How can the binding energy per nucleon graph be useful if you can't compare "all" products with "all" reactants?

Question

How can the binding energy per nucleon (BEN) graph be useful if one can't compare "all" products with "all" reactants?

Take the fission process of Uranium to Thorium: $${}^{238}U \to {}^{4}He + {}^{234}Th $$

Other than Uranium's and Thorium's binding energy per nucleon, Helium's binding energy per nucleon is also important as helium is also one of the products but this usually is ignored, and generally when there is a decrease in the BEN graph one concludes that energy is released.

Furthermore, I was wondering how the process of Uranium fission releases more energy than the fusion of hydrogen to helium when based on the BEN graph the difference in biding energy per nucleon between Uranium and Thorium is much less than between Hydrogen and Helium.

PS: I am a high school student and would really appreciate a simplified explanation

Answer 1

When calculating the energy released by the $\alpha$ decay of ${}^{238}\text{U}$ you cannot ignore the binding energy of the alpha particle or you will get a spectacularly wrong result.

You can finding binding energies on this web site. Using this data the calculation for the energy released in ${}^{238}U$ $\alpha$ decay is:

Nucleus	BE per nucleon (/MeV)	No. nucleons	Total BE (/MeV)
U-238	7.57015	238	1801.696
Th-234	7.59688	234	1777.670
He-4	7.07392	4	28.296

Then the energy released is the binding energy of the products minus the binding energy of the ${}^{238}\text{U}$:

$$ \Delta E = 1777.670 + 28.296 - 1801.696 = 4.270~\text{MeV} $$

Which reassuringly agrees with the value given by Wikipedia. If you ignored the binding energy of the helium nucleus your result would be wrong by $28.296$ MeV! If you'd like to give a link to the page where you saw the He binding energy apparently being ignored we can have a look at it and see what's wrong.

You ask about the hydrogen fusion process. If we take the commercially significant process $D + T \to He + n$ then the energies for this are:

Nucleus	BE per nucleon (/MeV)	No. nucleons	Total BE (/MeV)
D-2	1.11228	2	2.225
T-3	2.82727	3	8.482
He-4	7.07392	4	28.296

And again the energy released is the BE of the products minus the BE of the two initial nuclei so it's:

$$ \Delta E = 28.296 - (8.482 + 2.225) = 17.589~\text{MeV} $$

So the fusion releases more energy than the fission.

In a comment you ask about the reaction ${}^2D + {}^2D \to {}^3He + n$ and for this the energy is:

Nucleus	BE per nucleon (/MeV)	No. nucleons	Total BE (/MeV)
D-2	1.11228	2	2.225
He-4	2.57269	3	7.718

$$ \Delta E = 7.718 - (2.225 + 2.225) = 3.268~\text{MeV} $$

and this is indeed less than the energy released in ${}^{238}U$ fission. However I don't think there is a way to see this just by looking at the binding energy per nucleon graph. You need to do the calculation. The trouble is that for the calculation we need the total energy, and that's the BE per nucleon multiplied by the number of nucleons.