Beta Minus and Beta Plus Decay Disintegration Energies

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SUMMARY

This discussion clarifies the disintegration energies involved in beta minus and beta plus decay, specifically addressing the confusion surrounding the inclusion of electron masses in calculations. In beta minus decay, the mass of the electron is accounted for in the atomic mass of the reactants and products, as indicated in Equation (12.38). For beta plus decay, the disintegration energy calculation incorporates the mass of the positron, which is twice that of the positron due to the additional electron in the atomic mass of the resulting atom. The distinction between atomic and nuclear masses is crucial for accurate energy calculations.

PREREQUISITES
  • Understanding of beta decay processes
  • Familiarity with atomic and nuclear mass concepts
  • Knowledge of binding energy and its implications
  • Ability to interpret equations in nuclear physics
NEXT STEPS
  • Review the derivation of disintegration energy in beta decay using nuclear mass equations
  • Watch the referenced video starting at 11:08 for a detailed explanation of mass considerations
  • Study the differences between atomic mass and nuclear mass in decay processes
  • Explore the implications of binding energy in nuclear reactions
USEFUL FOR

Students and professionals in nuclear physics, educators teaching decay processes, and researchers focusing on particle interactions and energy calculations.

Mark Zhu
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Homework Statement
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Relevant Equations
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I am confused about the disintegration energies of beta minus and beta plus decay. Regarding beta minus decay, the textbook says that "the number of electron masses has been accounted for in Equation (12.38)." What does that mean? Usually the disintegration energy is simply the mass of the reactants minus the mass of the products, but as you can see the electron (Beta minus) mass is not accounted for in equation 12.38. Also, the disintegration energy for beta plus decay is shown in the second attachment Capture1.PNG. It accounts for twice the mass of the positron. I am confused. Thank you.
 

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If we compute the energy mass from the value of A, it does not depend on how the A is made up of protons and neutrons. So the right hand M() term, in comparison to the left hand M() term, overestimates by the difference in mass between a proton and a neutron, or about the mass of an electron.
 
1605021403893.png

The masses in the above equation are atomic masses, rather than nuclear masses. Thus, ##M\left(^A_ZX\right)## represents the mass of the nucleus plus the mass of ##Z## orbital electrons in the atom plus the mass equivalence of the binding energy of the atomic electrons. The binding energy of the electrons is small enough to be neglected. So, you can take the atomic mass ##M\left(^A_ZX\right)## to be the nuclear mass of ##^A_ZX## plus the mass of ##Z## electrons. Likewise, ##M\left(^A_{Z+1}D\right)## equals the nuclear mass of ##_{Z+1}^AD## plus the mass of ##Z+1## electrons. The fact that the atomic mass of ##D## includes one more electron compared to the atomic mass of ##X## is why equation (12.38) is correct even though at first glance it seems to be missing the mass of the electron created in the beta decay.

If you want to see a careful treatment of this tricky point, see this video. You can start watching at time 11:08. The switch from nuclear masses to atomic masses starts at time 13:05. The case of ##\beta^+## decay starts at 15:45.
 

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