc2
B. ∆msystemc2 +½(∆mve+2 + ∆mve-2 - ∆mvη2)
C. ∆msystemc2 + ½(mve+2 + mve-2 - mvη2)C is the best answer. There are three types of energy to consider in this nuclear reaction: the kinetic energies of the particles, the energy of any photons, and the energy associated with mass change. If Theory 1 were correct, then the energy of a single photon could be found by taking the initial energy of the system and subtracting the final energy of the system (energy of the neutrino and any energy gained by mass loss), not including the photon. The electron and positron both became energy, so all of their kinetic energy must be a part of the final system. In other words, we do not have to consider ∆mass for the electron, proton, or neutron, given that they are present on only one side of the reaction. The relationship is as follows:
KEe++ KEe- = Ehƒ + KEη - ∆msystemc2
½mve+2 + ½mve-2 = hνγ + ½mvη2 - ∆msystemc2
½mve+2 + ½mve-2 - ½mvη2 + ∆msystemc2 = hνγ
½(mve+2 + mve-2 - mvη2) + ∆msystemc2 = hνγ
which becomes:
hνγ = ∆msystemc2 + ½(mve+2 + mve-2 - mvη2)
The best answer is C.
D. ∆msystemc2 + ½(∆pe+2 + ∆pe-2 + ∆pη2)
What I don't understand is why do we subtract mc2 in the equation and not add??
