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Nuclear Binding, Masses, and the Semiclassical Mass Formula

Manual: General · Subject: Nuclear Physics

Analyze binding energies, saturation, and nuclear masses through microscopic and phenomenological lenses.

Binding energy systematics

Mass defect and binding energy

The nuclear binding energy is B=[Zmp+Nmn−M(A,Z)]c2B = \left[ Zm_p + Nm_n - M(A,Z) \right]c^2 where M(A,Z)M(A,Z) is the nuclear mass. Binding energy reflects the balance of attraction from the strong interaction and repulsion from the Coulomb interaction, plus shell and pairing corrections.

The semi-empirical mass formula

A useful global model is B(A,Z)=avA−asA2/3−acZ(Z−1)A1/3−aa(A−2Z)2A+δ(A,Z)B(A,Z)=a_vA-a_sA^{2/3}-a_c\frac{Z(Z-1)}{A^{1/3}}-a_a\frac{(A-2Z)^2}{A}+\delta(A,Z) where the volume, surface, Coulomb, asymmetry, and pairing terms encode major trends across the chart of nuclides.

Interpretation of terms

Volume term
Bulk saturation from short-range nuclear attraction
Surface term
Reduced binding for nucleons at the boundary
Asymmetry term
Penalty for neutron-proton imbalance
Pairing term
Even-even nuclei are generally more bound

Which term in the semi-empirical mass formula primarily penalizes neutron-proton imbalance?

What physical effect does the Coulomb term mainly represent?

ℹ️

Modern perspective

The liquid-drop picture is excellent for trends, but precision mass predictions require shell corrections, pairing correlations, and microscopic many-body calculations.

Typical mass-formula contributions

TermScaling
Volume∝A\propto A
Surface∝A2/3\propto A^{2/3}
Coulomb∝Z2/A1/3\propto Z^2/A^{1/3}
Asymmetry∝(A−2Z)2/A\propto (A-2Z)^2/A

Why does the surface term reduce the binding per nucleon for light nuclei?

State one reason the semi-empirical mass formula is not sufficient for high-precision nuclear mass predictions.