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 Incoming links: Uniform-sphere Coulomb displacement energy

Past exam of the particle and nuclear physics course of the University of Cambridge / 2015 / Paper 1 / 4 / ii / Solution 2026-10-20

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The mirror nuclei have interchanged neutron and proton numbers. Neglect their strong-interaction binding difference and describe them with the same uniformly charged sharp sphere of radius

R

. The nuclear Coulomb energy is

E_{C} (Z) = \frac{3}{5} \frac{Z ^{2} e ^{2}}{4 π ϵ _{0} R} = \frac{3}{5} \frac{Z ^{2} α ℏ c}{R} .

(1)

Changing one neutron into a proton decreases the nucleon rest-mass sum by

m_{n} - m_{p} = 1.294 MeV / c^{2}

, but raises the Coulomb energy. The diagram gives

(M_{F} - M_{O}) c^{2} = 2.25 MeV

for nuclear masses, so

2.25 = - 1.294 + \frac{3}{5} (9^{2} - 8^{2}) \frac{α ℏ c}{R} .

(2)

With the standard electromagnetic conversion

α ℏ c ≃ 1.44 MeV fm

, the uniform-sphere Coulomb displacement energy estimate is

R = \frac{( 3/5 ) ( 17 ) ( 1.44 )}{3.544} fm ≃ 4.14 fm .

(3)

This

R

is the uniform sphere's outer radius. If “charge radius” means the rms nuclear charge radius, the same distribution gives

⟨ r^{2} ⟩ = 3/5 R ≃ 3.21 fm .

(4)

The numerical result is an estimate within the charge-density model, not a model-independent radius measurement. A discrete-proton correction replaces

Z^{2}

Z (Z - 1)

, giving

Δ [Z (Z - 1)] = 16

and

R ≃ 3.90 fm

. The main answer above follows the literal continuum charged-sphere model. Strong isospin-breaking, nonuniform density and changes in the valence-nucleon distribution also limit the extraction.

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