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19

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 ...


11

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 ...


9

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 ...


8

$\newcommand{\d}[1]{\,\mathrm{d}#1}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}\newcommand{\p}{\psi_{100}}\newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$The hydrogen ground state is the following: $$\psi_{100}=Y_{00}\frac{2}{a_0^{3/2}}e^{-r/a_0}$$ The expectation value of the position operator on this state is the following: ...


7

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 ...


7

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

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

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

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 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 ...


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 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

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

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

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

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

For any sample of atoms in a gas they have a range of speeds. The likelihood of any atom having any particular speed is given by a probability distribution called the Maxwell-Boltzmann distribution. But the main point is that low mass atoms are more likely to have high speeds. For hydrogen and helium at temperatures typical of Earth's atmosphere there is ...


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

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 ...


2

I pulled most of this from Wikipedia here. A stationary state is called ''stationary'' because the system remains in the same state as time elapses, in every observable way. For a single-particle Hamiltonian, this means that the particle has a constant probability distribution for its position, its velocity, its spin, etc. A stationary state is not ...



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