# Positive binding energy shenanigans

In Introductory Nuclear Physics by Kenneth S. Krane, section 3.3 p.65 is presented the following formula for the binding energy of a nucleon: $$B=\left\{Zm_p+Nm_n-\left[m\left(^{A}X\right)-Zm_e\right]\right\}c^2 \tag{1}$$ where $$m_p,m_n,m_e$$ proton, neutron and electron mass respectively and $$m\left(^{A}X\right)$$ mass of the nucleon as a whole. I have two big problems with this definition:

1. If I want to find the interaction energy of a system I should subtract to the energy of the system the energies of the individual components and not the contrary as it's done in the formula, this is simple logic right? Am I mistaken here?
2. The binding energy has to be a negative quantity! But the quantity presented in the formula is clearly positive since experimentally we know that the mass of the components is always greater than the mass of the nucleus right? And this is not a simple matter of definition either: the binding energy has to be negative: think about two nucleons that fuse togheter to form a nucleus, for example a proton and a neutron to form deuterion, we know that when they fuse energy is released so the binding energy must be a negative quantity! Furthermore to separate a deuterion into a proton and a neutron we have to give energy to the system, like in the phenomenon of nuclear photodissociation!

All the clues seem to point in the direction of the formula being wrong, or even better: the formula is presented with the wrong order of subtraction, it should be the contrary: $$B=\left\{\left[m\left(^{A}X\right)-Zm_e\right]-Zm_p-Nm_n\right\}c^2 \tag{2}$$ but at the same time it's hard for me to believe that a book that's been around for 30 years still has such a big mistake inside it. Am I wrong? Is formula (1) right, the one in the book, or it's my formula (2) right? Is it simply a matter of definition? Based on my reasonings in point 2. I think not but maybe there is something that I am missing.

• So you propose the same definition except for the sign. Rejected. Jan 19, 2021 at 19:19

If $$2.2$$ MeV are necessary to break-apart the Deuteron, we could say "the Deuteron is bound by $$2.2$$ MeV", or that the BE of the Deuteron is $$2.2$$ MeV.