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23

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 ...


10

In a neutral hydrogen atom the ground state has the electron and proton spins anti-parallel i.e. lined up with each other but pointing in opposite directions. The state with the spins parallel and pointing in the same direction has a slightly higher energy, and transitions between these two states produce the notorious 21cm hydrogen line. Since the ...


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

The degeneracy of energy-levels can be traced to the fact that the hydrogen atom possesses an enhanced $SO(4)$ symmetry caused by (among other things) the conservation of the Laplace-Runge-Lenz vector operator, see e.g. this Phys.SE post and Ref. 1. References: G. 't Hooft, Introduction to Lie Groups in Physics, lecture notes, chapter 9. The pdf file is ...


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 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

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 ...


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 ...


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 ...


3

What you're trying to do is called radiometric calibration. The problem with doing it with an ordinary incandescent light bulb is that the bulb itself would have to be radiometrically calibrated to get a precise spectral calibration of the detector. That's because the bulb's spectral emissivity deviates from an ideal blackbody source with an emissivity ...


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

$r=n^2a_0$ 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

These relations are based on the fact that both the position and the momentum distributions are centred around zero, which is in turn due to the symmetry of the atom. Given that, the width of the position and momentum distributions ($\Delta x$ and $\Delta p$) is of the same order as a typical position or momentum within those distributions ($r$ and $p$).


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 ...


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 ...


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

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 ...


2

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 ...


2

If you assume perfect efficiency, then the energy of dissociation of a liter of water is computed as follows: 1 liter of water, molar mass 18g, => 55.6 moles The energy needed is 237 kJ per mole (from your link - see under "thermodynamics"). 237 * 55.6 => 31.7 MJ of energy for a liter of water. In terms of power, this is 3.67 kW for one hour. This ...


2

Tyson claims that an electron disappears from one orbital and appears in another and claims that this is like going from the second floor of a building to the fourth floor without existing in between. This doesn't actually happen. What happens instead is that each possible state of the system has a continuous amplitude associated with it. In a transition ...


2

To add to John Rennie's answer and Hennes's Answer, an answer to your last sentence: How will the aliens know what this means? is well given in a thoroughly interesting and enjoyable tale by Carl Sagan's novel Contact. I can't recall whether the following is made clear in the film with Jodie Foster. In the novel, signals received from the Vega ...


2

No, the difference in lifetimes shouldn't be due to energy difference. For one thing, these energy differences are nearly zero. The states of a given $n$ in a hydrogen atom should be nearly degenerate regardless of $l$-value. (I say "nearly" since there is some energy splitting due to the Lamb shift, spin-orbit coupling, etc., but these are expected to be ...


2

The shortest and correct answer: this degeneracy is determined by the symmetry of the system. The case of degeneracy in hydrogen atom is so-called "accidental degeneracy", when eigenfunctions belonging to different irreducible representations of the symmetry group of a Hamiltonian correspond to the same energy. This type of degeneracy can also occurs in ...



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