Energy and Human Ambitions on a Finite Planet, 2021a

15 Nuclear <strong>Energy</strong> 254
times more energy–dense than our customary ∼ 10 kcal/g chemical
energy density. See Box 15.3 for an example of how to compute this.
Box 15.3: Nuclear <strong>Energy</strong> Density
The example corresponding to Table 15.7 is said to correspond to
17 million kcal/g, but how can we get here? The mass change of
0.185 a.m.u. corresponds to a mass in kilograms of 3.07 × 10 −28 kg,
according to the conversion that 1 a.m.u. is 1.6605 × 10 −27 kg (Table
15.4). Multiply this by c 2 to get energy in Joules, yielding 2.76 ×
10 −11 J. 29 In terms of kcal, we divide by 4,184 J/kcal to find that this 29: This result, by the way, is the same as
fission event yields 6.6 × 10 −15 kcal.
We now just need to divide by how many grams of “fuel” we supplied,
which is 236.05 a.m.u. (Table 15.7), equating to 3.92 × 10 −25 kg, or
3.92 × 10 −22 g. Now we divide 6.6 × 10 −15 kcal by 3.92 × 10 −22 gto
get 16.8 × 10 6 kcal/g. Blows a burrito out of the water.
172.3 MeV in Table 15.7 using the conversion
that 1 MeV is 1.6022 × 10 −13 J.
Example 15.4.2 Considering that the average American uses energy
235
at a rate of 10,000 W, how much U per year is needed to satisfy this
dem<strong>and</strong> for one individual?
235
Since we have just computed the energy density of U to be 17 kcal/g
(Box 15.3), let’s first put the total energy in units of Joules, multiplying
10 4 Wby3.155×10 7 seconds in a year <strong>and</strong> then dividing by 4,184 J/kcal
to get kilocalories. The result is 75 million kcal, so that an American’s
annual energy needs could be met by 4.5 g 30 235
of U. That translates to
about a quarter of a cubic centimeter, or a small pebble, at the density
of uranium. Pretty amazing!
30: 75 million kcal divided by 17 million
kcal/g is 4.5 g.
We can take a graphical shortcut to all of Section 15.4.3, which hopefully
will tie things together in an instructive way.
Example 15.4.3 Refer back to Figure 15.10 (<strong>and</strong>/or Table 15.5)tosee
235
that U has a binding energy of about 7.6 MeV per nucleon. Where we
end up, around A ≈ 95 <strong>and</strong> A ≈ 140, the binding energies per nucleon
are around 8.7 <strong>and</strong> 8.4 MeV/nuc at these locations, respectively.
Multiplying the binding energy per nucleon by the number of nucleons
provides a measure of total binding energy: in this case 1,790 MeV for
235
U, about 825 MeV for the daughter nucleus around A ≈ 95, <strong>and</strong>
1,175 MeV for A ≈ 140. 31 Adding the latter two, we find that the fission 31: 7.6 × 235; 8.7 × 95; <strong>and</strong> 8.4 × 140
products have a total binding energy around 2,000 MeV, which is
greater 32
235
than the U binding energy by about 210 MeV—somewhat
close to the 172 MeV computed for the particular example in Table
15.7.
The graphical method got us pretty close with little work, <strong>and</strong> hopefully
led to a deeper underst<strong>and</strong>ing of what is going on. The rest of this
32: Binding energy reduces mass, so larger
binding energy means lighter overall mass.
© 2021 T. W. Murphy, Jr.; Creative Commons Attribution-NonCommercial 4.0 International Lic.;
Freely available at: https://escholarship.org/uc/energy_ambitions.