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29

The decay phenomenon is a purely quantum mechanical property. This problem is equivalent to a particle in a finite potential well, and a lower potential state that is available outside the well. Classically if the energy of the particle in the well is lower than the potential barrier - it will never get to the lower state. By quantum mechanics, the particle ...


15

Speaking loosely, each individual atom has a desire to become stable, but that translates into a probability of decaying. This means, since there are billions and billions of atoms in a macroscopically significat chunk of material, that there are always going to be unlikely holdouts, and these holdouts are responsible for radiation that after the initial ...


11

This would be a highly energenic event, a gravitational collapse in combination with the initial inward velocity of 1000km/s (which is greater than the escape velocity at the surface of the sun). There would be some type of nova event initially because the hydrogen already present in the sun would be compressed by the infalling new material, greatly ...


7

If ice is "all around the sun" I fail to see how it can be moving at a velocity of 1000 m/s inwards. The mass of the sun is $2\cdot 10^{30}\mathrm{\;kg}$ and the radius $7\cdot 10^{8}\mathrm{\;m}$. The thickness of a shell of ice with that inner radius and mass would be (assuming the usual density of ice of about 0.9x that of liquid water) approximately ...


6

The beryllium-7 nucleus is stable, but the beryllium-7 atom may decay by electron capture. This is because the reaction $$ \rm ^7_4Be^+ + e^- \to {}^7_3Li + \nu_e + 0.861\,MeV $$ is energetically allowed. The equivalent reaction, with the electron on "before" side replaced with a positron on the "after" side $$ \rm ^7_4Be \to {} {}^7_3Li^- + e^+ + \nu_e + ...


5

You can't have a "ball of ice with the mass of sun", because the ice in the middle of the ball wouldn't be strong enough to support the weight of the ice on top of it. Instead, the ice would collapse under its own gravity. This would cause the pressure and temperature inside the ball to increase until the water molecules that make up the ice would break up ...


5

Either. It's context dependent. Chemists generally mean the whole atoms, nuclear physicists usually mean the nucleus, and people not in those categories could mean either. And there are exception to all those rules or thumb. And the distinctions is important when people start throwing masses around because the mass of an electron is almost on the same ...


4

Quarks, the constituents of hadrons/mesons, interact via the strong, weak and electromagnetic force. So hadrons/mesons do interact via all this forces, too. Even if the total net-carge is zero. Take for instance the neutron, which has zero electric charge. Still it has a magnetic moment which gives rise to electromagnetic interactions. It can also decay via ...


4

Charged hadrons, and neutral hadrons with nonzero magnetic moment, interact electromagnetically. A spinless, neutral hadron would not couple to the electromagnetic field at tree level, but the most obvious example of such a particle is the $\pi^0$, which decays electromagnetically to two photons. All particles with flavor participate in the weak ...


4

No, Bismuth-218 forms Polonium-218 by beta decay. Assuming there aren't any nuclei big enough to fission into Bismuth-218, the only ways of forming Bismuth-218 are alpha decay, beta decay and beta plus decay. Beta plus decay would have to be Astatine-218 decaying to Polonium, but this decay mode doesn't happen (actually Polonium-218 beta decays to ...


3

Deuterium reacts with low energy neutrons to form tritium, though the cross section is very low. Tritium beta decays to $^3$He with a half life of about 12 years, so the process results in very slow production of $^3$He. The trouble is that $^3$He also reacts with low energy neutrons, but it forms tritium and a free proton rather than $^4$He. So the ...


3

The first thing you'll notice is that the Sun stops shining. It still produces heat and light, but everything is stopped by the thick layer of cold ice. The Sun however is not completely cold. The core is still active, even more than before. You doubled the mass, so the Sun has a higher pressure and thus can fuse easily hydrogen atoms together. The net ...


3

The important argument for this discussion is the Bethe Weizs├Ącker formula, which describes the binding energy of nuclei. I will try to give a cursory overview of the most important aspects. Not only heavy elements show fission and fusion. All elements up to iron-56 (one of the nuclei with the highest binding energy per nucleon) can create energy in ...


3

The electrons freed from the bounds of the fissioned nucleus will follow conservation of momentum and will move according to the kinetic energy they have. What will happen to them will depend on the medium they are in. Their kinetic energy will be too high for them to meet up with the fragments constructively, so there is very low probability the new nuclei ...


3

The electric potential between two point charges depends on the product of the charge of each. The charge in the nucleus is due to protons, and the count of protons is given as $Z$ for a total charge of $Ze$. The alpha decay removes two protons (and two uncharged neutrons). The charge of the alpha particle is $2e$, leaving a charge of $Ze - 2e$ in the ...


2

There are two general types of atomic warhead, the gun type where two sub-critical masses of Uranium 235 which when triggered one is fired at the other to create a critical mass, the other is an implosion device where a hollow sphere of Plutonium 239 is compressed by the detonation of an explosive jacket. Such devices comprise the trigger for themo-nuclear ...


2

In astrophysics, rates of beta decay and electron capture can be influenced by environment, specifically the ambient density of free electrons. If the gas is dense enough, the Fermi energy of the electrons could be higher than the maximum possible beta decay electron energy. This would suppress beta decay and enhance electron capture. Take the example of ...


2

The alpha particles are emitted as bare nuclei, with charge +2. This is how alphas were originally distinguished from betas and gammas back when radioactivity was being discovered: the three species bent in different directions in a magnetic field. It's possible to distinguish between a two-body decay (to alpha and negative ion) from a three- or four-body ...


2

There is a lighter nuclide which undergoes fission: $^8Be$. It fissions to two $^4He$ nuclei ($\alpha $ particles) with a lifetime on the order of $10^{-17}$ s. The binding energy per nucleon is much less for the beryllium than for the two $\alpha$s. It's important to note that $^8Be$ is an important link in the triple-alpha fusion process in older stars ...


2

I am addressing this part of the question: Also why do only neutrons show fission/fusion and why can't electrons preform fission/fusion? Nuclei with a large number of neutrons are unstable . It so happens for some of them that an extra neutron in a specific low energy range can be caught when impinging on that nucleus , but the resultant new isotope ...


2

what happened with the electrons in this process? After fission, the potential that bound the 92 electrons changes. The alpha has too much kinetic energy and cannot trap the two electrons it needs. It will pick up electrons at it comes to rest in the material or the air. The remaining, now Thorium, nucleus reorganizes and the electrons are bound in the ...


2

Remember that the rest mass of a proton is about 940MeV. So, a proton at 0.6c is pretty darn energetic. I will leave a precise energy up to the reader. However, there are a number of experiments in the physics literature of smashing protons into various things. At low (1MeV-ish) energies, you are looking mainly at classic Rutherford scattering cross ...


2

All the atoms have the same chance to decay at any given moment. If you have more of them at the same place you will simply have bigger chance of them decaying. It's like dice, you have 1 in 6 chance of getting a 6. If you have 100 dices you will have 100 times more chances of getting a 6. Thous you'll have more 6-es. Unless you cheat ;)


2

The total energy from the fission of one atom of U-235 is 202.5 MeV according to Kaye and Laby. Typical nuclear fuel is enriched to about 5% U-235, but never more than about 18% - higher than that and you are talking about material for bombs. It is very hard to define the "energy content" of nuclear fuel without knowing what kind of reactor you are using, ...


2

The notation is that of one specific isotope (isotopes are nuclides with the same number of protons) of the chemical element Pu. 94 is the number of its protons, which is also the total charge, 240 is the total number of nucleons (protons and neutrons). In a neutral Pu atom there will always be 94 electrons to offset the charge of the protons in the nucleus. ...


2

The process by which the lithium becomes fissile due to neutron capture is called neutron activation. The subsequent decay is simply a fission reaction. There seems to be a precedent on various sites for such a process to be called a 'neutron capture induced fission reaction', although most of the Google results for the term refer to the more usual fission ...


2

Decays happen to individual nuclei ( particles). When more than one nucleus(particle) are involved it is called an "interaction". In this case neutron Li scattering Neutron capture by a nucleus is a possibility, in this case there is an intermediate nucleus formed , which can then decay.


2

A ground state $^7\mathrm{Li}$ nucleus is stable, so this reaction is either direct or involves a unstable, intermediate, excited state of the lithium-7 nucleus. If you are studying that excited state1 then you consider this reaction as $$ ^6\mathrm{Li} + n \longrightarrow \, ^7\mathrm{Li}^* \longrightarrow \, ^4\mathrm{He} + ^3\!\mathrm{H} + \text{4.78 ...


2

It's basically got to do with the fact that nuclear decay is a quantum mechanical process, and quantum mechanical processes are not deterministic in the traditional sense, i.e. given a set of conditions you can't predict exactly what will happen in a particular process, only the probability of something occurring. In this case, nuclear decay occurs through ...


1

For a famous example of a nucleus with internal orbital angular momentum, consider the deuteron. Considerations of exchange symmetry, spin, and isospin demand that the deuteron have unit spin, rather than zero spin. However the pion-nucleon interaction, gleaned from neutron-proton scattering and deuteron formation, suggests that about 4% of the deuteron ...



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