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12

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 ...


10

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 ...


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

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 ...


5

A nice overview of the problem is given in arXiv:1205.3740. I'll summarise the most important points here. Let $d$ be the number of space dimensions. Then the Laplace operator is given by $$ \Delta=\frac{\partial^2}{\partial r^2}+\frac{d-1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\Delta_S\tag{1} $$ where $\Delta_S$ is the Laplace operator on the $d-1$ ...


5

The most common isotope of hydrogen has no neutrons. Other isotopes are deuterium with 1 neutron and tritium, with 2 neutrons. Since virtually all (99.98% according to wiki) naturally occurring hydrogen comes in the no neutron isotope, it seems reasonable that books show a schematic of that one when illustrating hydrogen. As a secondary motivation, the one ...


5

First, the idealized clean non-relativistic problem with the $1/r$ Coulomb potential is integrable: the exact wave functions and energy eigenvalues may be precisely written down using elementary functions. It is exactly true that the eigenvalues of the bound states are $-13.6\,{\rm eV}/n^2$ etc. Now, the first simple correction is the proton motion. 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

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 place;...


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 ...


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 ...


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

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

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 ...


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

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

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.


3

Let's think about a system that has a two-fold degeneracy for some given energy level. That is, two states $ \psi_{a} $ and $ \psi_{b} $, both of which correspond to energy $ E_{0} $. An example would be a spin-1/2 particle with a Hamiltonian that is spin-independent. Now imagine that when we apply a perturbation, H', to the system, the degeneracy breaks ...


3

The relativistic generalization of the formula; and the replacement of the electron mass by the reduced mass are clearly two basically independent steps (at least in the leading approximation), and both of them have to be applied to agree with the experiment. The most accurate experimental value of the energy is $-13.59844\,{\rm eV}$, see NIST: http://...


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\ \...


2

It is all over the place, but it's involved. It is normally calculated from the path integral propagator. The most concise source of the radial Green's function you are after is eqn (15) of Grosche 1998, in terms of modified Bessel functions, integral rep, $$ G_l^C(r'',r';E) = \int_0^\infty\dfrac{e^{i e^2s''/\hbar}ds''}{\sqrt{v'v''}} \dfrac{m\...


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

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 http://...


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

The instructor hasn't thrown out $r^4$. If you do the calculations properly you will get the desired result. What the instructor is telling is that while simplifying and taking commons out for $\frac{\text{dP(r)}}{\text{dr}}=0$ you get,$$(\text r^3)(6-\frac{\text r}{\text a_0})(f(\text r))=0$$which gives, $\text r=0$ and $6\text a_0$ from the first two terms....



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