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nv125
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Electrodynamics and Relativity. Please Help!
Consider the decay of a particle.
(a) Find the threshold kinetic energies of the incident particles in each of
the following reactions:
(i) ∏+ + p →∏++ ∏° + p
(ii) p + n →p + n + ∏° + ∏°
The incident particle is the first particle listed in the reactions; the target particle is at rest. The masses of the ∏+ and ∏° ions are 139.567MeV/c2 and 134.963 MeV/c2 respectively while the masses of the neutron and proton are respectively 939.566 MeV/c2 and 938.272 MeV/c2.
The teacher suggested that I will need to use the following equations to get the solution
(Here P is the momentum and Px is the momentum of the incident particle ∏+in the X direction, assuming it hits the target particle p(proton) horizontally, so no P(momentum) in y or z direction)
ƩPiu=0
PuPu=m2c2= E2/C2-P2
Also said that using this notations will be useful to solve the problem
For ∏+, the incident particle:
Pu=(E/C, Px, 0, 0)
Pu=(E/C, -Px, 0, 0) where Px is the momentum of the incident particle, and no need to worry about y and z direction so just put zeros.
This is what I have done so far to solve for part i, but I have no clue how to continue.
Since we are solving for the threshold kinetic energy, so assuming that ∏+, ∏° and p are at rest after the reaction.
(E/C, Px, 0, 0) +(mpC2, 0, 0, 0) - (m∏+C2,0, 0, 0 ) -(m∏°C2,0, 0, 0 ) - (mpC2,0, 0, 0 ) = 0
mp, m∏+, m∏° are just the rest mass.
PLEASE PLEASE help me to do this problem! thank you so much!
Homework Statement
Consider the decay of a particle.
(a) Find the threshold kinetic energies of the incident particles in each of
the following reactions:
(i) ∏+ + p →∏++ ∏° + p
(ii) p + n →p + n + ∏° + ∏°
The incident particle is the first particle listed in the reactions; the target particle is at rest. The masses of the ∏+ and ∏° ions are 139.567MeV/c2 and 134.963 MeV/c2 respectively while the masses of the neutron and proton are respectively 939.566 MeV/c2 and 938.272 MeV/c2.
Homework Equations
The teacher suggested that I will need to use the following equations to get the solution
(Here P is the momentum and Px is the momentum of the incident particle ∏+in the X direction, assuming it hits the target particle p(proton) horizontally, so no P(momentum) in y or z direction)
ƩPiu=0
PuPu=m2c2= E2/C2-P2
Also said that using this notations will be useful to solve the problem
For ∏+, the incident particle:
Pu=(E/C, Px, 0, 0)
Pu=(E/C, -Px, 0, 0) where Px is the momentum of the incident particle, and no need to worry about y and z direction so just put zeros.
The Attempt at a Solution
This is what I have done so far to solve for part i, but I have no clue how to continue.
Since we are solving for the threshold kinetic energy, so assuming that ∏+, ∏° and p are at rest after the reaction.
(E/C, Px, 0, 0) +(mpC2, 0, 0, 0) - (m∏+C2,0, 0, 0 ) -(m∏°C2,0, 0, 0 ) - (mpC2,0, 0, 0 ) = 0
mp, m∏+, m∏° are just the rest mass.
PLEASE PLEASE help me to do this problem! thank you so much!