Neutron Decay

Nuclei are made up of protons and neutrons. While neutrons are stable inside many nuclei, free neutrons decay with a lifetime of about 15 minutes. This makes them a radiation problem around nuclear reactors, since they can leak out of the reactor and decay. The neutron decays into a proton, an electron, and an antineutrino of the electron type.

\bgroup\color{black}$\displaystyle \mathrm{n}\rightarrow\mathrm{p^+e^-}\bar{\nu}_e $\egroup

The mass of the neutron is 939.57 MeV. The proton mass is 938.28 MeV. The mass of the electron is 0.511 MeV. The mass of the electron neutrino is nearly zero. (We think its not zero but we only measure it to be small, \bgroup\color{black}$ m<1$\egroup eV.)

This is a more complicated problem. Lets count variables. The three final state particles give us 9 variables with only 4 equations to solve. It is usual not to worry about the orientation of the whole decay (unless the initial state is polarized). The three final state particles will lie in a plane (for neutron decay in its rest frame). It takes two variables to define the orientation of the normal to the plane and the event can still be rotated inside the plane so like a rigid body, three angles are needed to describe the overall orientation of the event. Since the laws of physics are symmetric under these rotations, these three variables do not matter. So its fine to have 3 unknown variables at the end but here we have 5.

That means we can have some kind of interesting physics in this decay (that we won't understand here). That is, the final state is not completely given by kinematics. There are two variables undetermined that depend on the physics of the decay.

To make a problem we can solve, we ask the question ``what is the maximum electron energy possible?'' in this decay. The maximum possible energy will occur when both the proton and neutrino recoil directly opposite the electron's direction. If we pick the direction of the electron (2 variables), we have a one dimensional problem with three unknown momenta and just two equations, energy conservation and momentum conservation. We will still need to maximize the electron momentum to get the answer.

Lets work with the three independent variables \bgroup\color{black}$ p_e$\egroup, \bgroup\color{black}$ p_p$\egroup, and \bgroup\color{black}$ p_\nu$\egroup. The three energies are.

  $\displaystyle E_e=\sqrt{m_e^2+p_e^2}$    
  $\displaystyle E_p=\sqrt{m_p^2+p_p^2}$    
  $\displaystyle E_\nu=p_\nu$    

Defining all the momenta to be positive numbers, the two conservation equations are.

  $\displaystyle m_n=\sqrt{m_e^2+p_e^2}+\sqrt{m_p^2+p_p^2}+p_\nu$    
  $\displaystyle 0=p_e-p_P-p_\nu$    
  $\displaystyle p_\nu=p_e-p_p$    
  $\displaystyle m_n=\sqrt{m_e^2+p_e^2}+\sqrt{m_p^2+p_p^2}+p_\nu$    
  $\displaystyle \sqrt{m_e^2+p_e^2}=m_n-\sqrt{m_p^2+p_p^2}-p_e+p_p$    
  $\displaystyle \sqrt{m_e^2+p_e^2}+p_e=m_n-\sqrt{m_p^2+p_p^2}+p_p$    

To maximize \bgroup\color{black}$ p_e$\egroup, we must maximize \bgroup\color{black}$ p_p$\egroup. This will happen for \bgroup\color{black}$ p_\nu=0$\egroup. Then we have \bgroup\color{black}$ p_p=p_e$\egroup.

  $\displaystyle \sqrt{m_e^2+p_e^2}+p_e=m_n-\sqrt{m_p^2+p_e^2}+p_e$    
  $\displaystyle \sqrt{m_e^2+p_e^2}=m_n-\sqrt{m_p^2+p_e^2}$    

The proton is highly non-relativistic ( \bgroup\color{black}$ p<<m$\egroup), so we may approximate its energy by.

  $\displaystyle \sqrt{m_p^2+p_e^2}\approx m_p + {p_e^2\over 2m_p}$    
  $\displaystyle \sqrt{m_e^2+p_e^2}=m_n-m_p -{p_e^2\over 2m_p}$    
  $\displaystyle \sqrt{m_e^2+p_e^2}=\Delta m -{p_e^2\over 2m_p}$    
  $\displaystyle m_e^2+p_e^2=(\Delta m)^2 -\Delta m{p_e^2\over m_p}+{p_e^4\over 4m_p^2}$    
  $\displaystyle p_e^2\left(1+{\Delta m\over m_p}\right)=(\Delta m)^2 -m_e^2+{p_e^4\over 4m_p^2}$    
  $\displaystyle p_e^2\left(1+{\Delta m\over m_p}\right)\approx (\Delta m)^2 -m_e^2$    
  $\displaystyle p_e\approx \sqrt{(\Delta m)^2 -m_e^2\over \left(1+{\Delta m\over ...
...ight)} = \sqrt{(1.29)^2 -(0.51)^2\over \left(1+{1.29\over 938.28}\right)}=1.183$    
  $\displaystyle E_e=\sqrt{m_e^2+p_e^2}=1.289$    

When the neutrino momentum is zero, the electron takes nearly all the energy available. The proton, recoiling against it, takes very little energy since it is highly nonrelativistic even though it has the same momentum as the electron.

Jim Branson 2012-10-21