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Binding energy is the energy required to separate the nucleus into its constituent particles. The thing is that I read in a book that the binding energy is also the energy equal to the mass defect.

Does that mean that the increase in mass occurs through the energy we provide it? If so, I am confused that the energy we provided changed to mass then how is the strong force overcome?

Also I read online that for larger atoms, the number of neutrons increase much more than protons. Is it because they increase the strong force acting or they dilute the repulsion between the protons? And it said that the atoms become unstable because neutrons go to higher energy levels. What is that supposed to mean?

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    $\begingroup$ There are very good answers here: physics.stackexchange.com/q/64037 physics.stackexchange.com/q/104551 physics.stackexchange.com/q/169340 $\endgroup$ – jaromrax Jan 23 at 14:34
  • $\begingroup$ BTW mass defect is not the same as binding energy, different number with a bit different sense to ease up some calculations. $\endgroup$ – jaromrax Jan 23 at 14:42
  • $\begingroup$ I'm not so sure that "mass defect" is an official term in the lexicon of various organizations which tabulate these things. I couldn't find it. On the other hand, binding energy and mass excess are used extensively at National Nuclear Data Center (NNDC). When did "mass defect" start being used, and how is it different from BE? Mass excess is $\Delta = M_{atomic}-A$ in official use. $\endgroup$ – Bill N Jan 23 at 17:09
  • $\begingroup$ I still don't get it... $\endgroup$ – Megan mars Jan 23 at 17:55
  • $\begingroup$ Don't get what? The definitions of binding energy, mass defect, and mass excess or the concept of adding energy raising particles out of a potential energy well? $\endgroup$ – Bill N Jan 23 at 20:07
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Does that mean that the increase in mass occurs through the energy we provide it? If so, I am confused that the energy we provided changed to mass then how is the strong force overcome?

If we consider the nucleus to be a system of particles ($Z$ protons and $A-Z$ neutrons) trapped in a bound system, then the mass of the system will include the (negative) potential energy of the net binding forces. So, simply put $$m_{\mathrm{nuc}}=\frac{E_{\mathrm{total}}}{c^2}.$$ When enough energy is provided to balance out the potential energy well, the system is no longer bound (conceptually). At that point we consider the masses of the individual particles.

So, in one sense the energy we provide simultaneously overcomes the binding force and increases the total mass of the system, allowing us to "view" the individual masses, rather than the single mass of a bound system.

By the way, we rarely ever actually separate nuclei into their individual particles. We can only do this for the lightest group of nuclei. That total individual nucleon separation of heavier nuclei is a thought experiment.

EDIT: After a little research in some older textbooks, I found some conflicting ideas between two "giants" in nuclear science regarding "mass defect" but there is no disagreement about the definition of binding energy. :

  1. Emilio Segre, in Nuclei and Particles (published 1963) (pg. 190),

The quantity $A-M$, where $M$ is the exact mass of the atom [A is the atomic number], is usually called the "mass defect," the quantity $M-A=\Delta M$ is usually called the 'mass excess," and the quantity $$\frac{M-A}{A}=f$$ is called the "packing fraction."

  1. Irving Kaplan of MIT in Nuclear Physics, 2nd edition (1962) (pg. 221)

The difference in mass, $\Delta M$, is called the mass defect; it is the amount of mass which would be converted to energy if a particular atom were to be assembled from the requisite number of protons, neutrons, and electrons. The same amount of energy would be needed to break the atom into its constituent particles, and the energy equivalent of the mass defect is therefore a measure of the binding energy of the nucleus...The mass defect can then be written $$\Delta M = Zm_H + (A-Z)m_n - M_{Z,A},$$ where $m_H$, the mass of the hydrogen atom is 1.0081437 mass units and $m_n$, the mass of the neutron, is 1.0080830 mass units.

Kaplan doesn't mention mass excess.

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