Relative stability of four nuclei

Problem

Consider four nuclei: (A) deuterium $^{2}\text{H}$, (B) helium-4 $^{4}\text{He}$, (C) iron-56 $^{56}\text{Fe}$, (D) uranium-238 $^{238}\text{U}$. Their binding energies per nucleon are approximately: A $1{,}1$; B $7{,}1$; C $8{,}8$; D $7{,}6\ \text{MeV/nucleon}$. Rank the four nuclei from most to least stable per nucleon.

Solution

The stability of a nucleus, referred to the single nucleon, is measured by the binding energy per nucleon $B/A$: the higher $B/A$, the more energy is needed to pull out a nucleon, and the more stable the nucleus.

Comparing the given values:

$C\ (8{,}8) > D\ (7{,}6) > B\ (7{,}1) > A\ (1{,}1)$

Iron-56 is the most stable: it sits near the peak of the binding energy curve, the region of most tightly bound nuclei. Deuterium, which has only a single proton-neutron bond, is by far the least stable per nucleon.

$\ev{C > D > B > A}$

Connections

Topics: Nuclear physics Concepts: Nuclear binding energy Skills: Ranking reasoning