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Heavy water is easy to separate from regular water because the difference in mass is quite large. The molar mass of heavy water is 11% heavier that regular water. However if we take uranium separation, then the percentage weight difference between $^{235}$UF$_6$ and $^{238}$UF$_6$ is only 0.9%, so the relative difference is far smaller. So it's a lot ...

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As dmckee says in his comment - Population III stars have no metals (a tiny bit of lithium and beryllium), but they are not "pure hydrogen stars", they still have the big bang fraction of Helium. Taking the second part of your question first. These "stars" will last for ever. Their final fate is to become a completely degenerate ball of helium, supported by ...

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This is a tricky bit of intuition to get right. In essence, having a lower angular momentum expands the radial range that the electron is allowed to span - the inner turning point moves inward and the outward turning point moves outward - but the electron is moving much slower at the outward turning point, which means that it spends more time there and ...

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The graph shows the probability of finding the electron between the distances $r$ and $r + dr$. This probability is given by: $$P = \psi^* \psi dV$$ where $dV$ is the volume element: $$dV = 4\pi r^2 dr$$ So we get the probability: $$P(r,r+dr) = \psi^* \psi 4\pi r^2 dr$$ and therefore when $r = 0$ the probability $P = 0$. It isn't that the ...

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By "consume" we mean "convert into helium." That $6\times10^{11}\ \mathrm{kg}$ of hydrogen is part of the Sun (specifically it is found in the core of the Sun), and it is converted into $6\times10^{11}\ \mathrm{kg}$ of helium. The Sun doesn't need to suck up material from space. Note that this amount of material is miniscule compared to the $2\times10^{30}\ ... 6 In fact hydrogen is an old idea to get a high temperature superconductor, based exactly on the idea of its light mass. The problem is that one has to start from metallic hydrogen, which is a problem on itself. It has not yet been fully experimentally confirmed in the lab. You need pressures of several hundreds of GPa to achieve that (100 GPa is about 1 ... 5 Is the Coulomb potential also used to solve the hydrogen atom in relativistic quantum mechanics? Yes, the Coulomb potential is there in the solution of the hydrogen atom with the Dirac equation, which is formulated in the relativistic framework. Now it is time to specialize to the hydrogen atom for which $$\frac{V}{\hbar c}=-\frac{Z\alpha}{r}$$ ... 5 You say: The Coulomb potential comes from classical electrodynamics but actually the Coulomb potential is predicted by quantum electrodynamics as a low energy limit. Quantum field theory describes the interactions between charged particles as the exchange of virtual particles, and it's not immediately obvious that it would lead to an inverse square ... 5 The answer is that the premise is wrong. There can't be a hydrogen wave function with the coefficients you have written. Even if there was no$| 1 0 0 \rangle$state present, the state isn't normalized. That means that it isn't physical. However, remember that the coefficients are somewhat arbitrary, that is, we're allowed to multiply the whole wavefunction ... 4 It is a misconception to say that In QM [...] we know that the the electron does not radiate EM-Waves because it is not actually circling around the nucleus. It is sometimes here and there. In QM the notion of "circling round the nucleus" does indeed fail to make sense, but this is not why electrons don't radiate EM waves. Instead, an atom in its ... 4 Consider two charges$q_1$and$q_2$kept at some separation. Suppose we want to calculate the potential energy of the system. By definition, potential energy is the work done to assemble such distribution. We can assemble the system in two ways: Bring$q_1$to its place; no work done during this as there is no field present. Then bring$q_2$to its ... 4 When labeling states of the hydrogen atom, one doesn't refer to the z component of the angular momentum, but rather to the total angular momentum. The total angular momentum is positive, but, as you've stated, there are two states for$J=\frac{1}{2}$with$L=0$, and those are$J_z=\pm\frac{1}{2}$(Or some linear combination of them) As to why this is, ... 4 The energy is defined as $$E = \frac{p^2}{2m} + V(\vec r)$$ where the first term is the kinetic energy and the second term is the potential energy calibrated so that$V(\vec r)=0$for$|\vec r|\to\infty$. Consequently, you may say that the energy in a given state (an analogy of an orbit in classical physics) is equal to the kinetic energy$T_\infty$... 3 First, note that Population III stars are expected to be massive, not tiny, with masses upwards of$10^6\,M_\odot$. The reason for this is due to the Jeans criteria, where the mass follows $$M_J\propto T^{3/2}$$ In the early universe (~ 1 Myr), the temperature was around 10,000 K; so in a pure-hydrogen environment, no cloud with a mass less than about a ... 3 The term to look for is Coulomb wave. These wavefunctions are well explained in the corresponding Wikipedia article. Depending on your mathematical background, you should be ready for a bit of a formula jolt, as these wavefunctions rely very intimately on the confluent hypergeometric function. If you want the short of it, then I can tell you that the ... 3 The hydrogen atom has an infinite number or quantum mechanically allowed energy levels, as explained on this web page. Using that same link, scroll up the page a bit to better understand how transitions between these energy levels give rise to absorption or emission of photons of very specific frequencies. Then scroll further down to see how the hydrogen ... 3 Hydrogenic wavefunctions (as well as anything with well-defined angular momentum about a given axis) come in two flavours. The first set is 'cylindrical', and has wavefunctions$\psi\sim e^{\pm i|m|\phi}$. The second set is 'cartesian', and has wavefunctions$\psi_\text{even}\sim\cos(m\phi)$and$\psi_\text{odd}\sim\sin(m\phi)$. Both sets are perfectly ... 3 Based on comments, let's clarify things first. The H-atom (or any other quantum system) is not in an "eigenvalue" of a measurable and observable quantity (call these observables from now on). Observables can be position, momentum, energy, angular momentum, etc. A quantum system may be in the eigenstates of these observables, to which certain eigenvalues ... 3 This won't work, though possibly not for the reason you think. High energy protons will go straight through a turbine blade without transferring any significant amount of momentum to it. The LHC uses a seven metre long block of graphite to catch the proton beam if there's a beam dump. Steel has greater stopping power than carbon, but even so a turbine blade ... 2 Yes, the question is basically asking which are the most abundant elements, though it's specifically asking about elements that form molecules and this is presumably why the second most abundant element helium is excluded (helium being chemically unreactive). Big Bang nucleosynthesis produced mainly hydrogen and helium, with small amounts of lithium and ... 2 Generally, you are correct, and trying to store gaseous hydrogen for long periods of time without significant losses doesn't work. The usual way around this is use lithium deuteride, a solid compound, as the main fuel in the fusion portion of the device (the secondary). Deuterium/tritium is also used in the fission primary. According to ... 2 Look up Lithium deuteride: a form of lithium hydride where the hydrogen is all deuterium: this will give you most of your answers. The main fusion[1] reaction that lets slip most of the energy and thus the horrendous blast in a fusion bomb is: $$_1^2 D + _1^3 T \rightarrow _2^4 He (3.5{\rm MeV}) + _0^1 n (14.1{\rm MeV})\tag{1}$$ Here I've written$D$for ... 2 The key difference is the complexity scale. In a typical every day reaction involving water, the process is thermodynamically driven; the difference in free energy between the reactants and products is much greater than any effect the extra neutron may have. In short, things happen mostly because there is a loss of energy or gain of entropy; and all the ... 2 The state you have given is not normalisable as a consequence of the results of the calculations you have done. Even if the first state (with coefficient$A$) was not present it would not be normalised. To normalise what you have given, another constant needs to multiply everything through (so that the relative proportions are unchanged 2 The maximum$s_z$eigenvalue an electron can have is$\hbar/2$. Therefore, the only way that a quantum state can have$\langle s_z \rangle = \hbar/2$is if the state lies completely within the$s_z = \hbar/2$eigenspace. Thus your answer may only include pure states with$s_z = \hbar /2$. 2 In nonrelativistic quantum mechanics you set up a wavefunction that is a function of configuration space,$\Psi(x_1,y_1,z_1, \dots , x_n,y_n,z_n,t)$. Where$(x_i,y_i,z_i)\$ is the position of the i-th particle. If any particles are identical then you make sure that the wavefunction is (anti)symmetric in those coordinates. Then if you are doing a simple ...

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The main fusion reaction in the sun is the proton-proton chain reaction, which takes six protons and produces two protons, one alpha particle, two anti-electrons, and two electron neutrinos. The deuterium nucleus is only barely bound and can be destroyed — dissociated into a proton and neutron — by absorbing a gamma ray with energy more than 2 MeV. This ...

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Your main question was "Is there an example where the overlap between a bound state and the scattering states makes a measurable contribution to the energy in the perturbative regime?" Actually, I disagree with the statement of the other answer that the scattering states must be included in the perturbative calculations only if the result is to be highly ...

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