0
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ So you propose the same definition except for the sign. Rejected. $\endgroup$
    – my2cts
    Jan 19, 2021 at 19:19

2 Answers 2

1
$\begingroup$

First of all, I'm assuming you mean nucleus instead of nucleon in the very first line. Secondly, the 2 problems that you list are interrelated, and the matter is, indeed, semantical in nature.

Both the problems are resolved by realizing that as a result of the very fact that we're looking at a bound system, the net energy of the individual components is going to exceed that of the bound system (in fact, this could serve as a definition of a bound system). With this in mind, we see that the following is an equivalent way of thinking about binding energies:

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.

$\endgroup$
0
$\begingroup$

You give the correct answer in one of the options you propose: it is simply a matter of definition. It is not uncommon in physics (it is part of the culture) to include negative signs explicitly into formulas or definitions, so that the quantities themselves end up having the sign the person introducing the definition felt made the most sense or was most convenient (very often it is done so that the quantity in question will always be positive, if that is a possibility). Examples abound: it makes sense think as the drag force as negative, but a negative sign is often introduced explicitly into the formula to make the drag coefficient itself positive. Similarly for elastic constant, isothermal compressibility, etc.

The assertion you make about the biding energy having to be negative makes sense only after we have agreed on how we want to set the sign in the definition. If we agreed on the opposite definition (as in the book you quote), then the assertion would become that the binding energy must be positive. The physical content of the assertion would be exactly the same in both situations.

It is worth mentioning that the binding energy is often reported per nucleon (so, it would be the formula you present, but divided by A). Whether there can be isotopes with negative binding energy (more energy for the nucleus as a whole than for the constituent nucleons) is an interesting question. They would not be stable, of course, but could exist as metastable or instable states. Lithium 3, for example, is reported to have an atomic mass of 3.03078 amu (Wikipedia, Isotopes of lithium). After subtracting the 3 electrons you still have about 3.02913 amu, whereas 3 protons give you only 3.02183 amu. So, this isotope seems to have a negative binding energy, according to the formula you quote, which is also the one I am familiar with. This isotope is very unstable (very short half-life) I have seen its binding energy quoted as positive somewhere else, which I do not quite understand, but that would be a separate question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.