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22

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 ...


22

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 ...


20

As far as the theory goes, you are absolutely correct, the (negative) binding energy between atoms in a molecule contributes to the total mass of that molecule, so a stable molecule is less massive than the sum of the masses of its constituent atoms. However (as you yourself calculated), the mass difference is absolutely tiny, and as far as I know, it has ...


18

A lot of different forms, but mostly kinetic energy. A good table is given at Hyperphysics. The energy released from fission of uranium-235 is about 215 MeV. This is divided into: Kinetic energy of fragments (heat): ~168 MeV Assorted gamma rays: ~15-24 MeV Beta particles (electrons/positrons) and their kinetic energy: ~8 MeV Assorted neutrons and their ...


11

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 ...


10

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 ...


9

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 ...


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 ...


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 ...


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

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. ...


6

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 ...


6

It is not true that in all fusion and fission processes the mass of the products is less than the mass of the reactants. This is only valid for exothermic reactions. The change of mass is due to the change in the binding energy of the nucleons (note that the change in binding energy is in the order of 1 MeV, while the mass of the nucleons is around 940 ...


6

Great question! My answer would be that in order to get a bound state, we need to have a potential that is deeper than the kinetic energy the two particles have. We have a better chance of getting a potential of the right depth to bind two nucleons if: (1) They don't repel charge-wise. Compared to nucleon interaction the Coulomb force isn't that strong, ...


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

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, ...


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

Binding energy simply isn't the right metric (because it is calculated from different starting points on account of the differing masses of the constituent nucleons). Proper energy (AKA mass) of the states is the right metric. Wolfram Alpha gives the masses as $$M_{\mathrm{T}} = 2809.432 \,\mathrm{MeV}$$ $$M_{^3\mathrm{He}} = 2809.413 \,\mathrm{MeV}$$ ...


5

Very interesting question! In chemistry you spend lots of time discussing exothermic and endothermic reactions: when you put your reagents together, sometimes the reaction heats things up, and sometimes the reaction cools things down. Nuclear reactions are very different, in that essentially all spontaneous reactions studied in laboratories are exothermic. ...


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 ...


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 ...


4

The final stage of nucleosynthesis at the core of a massive star involves the production of iron-peak elements, mostly determined by competition between alpha capture and photodisintegration. The starting material is mostly Si28 and weak processes are unable to significantly alter the n/p ratio from unity on short enough timescales. Thus the expected outcome ...


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

The nucleon-nucleon interaction has a short range, roughly 1 fm. Therefore if there were to be a bound dineutron, the neutrons would have to be confined within a space roughly this big. The Heisenberg uncertainty principle then dictates a minimum uncertainty in their momentum. This amount of momentum is at the edge of what theoretical calculations suggest ...


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

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 ...


4

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 $Z$ ...


4

One can answer this question by calculating the energy needed to shift half the Earth's mass so that it is infinitely far from the other half. Let's calculate the gravitational potential energy released as we create a planet: assuming a constant density $\rho$, when the planet is growing and of radius $r$ and thus of mass $M(r)=\frac{4}{3}\pi\,r^3\,\rho$, ...


4

The force between nucleons is rather complicated, so let's consider the simpler example of assembling a hydrogen atom from a proton and an electron. We start with the proton and electron at rest and a long way apart. The kinetic energy is zero (because they're at rest) and the potential energy is given by: $$ V = -k\frac{Q_1Q_2}{r^2} \tag{1} $$ where ...



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