neutron's decay, kintetic energy

[rayman123]

1. The problem statement, all variables and given/known data
What is the kinetic energy given to the proton in the decay of the neutron when a) the electron has negligibly small kinetic energy; b)antineutrino has small negiligibly energy




2. Relevant equations
The decay reaction
n\Rightarrow p+e^{-}+\bar{\nu}
now I try to work out the Q-value (I assume that neutron is at rest)
m_{n}c^2=m_{p}c^2+T_{p}+m_{e}c^2+T_{e}+m_{\bar{nu} }c^2+T_{\bar{\nu}}
then movin all the therms with c^2 to the left side I get
(m_{n}-m_{p}-m_{e}-m_{\bar{\nu}})c^2=T_{p}+T_{\bar{\nu}}
so
Q=T_{p}+T_{\bar{\nu}}

the Q value can be easily calculated but what with the other kinetic energies on the right side? It this correct?


3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[lanedance]

sorry what is the T?

is your decay is as follows:
n\Rightarrow p+e^{-}+\bar{\nu}

when one particle has negligible KE, then you can just consider a momentum energy balance for the other 2 particles

so, say there is excess energy E, and the 2 particles of interest (with masses m1 & m2) will move in oppposite dierctions along a single axis (with speeds v1 & v2), then they must satisfy:

conservation of energy
E = m_1 v_1^2 + m_2 v_2^2

conservation of monemtum
m_1 v_1 + m_2 v_2 = 0

[rayman123]

T stands for kinetic energy

[vela]

For part (a), you're assuming the electron is at rest, so what does that tell you about the momenta of the antineutrino and proton?

Instead of separating out the kinetic energy at the beginning of the problem, I'd try solving for the total energy of the proton and then subtract off the rest energy at the end to get the kinetic energy.

[rayman123]

that their momenta are equal?

[vela]

Yes, equal in magnitude but opposite in direction.