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No, if the spin were classical, as any classical angular momentum it weren't quantized. It's easier to take as an example the orbital angular momentum. It is quantized because of the wave-nature of the quantum particle. To put it in an intuitive form, not every function can represent the wave-function of the electron in the atom. Imagine an oscillating ...


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A single photon cannot decay into anything with mass because the impulse-energy relations for massive and mass-less particles differ by the rest mass, so you cannot conserve both impulse and energy. If you add a particle to absorb impulse and energy, you add enough freedom through new variables that this system of 4 equations (energy and impulse in three ...


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Yes, you have very likely done what the teacher wanted you to do. But you should maybe mention to teacher that he has asked a poor question (actually... maybe you shouldn't mention it... or at least be diplomatic about it...) The reason the question is dumb is because, as pointed out in the comments, you could (reasonably) use the relativistic expression ...


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I am replying to this because you seem to be a student, and not so clear on the statements. I have read that during fission and fusion processes, there is some kind of equilibrium between the single nucleus and the disintegration products, so they are constantly being converted into each other. " I have heard" is not enough, you should give a quote or ...


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The law of conservation will always hold. The Fist Law of Thermodynamics: The increase in internal energy of a closed system is equal to the difference of the heat supplied to the system and the work done by it: ΔU = Q - W Law of Conservation of Energy: The total amount of energy in any isolated system remains constant, and cannot be created or destroyed, ...


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I can answer the question partially. Energy conservation can be broken in time intervals less than plank time http://en.m.wikipedia.org/wiki/Planck_time (approximately 10^-44s). In time intervals longer than that, energy must be conserved.


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You're right that1+, 2+, 3+, 4+, and 5+ are all possible values for the final state following an E3 transition. However, if the final state were 1+,2+, or 3+, then an E1 transition would be possible. If the final state were 4+, then an M2 transition would be possible. So if the lowest multipole is E3, that leaves just one option.


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Maybe I should add that nuclei which are unstable to beta decay can still exist for quite some time. However, beyond the limit of beta-stability, there comes a point at which it is energetically favorable for a nucleus to decay by emitting a neutron. Another way of putting this is that the last neutron is not bound within the nucleus. This point in an ...


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The decay $$D_s^{**+} \rightarrow D_s^+ \pi^0 $$ is a strong decay (flavors charm and strangeness being conserved). Thus, parity must be conserved. The final state particles are both pseudo-scalars, and hence the product of their parities is positive. Thus, the final state parity equals $(-1)^{L_f}$, where $L_f$ is the final state orbital angular momentum. ...


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Effectively there is a maximum number, or rather producing more and more neutron-rich isotopes requires energy. You can think of it this way. Identical particles are affected by the Pauli exclusion principle; this applies to neutrons in the nucleus. Therefore a stack of neutrons will fill up the lowest energy states, but will then have to occupy higher ...


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A thermonuclear weapon requires extreme temperatures in order to ignite and efficiently produce and consume fusion fuel through the Jetter cycle. Unlike most terrestrial thermal processes, the speed of heat transfer through radiation transport is very slow in the thermodynamic regime of the weapon during disassembly. The temperature of the weapon becomes so ...


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When a nuclear bomb goes supercritical, it has entered a state where the neutron population will grow unbounded by the feedback factor of temperature. I'm not saying that it's insensitive to the temperature, it's just that the mechanism is insufficient to stop the reaction. Because of that, the temperature of the nuclear core increases until some other ...


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Yeah, you've basically nailed it with the first one. During the chain reaction you have a huge amount of energy immediately dumped into a tiny space, but there's no physical mechanisms containing it. All of these high-v particles immediately collide with other particles, bouncing them so violently that they collide with other particles, and so on, creating ...


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First, the $k$ in the first region is incorrect. Check your signs. You should have $$k_{I} = \sqrt{2M(E-V_o)}/\hbar.$$ Second, your solution to the second region will have sine and cosine solutions with $$ k_{II} = \sqrt{2ME}/\hbar.$$ Real exponential solutions will occur in each region if $E<V_o$ in region one and/or $E<0$ in region II. Look at ...


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What you're missing is that you're interested in $E < 0$. Such states are "bound" in the square well. The equation: $ \frac{d^2 f}{dx^2} ~=~ - k^2 ~ f(x)$ is solved by $f(x) ~=~ A e^{i k x} + B e^{-i k x}$. If $k^2 < 0$ then you get another $i$ in this picture and you switch from sinusoidal behavior to plus/minus exponential behavior.


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Normal ordering is an ordering of a product of field operators, in which all the annihilation operators are placed in to the right of all creation operators. This mean that the expectation value of a normal ordering in relation to the vacuum state is zero. If $c_a^\dagger c_b = \rho_{ba} + :c_a^\dagger c_b:$ and $|\Phi\rangle$ is the vacuum state, then ...


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One can write the magnetic momentum operator in the following form: $$\mu=g_p\bf{s_p} +g_n\bf{s_n}+\frac{l}{2}$$ where $\bf l$ is a angular momentum, $g_p$ and $g_n$ are gyromagnetic ratio for proton and neutron respectively, $\bf{s}_p$ and $\bf{s}_n$ are sin operators.The coefficient $\frac{1}{2}$ in front of $\bf l$ is appear because of the contribution in ...


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Certainly not. You have extracted only part of the full equation, which actually looks Pythagorean in structure: $E^2 = (mc^2)^2 + (pc)^2$ This relates the energy of an object to its mass and momentum. Its more famous cousin, $E = m c^2$ is simply the limit where $p=0$, or the energy of an object in a reference frame in which it is at rest. On the other ...


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You may wish to look at http://arxiv.org/abs/nucl-th/0104037 (Nucl.Phys. A694 (2001) 295-311) - they give figures for Th-224. For low deformation, the moment of enertia is about 5000 m fm^2, where m is the nucleon mass and fm is, I guess, Fermi (femtometer). I would think the moment of inertia of Th-232 should not be much different.


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In a sense, this question is unanswerable. (Still a good question!) } Consider the natural decay chain of $_{92}U^{238}$. This isotope goes through 14 steps, along various routes, to decay to $_{82}Pb^{206}$. This $4n+2$ series is the longest naturally occurring one. See http://en.wikipedia.org/wiki/Decay_chain However, someone could immediately claim ...


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We know/assume that the mass of the particle is constant. Elastic scattering means no change in kinetic energy (in the center-of-mass frame). Since the mass of the particle is constant, this means there is no change in the speed (not velocity; speed) of the particle during an elastic collision. Torque is $\vec{\tau} = \vec{r} \times \vec{F}$. For a ...


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Short answer: you don't. Slightly longer answer: You're using beams of particles, and you focus each of them as much as you (practically1) can so that the particles in each beam are reasonably close together. The result is a wide variety of interaction distances from far apart through near misses to closer interactions still. You mentioned electrons ...


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I think you've misunderstood what the uncertainty principle tells us. The electrons in an atom do not have a position because they are delocalised over the whole atom. So two atoms can't behave differently because their electrons are in different positions - all atoms of the same element/isotope have thir electrons delocalised in an identical way. We can ...


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All bound states (groups that are not decaying away, they're more "orbiting" in some way) of nuclei and electrons may be written as a (linear combination of) energy eigenstates – eigenstates of the Hamiltonian $H$. The Hamiltonian (energy operator) is always the same – it captures all the information about the laws of Nature and there are the same. The ...


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If you're imagining some kind of colossal explosion from the nuclear reactor, then this won't happen unless some kind of "make me into a bomb if you throw me at alien ships" feature is expressly designed into the reactor. The reason is that bombs and reactors work very differently. The main problem that must be solved in a bomb is that as the chain ...


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Access to lots of water- water cools down significant amounts of heat Usually located away from populated areas They don't produce smoke or Carbon Dioxide, will not pollute wildlife (no greenhouse effect)


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Just to offer a random benchmark--because I haven't made a study of this question--$^5\mathrm{He}$ is described as having a linewidth of $0.60 \,\mathrm{MeV}$, which corresponds to a lifetime in the single digits of nanoseconds. Refernces: http://ie.lbl.gov/toi/nuclide.asp?iZA=20005 The data for which is taken from the ENDFs. ...


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About short-decays it is not easy to measure resonance widths. There are many reasons for this. Short-lived resonances are wide and their ranges of energies overlap at least partially with other resonances. I mean, neighbor resonances form a strong background. Therefore, comparisons with theoretical predictions are very difficult. The ideal case for ...


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The intrinsic width of a line due to a decay process is related to the lifetime $\tau$ (not the half-life) by the energy-time version of Heisenberg's Uncertainty Principle $$ \sigma_E \, \tau = \frac{\hbar}{2} \,.$$ Recall that the standard deviation of an exponential decay is the lifetime. Be careful that you do not confuse the experimental resolution ...


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It is worth realising that protons are colorless; they have no overall `strong force' charge. This is similar to the atom having no electric charge. Nucleon-nucleon interactions, such as proton-proton scattering, are mediated by pions. This is the Yukawa theory, which treats the proton and neutron to be components of an $SU(2)$ doublet. Without ...


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The electromagnetic repulsion between two protons is a long-range force, depending on $1/r^2$, where $r$ is the separation of the two protons. The electromagnetic repulsion between two protons is not the reason that they do not stick together; if they are forced together (or can tunnel through the Coulomb barrier) then short-range strong nuclear forces are ...


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Yea, a proton and neutron stick together, but two of the sam kind don't. You don't get neutron balls even with only the interneucleon force and no electric repulsion. I asked about it some years ago in a physics on-line forum, long before StackExchange. Ended up getting a textbook and eventually learning that "the force is largely insensitve to species ...



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