# Tag Info

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One approach: the liquid-drop model says they should have the same $R\propto A^{1/3}$, but different Coulomb energies proportional to $Q^2/R = Z^2e^2/R$.

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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 ...

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Nuclear weapons work precisely. The removal of the lead is only one step. A controlled explosion launches one subcritical mass at the other. An anti-missile strike would not trigger such a precise mechanism, but instead, would break it.

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With multiple nuclide types, the radioactivity is no longer described by a simple exponential, and as a result no longer has a simple half-life. Consider a rock which has equal numbers of two isotopes A and B, A with half-life of 1000 years and B with 1 million years, each with the same emission energies. The immediate decay level of A will be 1000 times ...

1

There are several things to consider. The amount of each nuclide in the mixture will determine the overall activity at any particular time. The total activity of a mixture of 3 nuclides would be $$A = \lambda_1 N_1 + \lambda_2 N_2 + \lambda_3 N_3$$. Because the $N$s are changing at different rates, there will not be a single resultant halflife for the ...

1

You can see that U(235) will have more mass than the particulate elements that are blown off if you explicitly write the equations for energy conservation. If you consider that Xe(140) Sr(94) and n(1) have kinetic energy when they are blown off then you can write the following equation for the initial and final states: $$M_\text Uc^2 ... 2 The alpha particles are emitted as bare nuclei, with charge +2. This is how alphas were originally distinguished from betas and gammas back when radioactivity was being discovered: the three species bent in different directions in a magnetic field. It's possible to distinguish between a two-body decay (to alpha and negative ion) from a three- or four-body ... 2 In astrophysics, rates of beta decay and electron capture can be influenced by environment, specifically the ambient density of free electrons. If the gas is dense enough, the Fermi energy of the electrons could be higher than the maximum possible beta decay electron energy. This would suppress beta decay and enhance electron capture. Take the example of ... 3 The electrons freed from the bounds of the fissioned nucleus will follow conservation of momentum and will move according to the kinetic energy they have. What will happen to them will depend on the medium they are in. Their kinetic energy will be too high for them to meet up with the fragments constructively, so there is very low probability the new nuclei ... 6 The beryllium-7 nucleus is stable, but the beryllium-7 atom may decay by electron capture. This is because the reaction$$ \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 + ...

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 ...

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To get Ni62 requires the production of nuclei heavier than Fe56. The problem is that these iron-peak elements are mostly produced in rapid nucleosynthesis reactions in the centres of stars (either massive stars, or in type Ia supernovae). The iron-peak elements are produced in a nuclear statistical equilibrium by burning Silicon. Rapid alpha capture ...

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Aside from the obvious answer of randomness in probability distribution, each atom's decay event does depend on its overall energy, which cannot be measured individually by current technology. For example it may be contained in a gas where the density in one area of the gas is slightly higher than other areas and may be undergoing more "collisions" ...

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I believe the first formula you give is obtained by approximating the integrand $S(E) e^{-(E/k_BT-\sqrt{E_G/E})}$, where $E_G$ is the Gamow energy, by a Gaussian centered about the maximum value $E_0$. However, if you take $T\rightarrow \infty$, then the integrand becomes $$e^{-\sqrt{E_G/E}}\rightarrow 1,$$ and the integrated cross section blows up, as ...

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 ...

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Short term, the ice would be vapourised as it fell. It would mix with the sun and form a bizarrely metal-rich star of twice the mass. Such a star would have a much more opaque envelope. This leads to (once an equilibrium is reached) the final star being much less luminous and cooler than a 2 solar mass star of more normal composition. It would probably be ...

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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 ...

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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 ...

0

First and most important effect of having two sun sized bodies in our solar system very close to each other is that this new system will throw the planetary motion off its course, and there will be chaos(Noticeable chaos right at the moment when lets us assume the ice appeared 1000 km away from sun out of nowhere). Gravitational pull will be twice as much as ...

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 ...

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 ... 1 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. 1 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 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. ... 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 ... 1 The entire atom is referred to by Pu. 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 ... 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 ... 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 ... 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 ;) 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 ... 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 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 ... 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 ... 0 The iron-peak elements are mostly the product of alpha capture reactions onto nuclei that begin with a similar number of neutrons and protons (Z = N). The nuclear burning associated with carbon and oxygen (in type Ia supernovae) or silicon (in the cores of massive stars at the ends of their lives) is very fast or even explosive. The important reactions in ... 1 I think you're referring to the tunneling effect. If you have two states of low energy (here: single nucleus/two nuclei) with a high energy barrier in between (here: highly deformed nucleus), then it's possible to observe transitions from one state to another, even if there is insufficient energy in the system to climb the energy barrier. A non-mathematical ... 1 Most fissile materials have some probability of spontaneous fission. For example in uranium-235, seven out of every billion decays are fissions. These spontaneous fissions are the reason why a critical mass of fissile material may spontaneously develop a fission chain reaction. Alpha particles incident on beryllium-9 will break the Be nucleus into two ... 1 Uranium 235 is naturally radioactive, with a half life of 703.8 million years. So if you take a lump of uranium 235 there will be nuclei decaying and releasing neutrons just due to its normal decay. These neutrons will then cause other nuclei to decay, and off goes your chain reaction. So you don't need anything to start the reaction. All you need to do is ... 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, ... 1 You have to use the change in mass. First, figure out to mass of the element/fuel when it is whole. Then, find the mass of the result of the reaction: the two new elements and the neutrons. Take the result on step 1 and subtract the result on step 2. This is the mass that you plug into the equation. 0 Inao Shyamananda, Binding Energy of a nucleus normally increases with increasing mass number. In the liquid drop model of the nucleus, when a person talks about constant binding energy, he/she means constant binding energy per nucleon. More precisely neither is a constant. 1 1) There are four vectors and a system of particles has an invariant mass according to the addition of its four vectors. 2) each nucleus has a rest mass, M, when it is not moving. 3) each nucleus has a fixed number of protons and neutrons, protons are Z in the table, and neutrons are A-Z 4) Each proton and each neutron has a fixed mass in its center of ... 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 ... 0 In a Hydrogen atom, the "orbital angular momentum of electron" is in fact the relative orbital angular momentum. The nucleus (proton) turns around the common atomic center too, but in a smaller orbit. The atomic electron and the nucleus do not have certain individual orbital angular momenta. They are in mixed states. See my explanations here and here. 1 You are getting the right thing. This is the binding energy formula.$$E_{binding} = (M_{constituents}-M_{BoundState})c^2 When the constituents come together to form a bound state the total mass is lowered not raised. Binding energy is the energy corresponding to the mass lost by the constituents as a result of them entering the bound state.

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The electric charge difference between the earth and the atmosphere grows with altitude, at around 88 DC volts per meter. This electric potential may be shorted out when a thermonuclear explosion releases radiation which ionizes the atmosphere. About 5% of a nuclear explosion's energy is in the form of ionizing radiation. A study of lightning flashes ...

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