Calculating energy released from beta decay

Click For Summary
Beta decay involves a neutron transforming into a proton and a W- boson, which subsequently decays into an electron and an anti-neutrino. The energy released during this process is derived from the mass difference between the neutron and the decay products, with most kinetic energy appearing in the electron and anti-neutrino due to conservation of momentum. The confusion arises from the application of the energy-momentum relation, E = m²c⁴ + p²c², which can lead to incorrect mass calculations if not handled properly. The energy partition is influenced by the angles of the emitted particles, particularly in cases where the neutron is part of a nucleus. Understanding these dynamics is crucial for accurately calculating the energy involved in beta decay.
tyrant91101
Messages
13
Reaction score
0
I have been learning particle physics lately but it's been mostly from a theoretical perspective and not a mathematical one so I have yet to come across any such math but my curiosity is peaked.

From what I understand it, this is the process:

<br /> n \rightarrow p + W^{-}<br />

Followed by:

<br /> W^{-} \rightarrow e^{-} + \bar{v}_{e}<br />

Since m_{e} ~= .511 MeV/c^2 and m_{v_{e}} &lt;&lt; m_{e}, there is about 79.9995 GeV/c^{2} missing. I am unclear where this energy goes. Does it go into the momentum of the electron (since it is ultrarelitivistic during the decay)?

I've tried to solve the equation E = m^{2}c^{4} + p^{2}c^{2} but I get a weird mass for the electron (5.68 x 10^{-12} kg) and I am all around confused by the equation.

If I have the wrong equation, which other one do I use for this type of math? If I am using the write equation, what is the best way of dealing with the units?
 
Physics news on Phys.org
The kinetic energy of the proton (or nucleus), electron and anti-neutrino comes from the difference in mass between the products p + e^{-} + \bar{v}_{e} and the neutron. The partition of the energy is governed by conservation of momentum. If the neutron is in a nucleus, the nucleus acquires very little kinetic energy because it's some massive. Even the case of the bare neutron, the vast majority of kinetic energy is manifest in the electron and anti-neutrino, and the partition of energy is largely determined by the angle between the electron and anti-neutrino trajectories.

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta2.html#c1
 
Last edited:
  • Like
Likes Atomillo

Similar threads

  • · Replies 32 ·
2
Replies
32
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
21
Views
2K