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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 ...


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 ...


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 ...


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 ...


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. ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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

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 ...


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 ...


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 ...


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

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 ...


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 ...


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.


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 ...


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 ...


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

Facts : The properties depending of the atomic mass are different. fact 1 : The melting point depend upon the atomic mass. There is no need of a new experiment to relate the hardness to the temperature distance to the melting point : ie heat helps to bend metals or makes other matters more brittle. fact 2 : Moreover, stress diffusion depends upon the ...


2

The nuclear force is a contact force, with potential energy curve $$ V \propto \frac{e^{-r/r_0}}{r}. $$ The range parameter $r_0$ is roughly one femtometer. Nuclei in a solid are typically $10^5\rm\,fm$ apart, so the nuclear interaction between nuclei from different atoms is astoundingly suppressed. If you think of a solid as a lattice of atoms connected ...


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.


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 ...


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

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 ...


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 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 ...



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