What Is the Kinetic Energy of a Proton in Neutron Decay?

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Homework Help Overview

The discussion revolves around calculating the kinetic energy of a proton resulting from neutron decay, specifically considering scenarios where the electron and antineutrino have negligible kinetic energy. The subject area is particle physics, focusing on decay processes and energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the Q-value for the decay reaction and questions the treatment of kinetic energies involved. Some participants suggest using momentum and energy conservation principles to analyze the situation further, while others inquire about the meaning of terms used in the equations.

Discussion Status

The discussion is active, with participants providing insights into the relationships between kinetic energy and momentum in the context of the decay process. There is a focus on clarifying assumptions regarding particle motion and energy contributions, but no consensus has been reached on the approach to take.

Contextual Notes

Participants are navigating the complexities of the decay process, including assumptions about particle velocities and the implications of negligible kinetic energies. The original poster's assumptions about the neutron being at rest are also under examination.

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Homework Statement


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




Homework 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?


The Attempt at a Solution








 
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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
 
T stands for kinetic energy
 
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.
 
that their momenta are equal?
 
Yes, equal in magnitude but opposite in direction.
 

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