For the reaction $\rm ^{235}U + {}^1n \implies {}^{91}Kr + {}^{142}Ba + \text?$, what is the unknown product? Does $PE_{strong}$ and $PE_{electric}$ change, and how so / why?

I see that there are 236 nucleons on the left, and 233 on the right; also, the # of protons in each is the same ($92 + 0 \implies 36 + 56$). This leads me to believe that 3 neutrons are released from the reaction.

For the electrical potential energy, I figured that there was no change, because the # of protons doesn't change. This is wrong, but I can't figure why.

I also figured that the strong potential energy would decrease because the resulting nuclei are more stable (so their energy well is deeper, and more negative). This is also wrong, and again, I'm not sure why. So long as the nucleus is for an atom greater than Fe, I thought breaking to smaller pieces reduces energy (isn't this why these radioactive particles break apart, is to reduce energy to a more stable state?).

There's definitely a flaw in my reasoning, but I don't know what it is here.

  • It's the other way around. Smaller than Fe you release energy (as photons) through fusion (like in the sun), larger than Fe through fission. – OrangeDog May 30 '17 at 21:44
  • While the number of protons doesn't change, a third of them are a lot further away from the others. – OrangeDog May 30 '17 at 21:48
  • I understand that regarding total energy, smaller than Fe releases energy in fusion (because you're forming bonds), and larger than Fe releases energy in fission (only explanation I have is that it's more stable, meaning less energetic (though I do feel like breaking into two would require taking in energy, so can't reconcile that)). But the strong potential energy apparently doesn't decrease (even though total energy does), and I don't see why. – Alex G May 30 '17 at 21:52
  • The energy released is equal to the mass energy of the nuclear matter on the left hand side of the equation minus the mass energy of the nuclear matter on the right hand side. – WAH May 30 '17 at 22:02

It's not clear what level of complexity you have in mind here, but here are some comments you might find helpful.

First, I'm assuming you've counted the number of nucleons correctly.

To find the change in the electrical potential energy, remember that this energy is mostly stored as the self-energy of the positively charged nucleus. Each nucleus is roughly spherical with radius $R\simeq A^{1/3}\rm\,fm$, with positive charges smeared uniformly throughout. If you have a $1/r$ potential, the self-energy of a uniform-density sphere goes like $R^5$. So fissioning into two marginally smaller nuclei can still make a big difference in the electrostatic binding energy.

The reason you can think of the nucleus is a uniform-density sphere(oid) of charge is that the nuclear potential is quite different from the electrostatic potential. A useful approximation to the nuclear potential is to be more or less flat inside the nucleus, with a step function about a nucleon thick to zero potential just outside the nuclear surface. Reconfiguring a large spheroid into two smaller spheroids has a much smaller effect on this mostly-uniform potential than it does on the electrical self-energy.

Regarding electrical potential energy: you used to have all the protons very close together; now you have two smaller lots of protons with a lot of space between them. As the nucleus fragmented into two smaller charged fragments there will have been a very strong repulsive force until the electrons adjusted and started screening the respective nuclei. So you expect a lot of electrostatic potential to have been released.

  • So, when this electrostatic potential is "released," does this mean that PE is actually increased because as $r \implies \infty$ the energy goes to zero? As in, the magnitude of the repulsion is far decreased, but it is "less negative", correct? – Alex G May 31 '17 at 4:47

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