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24

One cannot tell by the light spectra. Hydrogen and antihydrogen would give the same lines in the spectrum. The prevalence of matter over antimatter from other evidence indicates matter is predominant in the observable universe, and here is a nice review. How do we really know that the universe is not matter-antimatter symmetric? The Moon: Neil ...


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

The probability density of the ground state is time independent, so there is no motion in this sense. Yet the expectation value of the kinetic energy is non-zero, so there is movement in this sense. How are these notions of movement reconciled? First off, classically, if we had a particle in a $1/r$ potential and released it from rest, it would indeed bob ...


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: ...


6

The orbitals, which recently have been observed for the hydrogen atom, are probability distributions. These probability orbital distributions have been calculated using quantum mechanical solutions of the Schrodinger equation which give the wave function, and the square of the wave function is the probability distribution for finding the electron at that ...


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 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 ...


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 time-independent Schrödinger equation for the hydrogen atom is $$-\frac{\hbar^2}{2m}\vec \nabla^2\psi-\frac{e^2}{4\pi \epsilon_0r}\psi=E\psi $$ If your aim is just to verify that the $1s$-wave function $$\psi_{100}=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}\hspace{2cm} a\equiv \frac{4\pi\epsilon_0\hbar^2}{me^2} $$ is indeed an eigenfunction, then your task ...


4

From the Virial Theorem we can say the total energy of the atom is propotional to the potential energy of the atom. The potential energy is given by a coulomb potential and so is it will be roughly proportional to $\frac{1}{\langle r \rangle}$ where $\langle r \rangle$ is the mean radius of the electron's orbital. For a hydrogen atom the energy $E\propto ...


3

$r=n^2a_0$ for the Bohr model. Similarly, in the Schrodinger model, the most probable radius is $r=n^2a_0$, when $n=l+1$


3

Yes, the mean radius grows as $n^2$ asymptotically, but to have $1-10^{-20}$ for sure you have to add some more ;-). Highly excited atoms are called Rydberg atoms and are dealt with in experiment.


3

The potential energy function is the same for both. The energy level solutions are the same for both. The key difference is that in (most modern interpretations of) the Schrodinger model the electron of a one-electron atom, rather than traveling in fixed orbits about the nucleus, has a probablity distribution permitting the electron to be at almost all ...


3

The electron is the lightest lepton and the proton is the lightest baryon, so it's hard to see what reaction could occur without violating lepton number or baryon number. I suppose if proton decay (to a pion and positron) occurs then there could be a reaction to give a pion and two photons.


3

The symmetries that you're missing are conservation of baryon number $B$ and lepton number $L$. We strongly suspect that baryon number is not an exact symmetry, because the universe appears to contain very many baryons and very few antibaryons. Actually, a better metric for the baryon asymmetry of the universe is to compare the baryon density to the density ...


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

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

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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