 Solution

ID: past-exam-of-the-particle-and-nuclear-physics-course-of-the-university-of-cambridge/2021/section-b/6/a/solution

There are three free-particle rest-energy terms and five nuclear corrections. In the paper's natural units, their meanings are:

$Z m_{p}$ : the rest energy of $Z$ free protons.
$Z m_{e}$ : the rest energy of the atomic Electrons; including this term makes the result an atomic, rather than bare nuclear, mass. Small atomic binding energies are neglected.
$(A - Z) m_{n}$ : the rest energy of the free neutrons.
$a_{1} A$ , with $a_{1} < 0$ : the attractive volume term. The short-range nuclear force mainly binds each nucleon to nearby neighbours, so the leading attraction is extensive.
$a_{2} A^{2/3} > 0$ : the surface correction. Surface nucleons have fewer neighbours and are less bound; surface area grows as $A^{2/3}$ at approximately constant nuclear density.
$a_{3} Z^{2} / A^{1/3} > 0$ : the nuclear Coulomb energy cost of repelling protons, scaling as $Z^{2} / R$ with nuclear radius $R \propto A^{1/3}$ .
$a_{4} (A - 2 Z)^{2} / A > 0$ : the nuclear symmetry energy, penalizing unequal neutron and proton populations. The Pauli exclusion principle forces an excess species into higher single-particle states; strong-interaction effects also contribute.
$δ (A) / A^{1/2}$ : nuclear pairing. It lowers the mass for even-even nuclei, raises it for odd-odd nuclei, and vanishes for odd $A$ in the stated model.

The sign convention is significant: the quantity labeled

E_{bind}

in the supplied mass formula is added to the free-particle masses and is normally negative for a bound nucleus. It is the negative of the conventional positive nuclear binding energy

B

M = Z m_{p} + Z m_{e} + (A - Z) m_{n} + E_{bind}, B = - E_{bind} > 0.

The coefficients must therefore retain their printed signs; using positive

B

while still adding it to the masses would be inconsistent.

 New to topics? Read the docs here!