Beta Threshold Energy: Electron vs Positron

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SUMMARY

The discussion centers on the energy conditions for beta decay, specifically comparing electron and positron emissions. For electron emission, the condition is defined as [M(A,Z) - M(A,Z+1)]c^2 > 0, while for positron emission, it is [M(A,Z) - M(A,Z-1) - 2m_e]c^2 > 0. The asymmetry arises because positron emission requires the loss of both the positron and an electron, whereas electron emission only involves the loss of an electron mass and an increase in atomic number. This highlights the different energy thresholds required for each type of beta decay.

PREREQUISITES
  • Understanding of beta decay processes
  • Familiarity with atomic mass and nuclear potential well concepts
  • Knowledge of energy conservation in nuclear reactions
  • Basic principles of particle physics, particularly electron and positron behavior
NEXT STEPS
  • Research the principles of beta decay in nuclear physics
  • Study the role of atomic mass in nuclear reactions
  • Explore the concept of nuclear potential wells and their impact on particle emissions
  • Investigate the differences between electron and positron interactions in particle physics
USEFUL FOR

Students and professionals in nuclear physics, particle physicists, and anyone interested in the mechanisms of beta decay and energy thresholds in atomic reactions.

Puffin
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This has been annoying me since I read this somewhere.

Beta decay is only possible if energetically allowed. For electron emission this means:

[M(A,Z) - M(A,Z+1)]c^2 >0

For positron emission this is

[M(A,Z) - M(A,Z-1) - 2m_e]c^2 >0

Why the asymmetry? I would naively think that both of them would need

[M(A,Z) - M(A,Z+/-1) - m_e]c^2 >0

Why isn't this so? If anything, I'd have thought that the electron would have had the higher threshold (if that's the right word in this context) because it still has to climb out of the nuclear potential well. Am I missing something really simple? Cheers.
 
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M(A,Z) is not energy of the nucleus, but energy of a neutral atom with Z electrons.

Eugene.
 
Supplementing meopemuk's comment, one is using atomic masses.

In the case of beta (e-) one losses an electron mass from the nucleus, but Z increases by 1, but then the atom gains an electron because of the increase in Z.

In positron emission, the Z decreases by 1, and one electron leaves the atom. So the masses of the positron and one electron are lost.
 

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