New answers tagged nuclear-physics
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Like dmckee says, the potential energy of electrons in an atom doesn't really compare to the energy of the nucleus. Since the nucleus is so tightly packed, and (in the case of Uranium) contains so many protons, they have a lot of potential energy—it takes a lot of work to "push" them together. The strong force holds protons and neutrons together when they're ...
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Has your instructor (or your book) mentioned how much bigger a atom is than a atomic nucleus? On order of 10000 times.
Moreover, except for the $s$-shell electrons, most electron never come very close to the center (the $p$, $d$, etc shells all have nodes at $r=0$) so at the moment of fission the nuclei are sitting roughly at the middle of a roughly ...
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The magnetic quadrupole moment tensor is given by
$$m_{ij}=\left\langle \frac{2}{3}\left(\mathbf{r}\times\mathbf{J}\right)_i r_j \right\rangle,$$
in analogy with the magnetic dipole moment vector
$$m_i=\left\langle \frac{1}{2}\left(\mathbf{r}\times\mathbf{J}\right)_i \right\rangle.$$
The magnetic field at a point $\mathbf{R}$ is then, up to quadrupole ...
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The primary purpose of the water moderator is not to shield the outside world from radiation (although this is one benefit). The water moderator has two primary tasks in a PWR or BWR:
High energy (~2 MeV) neutrons released in fission are down-scattered (or moderated) into the thermal energy range (~0.25 MeV). The cross section for fission is much higher ...
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While they work on the same principles, the detonation of an atomic bomb and the meltdown of a nuclear plant are two very different processes.
An atomic bomb is based on the idea of releasing as much energy from a runaway nuclear fission reaction as possible in the shortest amount of time. The idea being to create as much devastating damage as possible ...
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The "correct" half-life is the one that best fits all the experimental data available.
Start with your original equation,$$N(t)=N_{0}e^{-\lambda t}$$ Take natural logarithms of both sides and obtain:$$\ln (N(t))=\ln(N_0)-\lambda \ t$$Note that $N(0)$ is not the same as $N_0$. The first is an experimental point, with the same sort of random and ...
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Since radioactivity is a random process, you'd expect some fluctuation in the number of decays, i.e. if you wait for the half life, there's no guarantee that exactly half of the nuclei will have decayed!
Based on your 3 estimates of the half life, you could just take the mean and go with that.
And how to find the isotope? There's lists for that. I just ...
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Water cooled nuclear reactors slow neutrons down to "thermal" speed to increase the probability of their interactions with other nuclei. Thermal neutrons have a kinetic energy of about 0.025 eV (electron volt) or about 4.0 × 10−21 J, or a speed of 2.2 km/s. Neutrons released from a fission reaction have a mean energy of 2 MeV, or 20,000 km/s. Their speed can ...
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In a nucleus whose N/Z ratio is too large, the Pauli exclusion principle forces many of the neutrons to be in states with high energies. This makes the system less stable. For a fixed N, adding protons also makes such highly neutron-rich systems more stable, because the interaction between the protons and the neutrons is attractive, and the protons can go ...
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So here's what I did, I discretized the problem and deduced $N_x$ and $N_y$ at some $t_n$. These happened to include sums that could easily be turned into integrals. Namely:
$$N_x(t_n)=N_0e^{-\lambda t_n}+\Lambda\sum_{i=0}^n\Delta t_ie^{-\lambda(t_n-t_i)}$$
$$N_y(t_n)=N_0(1-e^{-\lambda t_n})+\Lambda\sum_{i=0}^n\Delta t_i(1-e^{-\lambda(t_n-t_i)})$$
As a ...
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The first of your equations is correct. You can see this in two ways. First, just look at the dimensions. In general, the argument of a logarithm should be dimensionless; only your first option is. Second, and maybe more convincingly, look at what you get when you take $\Lambda \to0$. You should be able to reproduce the standard decay equation:
...
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Normal ordering can indeed by defined with respect to any state. Normally, in a first QFT course, it will be introduced w.r.t. the non-interacting vacuum for scalar fields. However, I am doing a course in bosonization at the moment where we defined normal ordering w.r.t. the Tomanaga-Luttinger ground state, which is essential the filled Dirac sea, which is ...
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In an alpha decay no electrons are created or destroyed. There is a small correction needed for the Coulomb term when the alpha escapes without carrying two electrons with it, but that is at chemical, not nuclear energy scales and is (usually1) sorted out by chemical means in fairly short time scales.
So, no you do not figure the mass of any electrons into ...
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Conservation of energy and the electron-degenerate pressure.
For the neutron to decay you must have
$$ n \to p + e^- + \bar{\nu}$$
or
$$ n + \nu \to p + e^- \quad. $$
In either case that electron is going to stay around, but in addition to the neutrons being in a degenerate gas, the few remaining electrons are also degenerate, which means that adding a ...
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There is Beta decay in neutron stars. This is the simple answer. Since a neutron star is electrical neutral, there is the same amount of $\beta^+$ as $\beta^-$ decay, this is called the chemical equilibrium.
This means, every time when a neutron decays, a proton captures (in average) an electron and the star stays stable.
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Yes.
Not only can it, but the values are tabulated for unstable isotopes.
Compare the data provided for $^{206}\mathrm{Pb}$ (stable) with the provided for $^{210}\mathrm{Pb}$ (unstable). Look in the section headed "Decay properties". Moreover, note that there is a separate $Q$ provided for each decay mode (but not for each channel).
In this instance, ...
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Let me start off by saying that the "current" masses for the light quarks (up/down) are very small and about 1-4 MeV. The pions have masses of about 140 MeV and the proton/neutron have masses of about 1 GeV.
As you start at high energies and move to lower energies, the strong nuclear force increases in strength... till about $ \lambda_{QCD} \sim 250 MeV$ ...
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ITER is aiming for 150.000.000K. Please note that this temperature of the plasma, i.e. average kinetic energy of the ions is in electron volts
For example, a typical magnetic confinement fusion plasma is 15 keV, or 170 megakelvins .
15 KeV is enough to assure that the plasma does not neutralize itself and the bare nuclei have a high enough statistical ...
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Most of the resonances detected in particle physics scattering experiments are bound states of the strong force, bound for a time interval before decay .
These are created in the interaction and seen in invariant mass combinations of the interaction products, statistically.
The distinction with electromagnetic or weak decays comes from the widths of the ...
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The other answers already tell you what you need to know, but I'd like to add that this situation arises very often. One daily interaction with it is in car engines. They run at a certain amount of rpm, rotations per minute. Scientifically, these units would be $1/\mathrm{minute}$ or probably $1/\mathrm{second}$ because we hardly ever use minutes when ...
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Both you and freude are correct.
The units of Avagadro's constant are atoms / mole, but atoms is just a number and is dimensionless. That's why we write Avagadro's constant as mol$^{-1}$, and why the atoms units disappear from your final equation.
Response to comment:
You can write your expression for $\mu$ in various ways. In your expression:
$$ \mu = ...
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Avogadro constant is measured in units $mol^{-1}$ according to http://en.wikipedia.org/wiki/Avogadro_constant
What are units $atoms$? What do they measure? I have never heard about it.
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"Why is it that neutrons evaporate from nuclei more easily than protons do?"
This is not true in general. In depends strongly on where the nucleus is relative to the neutron drip line and the proton drip line. It also depends on Z and on how excited the nucleus is.
A cold nucleus that's beyond the neutron drip line will emit a neutron in a very short time, ...
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Your intuition about the charge repulsion and strong force acting on Protons more is less important that you think. The strong nuclear force is a few orders of magnitude greater than electromagnetism so coulomb repulsion just doesn't contribute much.
What matters most is the nuclear binding energy to separate a proton from the nucleus. If the resulting ...
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Ultra-dense deuterium exists. There are more than 20 papers now on this subject. There also exists a similar form of hydrogen named ultra-dense protium. Every state of a hydrogen atom is a Rydberg state, but the ultra-dense hydrogen materials do not contain ordinary Rydberg states.
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They are the same particles. When $\alpha$-radiation eventually gets stopped by an object (a sheet of paper or simply a meter or so of air will do the trick) it attracts two electrons and becomes elementary Helium.
Most of the world's Helium actually originates from reserves underground where Uranium and other $\alpha$ radiators have been creating He for ...
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They are exactly the same, with the different notations arising in different contexts. You could start with a bunch of helium gas and heat it up or shine UV light on it to turn it into a plasma, and then you'd probably say you have $\mathrm{He}^{2+}$ (or $\mathrm{He}\ \mathrm{III}$ if you are an astronomer). The symbol $\alpha$ is more often reserved for ...
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In this case, "size of atom" really means "size of the box that is holding the electron in its place". The "box" is provided by the confining electromagnetic force exerted by the nucleus on the electron. Similarly, the box representing the nucleus is provided by the strong nuclear force between protons and neutrons that hold them all in place.
In the ...
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