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$^1_1H$ is a special case as the proton does not have any other nucleons to bind onto. I suppose that you could call the binding energy per nucleon zero which means that you require no energy to split up the nucleus of $^1_1H$ into its constituent parts? Note also that the binding energy per nucleon is not necessarily the full measure of whether a nucleus ...


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[The figure shown in the OP question above ...] is experimental data for the ratio $R = $ [...] as a function of the centre of mass energy $\sqrt{s}$ The so called cross section ratio $$R[~\sqrt{s}~] = \frac{\sigma^{(0)}[~e^+~e^- \rightarrow \text{hadrons}, \sqrt{s}~]}{\sigma^{(0)}[~e^+~e^- \rightarrow \mu^+~\mu^-, \sqrt{s}~]}$$ ?? Actually, ...


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'Crossover diagrams' exist when the final products are indistinguishable. When using Feynman diagrams for computations, you need to add all distinct diagrams that depict the same process, i.e., have the same external lines. Both the crossed and uncrossed diagrams for the electron-electron process have the same final products, since you cannot tell the ...



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