# Tag Info

## Hot answers tagged binding-energy

23

There are a couple different meanings of the word that you should be aware of: In popular usage, "quantized" means that something only ever occurs in integer multiples of a certain unit, or a sum of integer multiples of a few units, usually because you have an integer number of objects each of which carries that unit. This is the sense in which charge is ...

18

A neutron is not a proton and an electron lumped together (as your question seems to suggest you think) A hydrogen atom is a bound state of an electron and a proton (bound by the electromagnetic force) whereas a neutron is a bound state of three quarks (bound by the strong force). You might be tempted to think that a neutron is also a bound state of an ...

8

The Sun obviously produces far more energy per second than is required to fuse an iron nucleus with some other nucleus. The problem is concentrating all that energy on the iron nucleus. It's not enough to known that it takes the energy from $n$ hydrogen fusions to fuse one iron nucleus, it's getting the energetic products from those $n$ hydrogen fusion ...

6

From the binding energy given experimentally, using precise QM calculations or using a given formula, one should first check for "stability in particles", if the binding is negative, you will of course not have stability. Then the next thing, if you have a formula, is to check for each type of stability. For example, to check for stability against a given ...

6

A neutron is a fermion, a hydrogen atom is a boson. This is related to the fact that a neutron decays into three fermions rather than two which is what you seem to think. A neutron is composed of three valence quarks, $u,d,d$, while a hydrogen atom is made out of $u,u,d,e^-$. The internal size of a neutron is about $10^{-14}$ meters while the internal size ...

6

When people say that the decay rate depends critically on the $Q$ value, they're talking about alpha decays compared to other alpha decays. When you compare alpha decay to emission of other small clusters, the dependence on the atomic number $Z_c$ of the emitted cluster is much more prominent. The reason is as follows. In the Gamow model of beta decay, we ...

5

The only difference is that the strong force is stronger; meaning that binding energy can be a notable fraction of the total energy. You write When it comes to things like gravity and the electromagnetic force, masses aren't reduced which is not quite true, but then let on that you know by continuing but with nuclei the mass difference is ...

5

There are actually several different "limits" one might encounter. The one everyone talks about is not fusing past iron. This comes from the fact that isotopes in the vicinity of ${}^{56}\mathrm{Fe}$ consist of the most tightly bound nuclei. See Wikipedia for a discussion and some binding energy curves. If you are interested in why there is a peak, it comes ...

5

To understand binding energy and mass defects in nuclei, it helps to understand where the mass of the proton comes from. The news about the recent Higgs discovery emphasizes that the Higgs mechanism gives mass to elementary particles. This is true for electrons and for quarks which are elementary particles (as far as we now know), but it is not true for ...

5

A bound system will indeed have a lower mass than it's constituent free parts. The binding energy is usually understood to be the energy it would take to separate the bound system to free constituent parts. Then we would say $M_{tot} = M_1 + M_2 - E_{bind}/c^2$ Or we could by convention make the binding energy a negative number and say it "increases" mass ...

4

The neutron decays into a proton, an electron and an antineutrino. So even the end components are different from Hydrogen which is just a proton with an electron orbiting around it. The binding forces are also different. The proton and the electron are bound by the electromagnetic force. The neutron by the strong to the rest of the nucleons in a nucleus. ...

4

A good question touching the deep physical concepts. I will try to give an answer in several steps. Before answer your question let me answer another one: what is a free particle? A simple (but complete) answer is that a «particle» is a long-living elementary excitation of a system (a quantum field). «Long-living» means living long enough to be observable, ...

4

As you correctly stated in normal situation the star cannot sustain the process. This doesn't mean that there are no such reactions going on in the core. The difference is that during the pre-supernova phase of the star the production of iron is negligible compared to the star. When it goes supernova, it produces a comparable amount of iron.

3

The Weizsäcker formula and all similar formulae in nuclear physics are formulae for the masses of the nuclei, not atoms. That's true by design: the models behind the individual terms (droplet, shells etc.) are models for the nucleus only. At the same moment, the accuracy of similar semiempirical formulae is not that marvelous which means that the errors are ...

3

The question is at least a little bit indeterminate because of "naturally occurring". For any given nucleus, the more highly ionized it becomes the greater the binding energy of the remaining electrons, culminating with getting the last one off of the hydrogen-like core, which runs roughly $13.6\frac{Z^2}{n^2}\text{ eV}$. Unfortunately for significant $Z$ ...

3

A nucleon in nuclear context is simply not the same as one in a free context. Not in mass nor in form factor. These corrections are not known in complete detail but there are parameterizations of them that are used in nuclear and particle physics experiments. In my disertation project we used a parameterization due to de Forest, which is a popular but now ...

3

That's because the mass of an object is the same as the energy the object possesses at rest. According to $E=mc^2$ e.g. a compressed spring has more mass than an uncompressed one, a charged battery has more mass than an uncharged battery, etc. Mass and (rest) energy are not just equivalent, they are the same thing. Energy bends space time which causes ...

3

To use a rather brain dead picture of the nucleus imagine that the nucleons can be modeled as a bunch of billiard balls zipping around (they can't but I'll discuss a way in which it the model is useful in a moment). In order for a large nucleus to emitted you have get all the nucleons that will make up that fragment moving in roughly the same direction and ...

3

If the kinetic energy of the object is exactly the same as the binding energy, the object will slow as it leaves the gravity well and will come to a halt at infinity (i.e. far enough away for the gravity to be insignificant). If the kinetic energy of the object is greater than the binding energy, the object will still slow as it leaves the gravity well, but ...

3

From a stat mech point of view one can view a star's goal as throwing off entropy over it's life span. Iron is as stable as things get from an entropy perspective. This results in the fusion limit. The entropy that comes along with enough temp and pressure to continue this process is not favorable for the continuation of fusion beyond iron. I believe ...

3

1) I gather you mean gravitation potential energy of the test particle. Well, any such thing is only useful in so far as it is related to a constant of motion throughout the geodesic--in the case of gravitational potential, being part of the conserved mechanical energy, kinetic + potential. (Another example could be angular momentum.) In GTR, these ...

2

I think just the opposite of David Zaslavski, and assert: The rest mass of particles is quantized,  being the spectrum of the component P_0 of 4-momentum in the Hilbert space of states where the particle is at rest. (For example, quarks and neutrinos have a 3-dimensional mass matrix, each eignevalue being infinitely degenerate.) This doesn't conflict ...

2

I'd just like to slightly refine John and Mike's answers. If the initial kinetic energy equals or exceeds the binding energy, then the particle speed asymptotically approaches a constant, never stopping and "turning around". Of course, in the very special case that the KE equals the binding energy, that constant speed is precisely zero. Note that though ...

2

Your basic nuclear reaction conserves the number of nucleons present.1 That is important, because at a bit less than 1 GeV each the mass of the nucleons dominates the total energy of all these states. So the only place available to get or lose energy in a reaction is by Changing the flavor of nucleons. Every neutron converted to a proton gets you a ...

2

Have a look at the binding energy per nucleon curve: There are many stable configurations below iron, so the binding energy is not the only criterion for stability. Graph of nuclides (isotopes) by type of decay. Orange and blue nuclides are unstable, with the black squares between these regions representing stable nuclides. The unbroken line passing ...

1

As you increase your energy, your mass increases as well, and the rest mass behaves as some potential. I assume you mean "as you increase your velocity"; correct me if I'm wrong. Rest mass does not depend on velocity. Be careful to distinguish between rest mass and relativistic mass. Rest mass, denoted as $m_0$, is the one that is equivalent to energy ...

1

It all depends on the scale at which you examine things. Just like thermal energy being resolved as the randomized kinetic energy of the parts if examined closely enough, so mass can be resolved into binding and kinetic energies plus some intrinsic masses of the bits if examined at the right scale.

1

Nuclear physics is in the realm of quantum mechanics. To first order one can think of "one nucleon" in the collective left over strong interaction field of all the rest, as a potential. As for strong interactions nuclei are indistinguishable to first order the binding energy per nucleon makes sense. The larger this energy the harder to extract a nucleon ...

1

Apparently what I'm suggesting is called cluster decay, and can happen. According to Wikipedia: Theoretically any nucleus with Z > 40 for which the released energy (Q value) is a positive quantity, can be a cluster-emitter. In practice, observations are severely restricted to limitations imposed by currently available experimental techniques which ...

1

You have to realize we are talking quantum mechanics here, and not just balancing energies. There exist quantum mechanical solutions of the strong potential binding nucleons ( protons and neutrons) that are more stable than others. In the shell model the so called magic numbers are well described . Nuclei which have neutron number and proton (atomic) ...

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