Is the Beta Decay Mass Condition Dependent on Atomic Masses?

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SUMMARY

The condition for beta decay via positron emission is defined by the equation M_p > M_d + 2m_e, where M_p is the atomic mass of the parent nucleus, M_d is the atomic mass of the daughter nucleus, and m_e is the mass of an electron. This equation accounts for the emitted positron and the necessity of an additional electron to maintain atomic stability. The discussion also clarifies that the most stable isobar on an atomic mass parabola is not necessarily the most stable overall, as stability can depend on various factors beyond atomic mass alone.

PREREQUISITES
  • Understanding of beta decay processes, specifically positron emission.
  • Familiarity with atomic and nuclear mass concepts.
  • Knowledge of lepton number conservation in particle physics.
  • Basic grasp of isobars and their stability in nuclear physics.
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  • Study the principles of beta decay, focusing on positron emission and its implications.
  • Research atomic versus nuclear mass differences and their significance in decay processes.
  • Examine the concept of isobars and their stability in relation to atomic mass parabolas.
  • Explore lepton number conservation and its role in particle decay reactions.
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Students of nuclear physics, physicists specializing in particle interactions, and educators seeking to deepen their understanding of beta decay mechanisms and atomic stability concepts.

Master J
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In beta decay, positron emission, how come the condition for decay is:

M_p > M_d + 2m_e

Thats: atomic mass of parent > "daughter + twice the mass of an electron.

I'm sure there is some simple way of showing it, but I can't seem to find it!

Also, is the most stable isobar on an atomic mass parabola the most stable one? It's a question my lecturer posed to us, and I have been thinking about it a while.

Thanks!
 
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It is 2m_e because it is the atomic and not the nuclear mass.
One m_e is because of the electron emitted by the nucleus.
The second m_e is the extra electron that must be added to the outer shell of electrons in the atom.
 
that is since the daughter nucleus will have one electron less than the parent, and what is listed is ATOMIC masses. So if you calculate it with NUCLEAR masses it should of course be:

M_p > M_d + m_e

if now M is nuclear masses.

The other question is a homework - course work question, and should not be adressed in this forum but in the homework forum, with an attempt to solution.
 
"Also, is the most stable isobar on an atomic mass parabola the most stable one?"
As you write this question it is a tautology: Is the most stable the most stable?
 
For the parent p to decay to the daughter d by positron decay, the parent would have to create both a positron and electron (lepton number is conserved) before having sufficient energy to decay by positron emission. Example: could a proton decay to a neutron plus positron? What would the minimum proton mass be?
 
Bob S said:
For the parent p to decay to the daughter d by positron decay, the parent would have to create both a positron and electron (lepton number is conserved) before having sufficient energy to decay by positron emission. Example: could a proton decay to a neutron plus positron? What would the minimum proton mass be?

nope, the electron number is conserved by emitting a neutrino

beta+ decay on subnuclear level is:

p -> n + e(+) + neutrino

You are proposing:

p-> n + e(+) + e(-)

which violates electric charge conservation.
 
malawi_glenn said:
nope, the electron number is conserved by emitting a neutrino

beta+ decay on subnuclear level is:

p -> n + e(+) + neutrino

You are proposing:

p-> n + e(+) + e(-)

which violates electric charge conservation.
Mea Culpa. There are three types of beta decay exibited by nuclei: Here are 3 examples, all from Cu64
A) Positron emission (by proton), with an antineutrino
B) Electron emission (by neutron) , with a neutrino
C) K-shell electron capture (by proton) with only an antineutrino emission

(From Wiki):
64Cu has a half-life of 12.701 ± 0.002 hours and decays by 17.86 (± 0.14)% by positron emission, 39.0 (± 0.3)% by beta decay, 43.075 (± 0.500)% by electron capture
 
where does the electron that must be added to the shell then come from? if its from outside the atomic system in question, then surely it has no place in the mass condition?
 
Master J said:
where does the electron that must be added to the shell then come from? if its from outside the atomic system in question, then surely it has no place in the mass condition?


WHERE it comes from is not important, the issue is that you are using atomic masses.. as I explain to you.
 

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