Binding energy of the alpha particle

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SUMMARY

The discussion focuses on calculating the binding energy of a carbon-12 nucleus compared to the binding energies of three alpha particles. The user employed the liquid drop model and the mass-energy equivalence formula BE = (A-Z)mn + Zmp - ma to derive binding energies of approximately 23 MeV and 28 MeV, respectively. The user questions the discrepancy between these values and the assumption that the total binding energy of three alpha particles equals that of carbon-12. The conclusion emphasizes that while the liquid drop model provides an approximation, using mass differences yields exact results, contingent upon accurate mass measurements.

PREREQUISITES
  • Understanding of nuclear physics concepts, particularly binding energy
  • Familiarity with the liquid drop model in nuclear physics
  • Knowledge of mass-energy equivalence and its application in nuclear calculations
  • Basic proficiency in using equations involving atomic mass and binding energy
NEXT STEPS
  • Research the liquid drop model and its limitations in nuclear binding energy calculations
  • Study the mass-energy equivalence principle and its implications in nuclear physics
  • Explore advanced nuclear models, such as the shell model, for a deeper understanding of binding energy
  • Investigate experimental methods for measuring atomic masses to improve binding energy calculations
USEFUL FOR

Students and educators in nuclear physics, researchers focusing on nuclear binding energy, and anyone interested in the theoretical and practical aspects of nuclear structure and stability.

wakko101
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Homework Statement


Calculate the difference between the binding energy of a nucleus of carbon 12 and the sum of the binding energies of three alpha particles. Assuming the carbon 12 is composed of three alpha particles in a triangular structure, with three effective "alpha bonds" joining them, what is the binding energy per alpha bond?


Homework Equations


I have calculated the binding energy for the alpha particles using the liquid drop model and using BE = (A-Z)mn + Zmp - ma (where mn is the mass of the neutron, mp that of the proton and ma that of the atom in question). When I do it the latter way, I get around 23 MeV, and when I do it the former way I get around 28 MeV which according to my googling is the corrent answer. I'm wondering why this would be so?


The Attempt at a Solution


Also, assuming we use the first equation for BE, then the different between the two binding energies is 0 (since we would have 3 times the BE for an alpha particle equals the binding energy of carbon 12). This doesn't seem right to me.

Any comments or advice would be appreciated.

Cheers,
W. =)
 
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The liquid drop model is an approximation. Using the difference in masses to calculate the binding energy is exact (assuming you know the masses exactly).
 

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