Energy and Human Ambitions on a Finite Planet, 2021a

15 Nuclear <strong>Energy</strong> 247
to 1.8 × 10 −9 J of energy. Since 1 MeV is 1.6 × 10 −13 J, we end up with
11,200 MeV corresponding to 12 a.m.u. (1 a.m.u. is 931.5 MeV).
In practice, <strong>and</strong> perhaps surprisingly, atoms (nuclei) weigh less than
the sum of their parts due to binding energy. In order to rip a nucleus
completely apart <strong>and</strong> move all the nucleons far from each other, energy
must be put in (left part of Figure 15.9). And any change in energy is
accompanied by a change in mass, via ΔE Δmc 2 . All the energy that
must be injected to completely dismantle the nucleus weighs something!
So the mass of the individual pieces after dismantling the nucleus is
effectively the mass of the original nucleus plus the mass-equivalent of
all the energy that was put in to tear it apart (middle panel of Figure
15.9). Therefore, binding energy effectively reduces the mass of a nucleus,
which we will now explore quantitatively.
+ <strong>Energy</strong>
+ <strong>Energy</strong>
<strong>Energy</strong>
Figure 15.9: One must add energy to overcome
nuclear binding energy in order to
bust up a nucleus into its constituent nucleons
(left). Thus, the collective mass of
a nucleus plus the mass associated with
the energy it takes to break it apart (via
E mc 2 ) must be equal to the sum of the
masses of the constituent parts (middle).
Therefore, if we compare the mass of the
nucleus alone (removing the energy’s mass
from the scale) it must be less than the mass
of the loose collection of nucleons (right).
A careful look at Figure 15.4 reveals that lighter stable nuclei (graysquares)
at the lower left of the chart have a mass a little larger than the
corresponding mass number, but by the upper right—around oxygen—
the mass has edged just lower than A. Table 15.3 shows this trend,
confirmable in Figure 15.4 for the first four nuclides in the table. The
difference between mass <strong>and</strong> A is most negative around iron, then turns
around <strong>and</strong> becomes positive again for heavy elements like uranium.
What is going on here? If the mass of a nucleus were just the sum of its
parts, we would expect the total mass to just track linearly as we add
more pieces. In fact, if we try to build a neutral carbon atom out of 6
protons, 6 neutrons, <strong>and</strong> 6 electrons, the sum, according to Table 15.4,
should be 12.099 a.m.u., not 12.000. The discrepancy is due to nuclear
binding energy, as was introduced in Figure 15.9.
Particle a.m.u. 10 −27 kg MeV/c 2
proton 1.0072765 1.6726219 938.2720882
neutron 1.0086649 1.6749275 939.5654205
electron 0.00054858 0.000911 0.510999
(a.m.u.) 1.0000000 1.660539 931.494102
Table 15.3: Example mass progression.
Nuclide A mass (a.m.u.)
2H
2 2.014
4He
4 4.003
12C
12 12.000
16O
16 15.995
56Fe
56 55.935
235U
235 235.044
Table 15.4: Constituent masses of atomic
building blocks, expressing the same basic
thing in three common units systems.
Nuclear binding energy is incredibly strong 15 <strong>and</strong> is able to overpower
the natural electric repulsion between positively charged protons <strong>and</strong>
stick them together in an unwilling bunch. The strong nuclear force
15: . . . relating to what we call the strong
nuclear force
© 2021 T. W. Murphy, Jr.; Creative Commons Attribution-NonCommercial 4.0 International Lic.;
Freely available at: https://escholarship.org/uc/energy_ambitions.