# 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 ...

19

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 ...

10

If they didn't release energy, they wouldn't happen. The alternative, nuclear reactions that require energy, clearly need said amount of energy, which has to come from somewhere, e.g. kinetic energy involved in the collision of two nuclei (even ones that release energy usually have a "barrier" and some amount of initial kinetic energy is needed to overcome ...

9

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 ...

7

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 ...

7

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

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 ...

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 ...

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

It's unusually symmetric. All four nucleons are in 1s spatial orbitals, in singlet pairs of spin and of isospin. The more symmetry as system has, the lower its energy.

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 ...

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

It isn't possible to measure potential energy because it has a (global) gauge symmetry. It's like trying to measure the height of a mountain - this could be the height above sea level, the height relative to the deepest sea trench, the height relative to the centre of the earth and so on. Any measurement can only measure the change in potential energy, and ...

5

No, quantum chromodynamics is very non-linear and quantum effects are very strong. What you have is an approximation. A better approximation is given by the Semi-emprical mass formula.

5

The relationship between nuclear masses and mass differences and binding energies has been confirmed by many decades of careful nuclear spectroscopy. It's possible to measure an atom's mass by purely mechanical means: you ionize the atoms, accelerate them to a known energy, and use a magnetic field to measure their momentum. This lets you come up with an ...

4

Rob's explanation of how we know is bang on, but I wanted to address a part of your question that might point to a basic misunderstanding. What is special relativity inside the nucleus? Everything is always relativity. Everything. Always. All those Newtonian equations like $T = \frac12 m v^2$ for the kinetic energy can be properly understood as ...

4

No, it is not correct because of relativity. Because you use $E=mc^2$, clearly, you are using relativistic effects so you need to take the full theory of relativity into account. While you might think that you have only used the special theory of relativity, you are also considering gravitational effects – the gravitational binding energy – and the general ...

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.

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

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. ...

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

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

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 ...

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

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

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

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

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

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 ...

3

If you use relativity (which the use of $E = mc^2$ implies), we cannot choose the potential arbitrarily, because the relation between energy and mass makes absolute values of the energy measurable through the gravitational forces exerted by stored energy. EDIT: Since you asked, I will explain it in somewhat more detail: Let \$V : \mathbb{R}^4 \rightarrow ...

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