Question: Particle Physics question: Why can we not create a nucleus for an atom with greater than 118 protons? Strong nuclear force only has a range of a few fm range however while the two protons on opposite ends of the nucleus may not experience attraction the one on the far end would surely experience attraction to the next nucleon and the next until the chain reaches the other side.

Jonathan Edward Davies answered on 12 Oct 2023:

So the simple answer is that some nuclei are more “tightly” bound than others, due to a variety of effects which I explain a bit more below. This means that it can be energetically more favorable to be radioactively decay into another element. What I mean is things always want to minimise their energy. It turns out that lead is the most stable of these and ultimately this is why things like Uranium will decay into other elements and why you can create a bomb (release a lot of energy) by either splitting apart heavy nuclei (fission) or fusing together lighter ones (fusion).

There’s a good graph demonstrating this here:

https://www.britannica.com/science/nuclear-binding-energy

Binding energy per nucleon measures how “tightly bound” a nucleus and energetically we want to maximise this because binding energy is essentially “negative energy”- energy that you have to add to pull apart the nucleus. Really in nuclear physics we think about the “nuclear” force rather than the strong force which acts between the quarks inside the protons or neutrons and is diluted and screened somewhat by the time we get to the dimensions of the nucleus.

There are a number of contributions to this overall binding energy. There is a surface tension effect and is proportional to the number of nucleons that are situated on the nuclear surface- it is largest for light nuclei. There is electrostatic repulsion between the positively charged protons- this becomes more important as the number of protons increases. There is a symmetry correction term taking into account that generally the most stable arrangement has equal numbers of protons and neutrons (because the n–p interaction in a nucleus is stronger than either the n−n or p−p interaction). There’s also a final term which depends on whether there are even or odd numbers of protons or neutrons. I believe the effect of nuclear spins of the protons and neutrons will feature in these last terms somewhere.

Sorry, maybe a bit of a technical answer to your question but you can maybe see that the situation is quite complicated, with lots of effects going on, and we can’t solve this system exactly.