# Tag Info

35

It all comes down to a balance between a number of different physical interactions. The binding energy of a nucleus is commonly described with the semiempirical mass formula: $$E(A, Z) = a_V A - a_S A^{2/3} - a_C \frac{Z(Z-1)}{A^{1/3}} - a_A \frac{(A-2Z)^2}{A} + \delta(A,Z)$$ where $A = Z + N$ is the total number of nucleons, $Z$ the number of protons, ...

27

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 ...

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

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 ...

20

The existence of nuclei is dependent on a number of quantum mechanical boundary conditions. They appear as solutions to a problem where there is a balance of: a) the attractive spill over color force that binds the quarks into a proton or a neutron, b) the repulsive electromagnetic force between protons, c) the Pauli exclusion principle, d) the instability ...

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 ...

16

Iron is a "special" element because of its nuclear binding energy. The very basic idea is that when you fuse two light elements together, you get a heavier element plus energy. You can do this up to iron. Similarly, if you have a heavy element that undergoes fission and splits into two lighter elements, you also release energy. Down to iron. You can see ...

13

Heavier nuclei can also undergo fusion, but that's not very useful for energy production. One of the reasons is, as you've mentioned, the binding energy per nucleon. Let's have a look at the binding energy curve (image taken from Wikipedia): Iron-56 has the highest binding energy per nucleon, which means it is the most stable nucleus. Roughly speaking, ...

12

The bonding of nuclei is dominated by 2 main forces - the strong nuclear force, and the electromagnetic force. The strong nuclear force is much stronger than the electromagnetic force, but acts over much shorter distances. For small nuclei (eg hydrogen and helium), if you're able to add more nucleons, they are likely to stick due to the attraction of the ...

12

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 know 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 events ...

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

Energy of a fission nuclear bomb comes from the gravitational energy of the stars. Protons and neutrons can coalesce into different kinds of bound states. We call these states atomic nuclei. The ones with the same number of protons are called isotopes, the ones with different number are nuclei of atoms of different kinds. There are many possible different ...

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 ...

8

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, ...

8

What determines the most stable element (Fe) is the trade off between the nuclear binding (attractive) and the coulomb repulsion between protons. Nucleons feel binding forces that can be described as bulk and surface forces. The bulk forces are those associated with the saturation of nuclear forces (nuclear density in the interior of heavy atoms is ...

7

A back of the envelope calculation suffices, meaning all factors of $2,3, \pi$ etc. have been ignored. The residual strong force is mainly due to pion exchange and can be modeled by this plus a short range repulsion due to exchange of $\omega$ mesons. The Yukawa potential from pion exchange is $e^{-m_\pi r}/r$. The weak interactions arise from $W^\pm$ and ...

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

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 ...

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

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 $M(r)=\frac{4}{3}\pi\,r^3\,\rho$, ...

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

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

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, ...

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

An analogy: nucleons stick together because of some powerful glue (because otherwise they would fly apart, especially protons because they have positive charge and thus repel each other). Bigger nuclei need more glue, but, until you reach 26 protons or so (i.e. iron), the amount of glue for each extra nucleon is decreasing. Past that point, it starts ...

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

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

Yes, matter of mass $m$ is directly converted to energy $E=mc^2$ which literally means that the weight of the remnants of the atomic bomb is smaller than the weight of the atomic bomb at the beginning. Fission reduces the mass by 0.1 percent or so; for fusion, you may get closer to 1 percent of mass difference. In principle, you may release 100 percent of ...

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 $$M_{\mathrm{T}} = 2809.432 \,\mathrm{MeV}$$ $$M_{^3\mathrm{He}} = 2809.413 \,\mathrm{MeV}$$ ...

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