Here's a question from TBR CBT 7:
28. 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)
D. ∆msystemc2 + ½(∆pe+2 + ∆pe-2 + ∆pη2)
Information from Passage:
Theory 1: e+ + e- --> η + γ
where η represents a neutrino, and γ represents a highly energetic photon (gamma ray). A neutrino is a particle with negligible mass and no charge, while a positron is an antimatter particle identical to an electron in mass, but opposite in charge.
Answer: 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- = Eh + KEη - ∆msystemc^2
½mve+^2 + ½mve-^2 = hνγ + ½mvη^2 - ∆msystemc^2
½mve+^2 + ½mve-^2 - ½mvη^2 + ∆msystemc^2 = hνγ
½(mve+^2 + mve-^2 - mvη^2) + ∆msystemc^2 = hνγ
which becomes:
hνγ = ∆msystemc^2 + ½(mve+^2 + mve-^2 - mvη^2)
The best answer is C.
I understand that C is the best answer. However, my question goes a bit beyond this. Given this formula in the explanation, KEe+ + KEe- = Eh + KEη - ∆msystemc2, why is ∆msystemc2 subtracted from the right side? Is it because the positron's and electron's masses dissipated into energy and formed a neutrino, which is essentially massless, but still has some tiny mass? Any help at all to clear this up would be greatly appreciated!
28. 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)
D. ∆msystemc2 + ½(∆pe+2 + ∆pe-2 + ∆pη2)
Information from Passage:
Theory 1: e+ + e- --> η + γ
where η represents a neutrino, and γ represents a highly energetic photon (gamma ray). A neutrino is a particle with negligible mass and no charge, while a positron is an antimatter particle identical to an electron in mass, but opposite in charge.
Answer: 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- = Eh + KEη - ∆msystemc^2
½mve+^2 + ½mve-^2 = hνγ + ½mvη^2 - ∆msystemc^2
½mve+^2 + ½mve-^2 - ½mvη^2 + ∆msystemc^2 = hνγ
½(mve+^2 + mve-^2 - mvη^2) + ∆msystemc^2 = hνγ
which becomes:
hνγ = ∆msystemc^2 + ½(mve+^2 + mve-^2 - mvη^2)
The best answer is C.
I understand that C is the best answer. However, my question goes a bit beyond this. Given this formula in the explanation, KEe+ + KEe- = Eh + KEη - ∆msystemc2, why is ∆msystemc2 subtracted from the right side? Is it because the positron's and electron's masses dissipated into energy and formed a neutrino, which is essentially massless, but still has some tiny mass? Any help at all to clear this up would be greatly appreciated!
