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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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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 $2\times10^{30}\ ... 6 In fact hydrogen is an old idea to get a high temperature superconductor, based exactly on the idea of its light mass. The problem is that one has to start from metallic hydrogen, which is a problem on itself. It has not yet been fully experimentally confirmed in the lab. You need pressures of several hundreds of GPa to achieve that (100 GPa is about 1 ... 5 From the horse's mouth: The proton source is a simple bottle of hydrogen gas. An electric field is used to strip hydrogen atoms of their electrons to yield protons. Linac 2, the first accelerator in the chain, accelerates the protons to the energy of 50 MeV. The beam is then injected into the Proton Synchrotron Booster (PSB), which accelerates the ... 4$SO(4,2)$is called the full dynamical group of the Kepler (or Hydrogen atom problem). The$SO(4)$,$SO(3,2)$and$SO(4,1)$subgroups of$SO(4,2)$are called partial dynamical groups. Unlike symmetry groups which commute with the Hamiltonian, dynamical groups do not. They have the following properties: The system's phase space is a coadjoint orbit of ... 4 The deuterium-tritium fusion reaction cross-section is highly temperature dependent and peaks at temperature of about$8\times 10^{8}$K, so I suppose these are the temperatures to aim for in a controlled nuclear fusion experiment. In fact according to this, the operating temperatures are at least$10^{8}$K. The density of the fusion plasma is a factor - ... 4 This is really just a footnote to Rob's answer. The Sun is an absolutely terrible fusion reactor. It uses a reaction$p + p \rightarrow d$that is hopelessly inefficient. The$d + t \rightarrow He + n$reaction that we use in fusion reactors is (up to) 26 orders of magnitude faster. As Rob says in his answer, the power produced per cubic metre in the Sun is ... 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$T_\infty$... 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 ... 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 ... 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 Probably, yes. 13.56MHz was the frequency used and it was a 300W field. http://www.rsc.org/chemistryworld/Issues/2008/April/ColumnThecrucible.asp The effect doesn't appear to be well described. So while the salt is necessary, the exact role it plays isn't detailed. It could be that some complex with the salt is resonating. 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 If you treat the 1s ground state's probability distribution as a classical charge density distribution (not really accurate, but I think the simplest way to interpret the problem), then there isn't one. This state is spherically symmetric, so the electric field is always radial and depends only on the radial coordinate and not on the angular coordinates. So ... 3 Perfunctory quantum comment #1: "orbital", not "orbit". There are no little balls moving in a circles in there. These things are quantum objects. But the sort answer is "yes", the proton has a non-zero momentum distribution that mirrors the electron's. Now, because the proton is nearly 2000 as massive, the proton's position wave-function is nearly 2000 ... 3 All matter that we know is made of atoms. All atoms derived from H and He. Therefore, if a banana exists, then a banana CAN be made. YOU cannot make a banana because technology. BONUS: Matter and energy are equivalent, therefore atoms can be produced from energy, therefore a banana can be made from energy. You cannot make a banana because technology. 2 "The question is about a fuel that would be available anywhere in space and not only in specific areas.": The problem is that the density in interstellar medium is extremely weak, less than the best void we can do on Earth. So you will need a long trip just to collect 1 liter at atmospheric pressure. Beside, you would need energy to collect and transform it ... 2 The only fuel we know of that is available anywhere in space is hydrogen. Unfortunately, there are 2 major stumbling blocks that prevent us from using it for propulsion. It is extremely sparse. The vacuum of space is a lot better than the best we can produce on earth. There's only about 1 atom per$cm^3$. On earth there are of the order of$10^{18}$atoms ... 2 It won't. The rms speed of an ideal diatomic gas is given by$(3kT/m)^{1/2}$, where$T$is the temperature and$m$the molecular mass. Were you thinking about non-ideal effects? These will start to become important if you increase the pressure enough. The details would depend on the interaction potential between the molecules. At (relatively) low densities ... 2 Emilio Pisanty has already given a good answer. Here we offer a qualitative (as opposed to quantitative) proof of the angular momentum dependence. Recall first of all that the energy-levels $$\tag{2} E_n ~=~-\frac{R_{\mu}}{n^2}$$ in the non-relativistic hydrogen atom without spin-orbit interactions are linked to the principal quantum number ... 2 The presence of gamma rays and positrons associated with lightning strikes suggests that the strikes are capable of accelerating particles with enough energy for D-H or D-D fusion. However I'd expect that even in a pure deuterium atmosphere the energy released by D-D fusion would be negligible compared to the energy released in the lightning strike itself. 2 As far as I know, the potential energy of the molecule could be minimum at a distance of approximately one molecular radius of the adsorbate. The energy minimum is in the order of something like$0.01–0.1\ \mathrm{eV}=1–10\ \mathrm{kJ\;mol^{-1}}$. Due to the weakness of the interaction, significant changes are only observed at low temperatures ($<273\ ...

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Since the formula you gave is real, the complex conjugate is the same as the wave function itself.

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