Derivation of the atomic nucleus formula

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SUMMARY

The heuristic derivation of the atomic nucleus formula, represented as $$R = r_{0} \cdot A^{\frac{1}{3}}$$, establishes a relationship between the radius of the atomic nucleus (R), the radius of a nucleon (r0), and the number of nucleons (A). The derivation begins with the concept that the total volume of the nucleus is equivalent to the sum of the volumes of individual nucleons. By cubing both sides of the equation, $$R^3 = A r_0^3$$, and applying the formula for the volume of a sphere, the relationship is confirmed. This theoretical approach provides a clear understanding of the geometric properties of atomic nuclei.

PREREQUISITES
  • Understanding of basic nuclear physics concepts
  • Familiarity with the volume formula for spheres
  • Knowledge of nucleon properties (protons and neutrons)
  • Basic algebra for manipulating equations
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  • Learn about the role of nuclear forces in determining nucleus size
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This discussion is beneficial for physics students, nuclear physicists, and educators seeking to deepen their understanding of atomic structure and the mathematical relationships governing nuclear properties.

Bestfrog
Does anyone know the heuristic derivation of this formula? $$R=r_{0} \cdot A^{\frac{1}{3}}$$ with ##R## the atomic nucleus, ##r_0## the radius of a nucleon (proton or neutron) and ##A## the number of nucleons.
I know that there is a sperimental derivation, but I would find a theoretic/heuristic derivation.
 
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I take it the "R" is the "radius" of the atomic nucleus. So all they're saying is that the total volume for radius R is the same as the the total of the volumes for each nucleon. Cubing both sides gives you ##R^3 = A r_0^3##. Then multiple both sides by 4/3 pi and you have volume=volume.
 
Thank you!
 

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