Tag Info

New answers tagged

1

This publication by IAEA is a good summary of considerations relating to control rod (CR or RCCA) material selection and lifetime management. I was wondering today, how long boron control rods remains in a nuclear power plant? As low as 4 years, as high as 30 years. Depends on the CR location in the core, since power varies with radial distance from the ...


3

Neither of the answers addresses what fissile really means, why to use fissile materials, and why not to non-fissile materials. Fissioning any fissionable isotope inherently releases a probabilistic number of neutrons, the average of which ranges from 2-4 per fission (bottom table (v bar)), and this number is not a major criterion for choosing an isotope. ...


0

It depends not on the number of neutrons it has but the number of neutrons that it releases in each step of the chain reaction. The isotope is chosen based on this.


4

The requirement for a material to be fissile (to be able to sustain a chain fission reaction) is not simply that it have neutrons, but that each fission releases enough neutrons of the right energy to trigger further reactions. As it so happens (and I know of no simple reason for this, it's just the way the equations work), U-238 isn't fissile. You can ...


2

The discrepancy between the predicted big bang nucleosynthetic abundance of Lithium 7 and the measured value can be summarised as follows. If we take what we know about the the baryonic mass density of the universe and the Hubble constant, we get a self-consistent picture between the cosmic microwave background, observations of galaxy recession etc. and the ...


0

In the abstract of this article Measurements of environmental background radiation at location of coal-fired power plants (PubMed, 2004) the terms are used in context. "Zero background radiation" - is the natural expected radiation measured far away of the specific pollutant site - the coal mine. It is a control or reference value in the study. ...


1

There is a large class of weak decays that can be predicted fairly well from theory (using the notion of weak universality). However, I have in mind the weak decay of heavy leptons and individual hadrons, so that is not quite what you were asking for as it occurs outside the nuclear context. You can tackle the nuclear problem too, but the phase space ...


0

In theory it could be done. The problem is that we're dealing with hundreds of entangled nucleons. A particle can be modeled as a wave in three-dimensional space. Two particles that aren't entangled can be modeled as two waves. But if they are entangled, you have to use one wave in six-dimensional space. In order to model an atomic nucleus, you'd need ...


1

You can find data of all (!?) current nuclear reaction experiments in the evaluated nuclear reaction databases (ENDF)[*]. Bibliographic information can be found in the experimental nuclear reaction database (EXFOR). The databases are maintained by the cooperation of nuclear data centres worldwide in the "International Network of Nuclear Reaction Data Centres ...


4

It is a prompt (immediate) reaction, and is more usually written something like N14(n,p)C14 to indicate that. It is far from the only such reaction. EDIT - To quantify my statement that there are many similar reactions, I went to the Evaluated Nuclear Data Files site hosted at Brookhaven (ENDF), entered 'n,p' for the reaction, 'sig' for the desired ...


2

Its the mass of the atom with atomic number Z and mass number A. If you study the right side of the equation you will see that the first two terms are the total mass of the protons and the total mass of the neutrons. All other terms are modifying this sum of the masses of the constituent particles. So to calculate the mass defect, simply take the first two ...


2

See the Wikipedia article on the semi-empirical mass formula. The formula is for the mass of the nucleus, so you would need to add on the mass of the electrons (and subtract their binding energy) to get the mass of the atom. However, as previous questions have mentioned, the mass of the electrons is a small correction and in fact generally smaller than the ...


3

Have a look at this table, which shows the binding energy (i.e. the ionisation energy) of electrons in various atoms. The highest energies are a few hundred eV, and those are for the core electrons (though admittedly it doesn't show $1s$ energies for the heavy atoms) so the average electron binding energy will be lower. By contrast, the average binding ...


0

The binding energy of the nucleus is an energy (or equivalent mass via $c^2$) which in some sense accounts for the "extra" (or lacking) energy which is not accounted for by the mass of its constituent particles. The binding energy per nucleon is exactly what it says, the binding energy divided by the number of nucleons (a nucleon is a proton or a neutron).


0

binding energy per nucleon means the nuclear energy between nucleons to attract each other.whereas binding energy of atoms is the bond energy of atoms to react with others.


0

First, you have to realise that in the nucleus there are two main forces that you have to consider (relevant to this question, at least): the electromagnetic force and the nuclear force. The former interests particles with an electric charge (protons, electrons), the latter holds protons and neutrons together - remember that protons would repel, so you need ...


8

It can't be solo neutrons, because they are unstable and decay into protons. So far as we know, there's not a stable configuration of mostly-neutrons that occurs in nature intermediate between heavy nuclei (uranium is roughly 3-to-2 parts neutrons) and neutron stars of 1-3 solar masses (which are about 90% neutrons). What you're describing would be the kind ...


1

The first formula is written in the Gaussian unit system, while the second one is in the SI system. In the Gaussian system, the unit of electric charge is $statC =g^{1/2}cm^{3/2}s^{-1}$. So, the Sommefeld parameter in the Gaussian unit system is dimensionless as it would be.


1

You need slow neutrons because if the neutrons are too "quick" then they scatter of the atoms instead of being captured by them. You can imagine a big lump of playdoh and a much smaller ball of playdoh. If you shoot the small ball with high velocity at the big lump then the ball is scattered - like two billiard balls. But when you put them slowly together by ...


3

Radioactive decay changes an atom from one that has higher energy inside its nucleus into one with lower energy. The change of energy of the nucleus is given to the particles that are created. The energy released by radioactive decay may either be carried away by a gamma ray electromagnetic radiation (a type of light), a beta particle or an alpha particle. ...


3

Why is energy released when an atom decays into another atom, when no energy is added? Atoms/nuclei are already created when we study them and organize them into the periodic table of elements. At the level of nucleons and elementary particles in general, special relativity holds. When we look at the periodic table of elements and count the number of ...


1

Nuclear-fusion experiments have been extensively performed with accelerators in the last decades of the 20th century reaching the proton drip-line. Today they are still object of interest allowing the study of superheavy elements. However the energy of the LHC is way too high. At that energy scale you go in the regime of quark-gluon plasmas and the nuclear ...


2

No, because the LHC puts too much energy into its particles for them to fuse. While we need enough energy to fuse particles, too much will stop it from happening.


3

The simple answer is No. Fusion happens at nuclear energies between particles to be fused, i.e. MeVs, because it is at the framework of nuclear bound states. LHC particles start with energies of TeV, so particle particle interactions are way over any nuclear bound state levels. Even if one accelerates deuterium nuclei the phase space is way over the ...


6

Let us calculate the energies. For the bomb, Google gives $3.8\times 10^{16}\,\text{J}$. For the stone, we use the Einstein equation $$KE=mc^2(\gamma-1)\qquad\gamma\equiv\sqrt{1-\beta^2}^{-1};\quad \beta=v/c$$ Plugging in the numbers, this gives a stone energy of $3.8\times 10^{18}\,\text{J}$. The stone is actually more powerful than the bomb. This is ...


2

The answer given in Walecka's 1974 paper is mostly correct. The one pion exchange contribution to the Hartree energy vanishes in balanced nuclear matter. The same point was made in the 1972 paper of Miller and Green (Phys Rev C5 241) where the same type model was used for doubly magic (finite) nuclei. If exchange is included (Hartree Fock as opposed to ...


3

The refractive index of air is about 1.0003, so the speed of light in air is about 0.9997$c$. You can work out the energy of the proton at this speed using: $$ E^2 = p^2c^2 + m^2c^4 $$ where the momentum is: $$ p = \frac{m_p v}{\sqrt{1 - v^2/c^2}} $$ I get the energy to be a shade over 38GeV.


0

$\hat{I}_D(k)={{g^2}\over4}\int_0^1d\alpha\int_0^\infty d\sigma\int\sigma\mathrm{e}^{-[q^2+\alpha(1-\alpha)k^2+m^2]\sigma}\vec{a}q$. Just a wild guess. [Oh, you asked for two formulae. Sorry.] Either that, or it is $\int_0^\infty\mathrm{e}^{-a\alpha}\sigma d\sigma=a^{-2}$.


2

Most atoms have an ionization energy of a few tens of electron volts. Beta decay electrons typically have a range of energies, with the mean and maximum energies typically a few million electron volts; the probability that the electron energy is small enough to be captured is pretty small. In addition, the daughter atom cannot capture the decay electron ...


2

No, Chemical Processes and Mechanical Processes can also release more energy than is input. See Exothermic Process and Catalysis. Think of an explosion of, say, dynamite: for the low energy input of lighting a wick which can be done with a lighter, you can output a large explosion. Also, Palladium is not Fissile, though it is a Fission product (aka ...


1

Yes, the atom (along with the surrounding material) may have a lasting electron deficit after the beta particle emission because harvesting this surplus charge is how betavoltaics work. Beta particles are more penetrating than charged nuclei like alpha particles, and because of that I believe they are better at carrying the charge away. The exact reasons ...


3

The semi-empirical mass formula is derived from a fit to the measured masses. If you know the numbers of protons and neutrons then the idea is that the SEMF should give you a good estimate of that mass (there are of course small residuals of the order 0.2 MeV to the fit). $$M(A,Z)c^2 = (A-Z)m_n c^2 + Zm_pc^2 - AE_b,$$ where $A,Z$ are the mass number and ...


2

The semi-empirical mass formula (SEMF) is much more involved than this, and tries to take into account various other factors that will affect the mass/energy of the nucleus. For any nucleus, of course if you KNOW its mass beforehand then simply subtracting the individual masses of protons and neutrons will give you ($\pm$)the binding energy, but the SEMF at ...



Top 50 recent answers are included