# Tag Info

1

While I have never conducted such an experiment (yet), I conjecture it will be the energy of the projectile what will determine the results. It will also bring momentum, but this would be comparable roughly to three A380 jets at cruise speed, so it probably would not change much on the result. The projectile would start evaporating already when passing ...

1

It depends a little on what you mean by "extreme" electric current, but the answer is probably no. The energy scales are wrong. Electric current in a metal is a sub-electron-volt process: a potential difference of much less than a volt can displace electrons all the way through a piece of metal. The weak interaction is a keV- or MeV-scale process. And ...

0

To make a bomb you need a nuclear fission chain reaction. In essence what you need on average slightly more than one neutron produced by a fission to produce further fissions. So of your 3 neutrons, if on average 1.1 neutrons initiated further fission then after 500 generations (one fission initiating others), which would not take very long, you would be ...

0

Energy releases for one fission is approximatively the same . So, what changes ? The number of fissions during one second ( for instance ) giving thermal power. Number of fissions = N * Sigma * Phi N : number of fissile atoms Sigma : probability , neutron cross section . Phi : neutron flux We suppose a bomb and a reactor with the same number N ...

2

Here's an ASCII energy-level diagram based on your description: . ------- 1000 keV . . . . . . . -------- 271 keV . 190 keV ----------- . . -------- 0 keV Ga-61 Zn-60 + p The key here (and the difference between my diagram and the way I read your question) is that gallium-61 is bound, which means that ...

1

It means that if $E_0(Ga-61)$ is the energy of the Ga-61 in rest, and $E_0(p)+E_0(Zn-60)$ the energy of the proton and Zn-60 in rest when both are separated by an infinite distance (so they don't interact), that: $E_0(Ga-61)-E_0(p)-E_0(Zn-60) = -190 keV$. The Galiumsystem is bound with respect to the proton and Zinksystem, so Galium is more stable than a ...

2

They usually don't hit and K for them is less then1. For the very reason we use moderaters like water to increase k greater than 1 for sustained reaction.

57

The answer is in wikipedia The photograph on the right shows two unusual phenomena: bright spikes projecting from the bottom of the fireball, and the peculiar mottling of the expanding fireball surface. The surface of the fireball, with a temperature over 20,000 kelvin, emits huge amounts of visible light radiation (more than 100 times the intensity ...

5

I don't have any special knowledge about this image or nuclear testing, but typically when we see an asymmetry in the evolution of a system, there should be an asymmetry in the laws governing it, or the initial conditions$^†$. There are three asymmetries that I can think of in the system that could explain the asymmetry in the evolution of the system. ...

7

Suppose you start with a linear solenoid. Due to the Lorentz force charge particles travel in circles (or helices) inside the solenoid so they can't reach the walls of the solenoid. But obviously the trouble is that they will leak out of the ends. Now we curve the solenoid round and join its ends together to make a torus so now the particles can't leak out ...

0

As written upper , Nb and Pr cannot be a pair of fission products . With a correct fission reaction , I can give a complete approximate calculation by hand . At the end , I obtain the energy of each fission product .

0

The sum of the actual atomic masses of the reaction products minus the sum of the actual atomic masses of the reacting species gives you the amount of mass converted to energy during the fission. That latter amount can be found by applying $E=mc^2$ to that mass deficit. As I wrote in my comment, protons cannot be destroyed in nuclear fission, so the atomic ...

2

Your ball of iron might reach as much as 1.2 solar masses before something drastic were to happen. Up until that point, the ball could be supported by electron degeneracy pressure. The iron ball would contract to about the size of the Earth or a little smaller, the interior would heat sufficiently to completely ionise the iron. The electrons would be so ...

1

Supernova happens when the core of a supermassive dying star starts fusing iron and heavier elements under massive gravitational pressure. The reaction is endothermic, unlike the fusion of lighter elements (iron is the peak) so the resulting outward radiation pressure stops apposing inward gravitational pull and the star collapses on itself. The pressure ...

2

As you say, in absence of moderator, light water in PWR, neutrons cannot reach thermal energy, no fission on uranium-235 appears, reactor begins not critical and it stops. For a neutron point of view, it is the same situation like scramming control rods.

0

The first is, is my understanding described above correct? Basically, yes. The second is, how does this affect us healthwise? We are stuck with these debris of cesium-137 and strontium-90 for the next 30 or more years, but what effect do they have on us? Stochastic health effects (induced tumors, cancers, leukemia,…) of ionizing radiation are ...

7

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 ...

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 ...

0

The mass defect is linked to the nucleus unbinding energy. It is equivalent to negative energy but it's not. Nucleons are composite particles. Most of their masses come from complex interactions between the heavy quarks and the massless gluons. In a nucleus, nucleons aren't individual particles, one may see them like a sea of quarks and gluons instead of ...

2

Excited nuclei don't get that way on their own: you have to hit them with a beam. Your reference starting on page 689, refers to beams of oxygen, silicon, calcium, and nickel, with energies of several MeV per nucleon, on stationary lead targets. Pick a target, beam, beam energy, and reasonable excitation energies for the target and beam nucleus, then use ...

1

Here is what wiki has to say about Polonium radiological toxicity: By mass, polonium-210 is around 250,000 times more toxic than hydrogen cyanide (the LD50 for 210Po is less than 1 microgram for an average adult (see below) compared with about 250 milligrams for hydrogen cyanide[66]). The main hazard is its intense radioactivity (as an alpha ...

2

Let's suppose you have swallowed one of the Po-210 sources from this student kit. Its activity is 3700 Bq (0.1 μCi). Based on the Table 6 in the meta-study [1], it is probably safe to ingest up to 0.02 MBq/kg of the Po-210. This means, that for 80 kg person, it is probably safe to ingest 1.6 MBq of the Po-210, so you "need" to eat approx. 400 of these ...

1

here's a view from a pure mathematician in an abstract ideal world, untroubled by real-world complexities. I hope it's helpful even if it is a gross simplification of reality. Atomic shells feel evenly balanced when they are of a size equal to twice a square number, i.e. 2, 8, 18, 32, 50, 72, 98, 128 etc. Every period in the periodic table has a number of ...

1

It's not (necessarily) a question of what's permitted so much as what's been measured, and published, and entered into the ENDF database. The reference for the 24Mg paper is 1984EL12, Nucl.Instrum.Methods 228, 62 (1984), D.Elenkov, D.Lefterov, G.Toumbev, "Two-Target DSAM following the (n, n'γ) Reaction with Fast Reactor Neutrons" and the abstract ...

2

The difference comes from the kind of force that holds the constituents together. The force on the quarks in the nucleon is the color force, (one of the four fundamental forces), between nucleons, it is the residual color force, which appears as the strong force that binds the protons and neutrons in a nucleus. The color force and the fact that the ...

2

Optical model potentials originally were invented to describe elastic scattering of a single nucleon (neutron or proton) off a complex nucleus. More recently, the formalism has been modified to include the scattering of heavy ions as well as nucleons. The assumption made in the model is that nuclear matter inside a heavy nucleus is at least partially ...

4

Making a bomb is not a simple matter, a lot of isolating the correct isotope with centrifuges is necessary , and it is called weapons grade ore. Natural deposits do not have weapons grade uranium. Interestingly enough there are signs that fission of the type happening in fission reactors has happened once on earth naturally: Oklo is the only known ...

0

No. A pair of neutrinos is pulled from the vacuum. One of them interacts with one of the quarks via the weak force, and they both change identity: the quark to another kind, thus changing the neucleon; the neutrino to an electron, which escapes. (The negative charge unit also moved from the quark to the lepton.) The electron escapes as the beta ...

0

It can only be E2. Your initial and final angular momenta are 0 and 2 respectively. You can not couple 0 and 1 to get 2 --- the Clebsch-Gordan coefficient is 0. But there are other transitions (like ^{135}Cs(5/2+) --> ^{135}Cs(7/2+) where both are possible because 5/2 can be coupled with 1 to get 3/2, 5/2 or 7/2, and with 2 to give 1/2, 3/2, 5/2, 7/2 or 9/2. ...

1

The probability of N atoms left is $$P(N) = \left( \begin{array}{c} 100 \\ N \end{array} \right) 0.5^{100} \approx \frac{1}{\sqrt{50\pi}}\exp\left(-\frac{(N-50)^2}{50}\right)$$ The expected value will be 50 which is also the most probable value. The standard deviation is 5, so we expect the number of atoms left to be 50 ± 5.

3

Basically Fission Bomb efficiency depends significantly on the neutron flux intensity before it blows itself apart hence the elements to be considered are the mechanism used to achieve supercritical mass the “neutron boosting” strategy Regarding the critical mass generation mechanism as far as I can remember Little Boy’s used a gun type one and it ...

1

The mass defect and the binding energy are not linear functions of the number of nucleons. They increase until the iron and then decrease. See the Figure 31-5 of The Mass Defect of the Nucleus and Nuclear Binding Energy The thorium Mass Defect for one Average Nucleon ( MDAN ) is higher than an uranium MDAN. Uranium decays in thorium and alpha particle ...

3

Think of a star as a big globe with ideal gas inside. Gravity acts as a force compressing the globe, the more it compress, the more energy goes to thermal part since $$dE = TdS-PdV$$ so a shrinking volume decreases $PdV$ term ($P$ is negative in this case, otherwise the system would be expanding), and for $dE=0$ because no reactions are occurring and no ...

3

I believe what he's pointing out is that energetic particles have sufficient kinetic energy to counter the pull of gravity, but as the star cools, the net kinetic energy decreases until the particles cannot go "up," or away from the centroid of mass. At that point, the star's mass collapses inward. The total mass is probably less than before, since the ...

0

In answer to your first question, the fuel in standard nuclear reactors is Uranium-235. The chain reaction that produces the heat is started by bombarding slow moving neutrons at U-235 nuclei - http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/fission.html The speed of these neutrons, and the choice of U-235 is very specific to the reaction. The spent fuel ...

1

a) To convert heat energy to electricity the heat energy has to be at sufficiently high temperature, as the latter limits the percentage of heat energy that can be converted to other energy forms. Very roughly the efficiency $\epsilon$ of an idealised heat machine is given by: $$\epsilon\:\text{(%)} \approx 100 \times \big(1-\frac{T_H}{T_L}\big)$$ Where ...

1

Short answer: turned ionization detectors are easy enough, but they are not "Geiger counters". The core of a Geiger counter is a gas ionization detector that runs in a saturated cascade mode. They respond to ionization in the gas and are very nearly digital in nature. Notice that Geiger counters are often rigged to emit an audible click when it respond, ...

2

To choose among different possibilities, theoretical calculations are made to evaluate the likelihood of a successful fusion between the candidates. This is not a simple evaluation because it relies on nuclear models, and several branches may occur. The FIAS institute in Germany collaborates in this subject with Dubna in Russia, and they have made ...

2

The gas inside the geiger tube will have a different response according to what kind of radiation is entering and what energy it has. For example, here's Helium's absorption for electrons , and here's its X-ray absorption . The tube only gives you a particle count, but if you have a fixed distribution of incoming particles then you can calculate/calibrate a ...

1

True, Geiger counters can be tuned to be more sensitive for certain types of decay and for the amount of radiation emitted. They can detect ionizing radiation in the form of alpha particles, beta particles and gamma rays. Given the different nature and energy of these types of radiations (alphas are Helium nuclei, betas are electrons or positrons, gammas ...

2

You have a two step process: the plutonium nucleus emits an alpha particle and forms a uranium nucleus the uranium nucleus may be formed in an excited state - if so it will relax to the ground state by emitting a gamma ray photon You are told that no photon is emitted when the alpha particle has an energy of 4.9 MeV, so the difference between the ...

1

The negative sign for the partial decay widths denote the sign of the corresponding reduced width amplitude in the R-Matrix formalism, which is what the author of this paper (and everybody else analyzing cross sections of low-energy nuclear reactions) is using to do his analyses. He's using the R-Matrix code SAMMY, which sadly has very little documentation. ...

3

What's "prompt" depends on just what you're doing. One second is a brief time interval if you're interested in radiological shielding, but an eternity if you're interested in the spectroscopy of a single nucleus. For instance, suppose you have neutrons capturing on some material. The neutrons typically capture in some very excited nuclear orbital and ...

2

The argument depends on a fact about photons that you may or may not have encountered yet: if you have a photon with energy $E$, that photon must carry momentum $p = E/c$ in some direction. If you change your opinion about the photon's momentum by observing it from some other reference frame, you also must change your opinion about its energy. For instance ...

1

Probably the best way to think of this, and the way I suspect it works in the minds of people who write exactly this sort of equation in physics textbooks, is that it's an equation describing a nuclear process, rather than a chemical process. The electrons initially bound to the nucleus are irrelevant spectators. A less ambiguous way to describe the ...

3

You don't need a calculation. Let's go backwards: consider the electron-positron pair. There is an inertial frame of reference (the "centre of mass" frame) where the total momentum of the pair is zero, i.e. the centre of mass is still. Now, keeping this frame of reference, rewind the movie: before the creation of the pair there was only a photon. But a ...

0

The energy of the decay has little to say about whether the covalent bond will remain after the decay. The reason is because the $\beta$-decay electron (or positron since the question doesn't specify) will be moving so fast (compared to the orbital electrons that the cross-section for scattering will be quite small. Since scattering off the orbital ...

4

If you want a more palatable assumption than a population of alpha particles dwelling inside each heavy nucleus waiting to escape, imagine this instead: Inside the nucleus that you have many protons and neutrons rattling about, and that pairing interactions cause alpha particles to form and disintegrate with some frequency which doesn't depend (much) on the ...

0

I think I found the solution. The decay energy of radiocarbon is the following $0.156476 MeV = 2.50702189 \cdot 10^{-17} kJ$ ${2.51 \cdot 10^{-17} kJ} \cdot {6 \cdot 10^{23} \cdot 1/mol} = 1.51 \cdot 10^7 kJ/mol$ If we compare this decay energy to the energy of the chemical bonds table (~ 200-400kJ/mol), we will see that it exceeds that 100000 fold, ...

3

First of all the $\beta$ particle emitting from nucleus is too energetic to be captured in atomic orbit to form atom.So it is definitely not the case. Infact thta's why we are sure that it is coming from nucleus not from atomic orbital. More over while writing for radio active decay we are looking for change in the nucleus as it is basically a nuclear ...

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