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I recently learned that binding energy is the energy released when nucleons are fused together to form a nucleus, as the bound state of the nucleons is more stable because of the strong attractive nuclear force. I was trying to comprehend this using the analogy of gravitational potential, but it seems I'm missing something.

Suppose a ball is dropped from a height 'h' to the ground. After a couple of bounces, the ball comes to rest on the ground. The initial potential energy of the ball is lost as sound and heat in each bounce before it came to rest.

Now compare it with the potential energy of the nucleons. Since the bound state is more stable, potential energy is lost in the fusion process. In the sun, this lost potential energy is converted to radiant energy (heat and light energy). But then why do we have a mass defect? The energy is already lost as radiant energy, and on top of that, if we lose mass in the formation, that does not seem to conserve mass or energy. (The ball didn't lose mass when it hit the ground, right?).

What is that I'm missing here?

Sasikuttan
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2 Answers2

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The ball didn't lose mass when it hit the ground, right?

It did. Or rather, the composite system of the Earth, the ball, and their mutual gravitational field lost energy, which is equivalent to mass.

The question is how much energy. Suppose the combined mass of the ball-Earth system is $m_\text{earth} + m_\text{ball} \approx 10^{24}\,\mathrm{kg}$, the mass of the Earth. The ball falls from the table and liberates one joule of energy from the gravitational field. Since the rest energy of the Earth-ball system is $m_\text{earth} c^2 \gg 1\,\mathrm{J}$, the constant-mass approximation is fine.

(Homework for you: compute the total gravitational binding energy of the Earth, and compare to Earth’s mass in energy units. The scale is $-G m_\text{earth}^2 / r_\text{earth}$ from dimensional analysis, but there’s a half or something out front from doing the integral of bringing all the masses together. The constant-mass approximation is good even if you’re assembling the entire Earth from asteroids and dust.)

In chemistry you come closer to caring about the mass stored in the electric field. For example, the ionization energy for ground-state hydrogen atoms, $13\,\mathrm{eV}$, is a part-per-billion correction to the mass of a hydrogen atom, $m_\text{H} \approx m_\text{proton} \approx 1\,\mathrm{GeV}/c^2$. The innermost electrons in heavy atoms can have binding energies of hundreds of keV.

It’s not that only nuclear interactions have mass changes associated with potential energy changes. But in nuclear interactions the energies involved are so large that the mass effects really can’t be ignored. A uranium fission liberates about $180\,\mathrm{MeV}$. A dozen uranium fissions is enough to produce a proton-antiproton pair from nothing. Nuclear binding energies are a part-per-thousand correction to the nuclear mass.

rob
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  • I see. So basically, what you are implying is that in any system, the energy release that is happening due to the lowering of potential energy is in fact mc^2, where 'm' is the loss in mass of the composite system. In the ball-Earth problem, this mass will be negligible because the loss in potential energy is small. – Sasikuttan Sep 29 '21 at 06:11
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    Yes, just so. The constant-mass approximation is valid when the energies involved in an interaction are small. One example is weak-field gravitation. – rob Sep 29 '21 at 12:12
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It would be better to picture any chemical or nuclear reaction as a ball rolling into a local valley (even if the surrounding region has higher hills and valleys). Each stable state the nucleons could be in corresponds to a local valley. It would need some other particle's kinetic energy to come knock it up and out in order to reach another local valley (fission, fusion, other isotope) Kinda like this picture https://xkcd.com/681/

See if this answer helps you Nuclear reaction: loss of mass or conservation of mass?

And the difference between heights of two adjacent valleys has to escape as kinetic energy of the products. That's where the unhelpful concept of "mass defect" comes from.

RC_23
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