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19

There is a rigorous formal analysis which lets you do this. The true problem, of course allows both the proton and the electron to move. The corresponding Schrödinger equation thus has the coordinates of both as variables. To simplify things, one usually transforms those variables to the relative separation and the centre-of-mass position. It turns out that ...

12

I assume you're talking of the hydrogen atom; the hamiltonian of the nucleus + electron system is $$H = \frac{p_e^2}{2 m _e} + \frac{p_n^2}{2 m _n} - \frac{e^2}{|r_e - r_n|}.$$ You can do a change of coordinates (center of mass coordinates) $$\vec{R} = \frac{m_e \vec{r}_e + m_n \vec{r}_n}{m_e+m_n} \\ \vec{r} = r_e -r_n$$ and find the conjugate momenta to ...

9

Bohr postulated that electrons orbit the nucleus in discrete energy levels, and electrons can gain and lose energy by jumping between energy levels, giving off radiation of frequency $\nu$ according to the formula $\Delta E = E_2 - E_1 = h\nu$ where $\nu = \frac{1}{T}$, where T is the period of orbit, as in classical mechanics. Now during the transition, ...

7

The problem with attempting to fuse two protons is that there is no bound state $^2$He, for the rather obvious reason that there are no neutrons present to hold the two protons together. The fusion of two protons requires one of them to undergo beta plus decay while the two protons are close, and the probability of this is vanishingly small. It happens in ...

7

The Hamiltonian for the hydrogen atom $$H = \frac{\mathbf{p}^2}{2m} - \frac{k}{r}$$ describes an electron in a central $1/r$ potential. This has the same form as the Kepler problem, and the symmetries are similar. There is an obvious $SO(3)$ generated by the angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. In other words, the components of ...

6

There is no integration of the radial part because, as you said yourself, we want the probability of finding the electron somewhere in the spherical shell between $r$ and $r+dr$ from the nucleus. (in a differential shell between $r$ and $r+dr$, and no need to integrate over $r$.)

6

With a potential $V(x) = - \frac{\alpha}{|x|}$, with the notation $a = \large \frac{\hbar^2}{m \alpha}$, solutions are : $$u^+_n(x,t) \sim x e^{ - \large \frac{x}{na}} ~L_{n -1}^1(\frac{2x }{na}) e^{ -\frac{1}{\hbar} \large E_nt}~~for~~ x>0$$ $$u^+_n(x,t) = 0~for~~ x\le0$$ and : $$u^-_n(x,t) \sim x e^{ + \large \frac{x}{na}} ~L_{n -1}^1(\frac{2x ... 4 There are two reasons: First, the expansion of space, which was rapid in the early universe, separated the initial density fluctuations into isolated potential wells. Dark matter and ordinary matter then accumulated into these local potential wells, which eventually become galaxy clusters. Second, the temperature of the early universe was very high, so ... 4 No, the radial parts of the wavefunctions are not orthogonal, at least not quite to that extent. The radial components are built out of Laguerre polynomials, whose orthogonality only holds when leaving the secondary index fixed (the \ell or 2\ell+1 or whatever depending on your convention). That is,$$ \langle R_{n'\ell} \vert R_{n\ell} \rangle \equiv ...

4

When you use the reduced mass, what you have first done is to go from the variables $(r_1,p_1)$ and $(r_2,p_2)$ to $(r=r_1-r_2, p/\mu=p_1/m_1-p_2/m_2)$ and $(M R=m_1 r_1+m_2 r_2,P=p_1+p_2)$, where $M=m_1+m_2$ is the total mass and $1/\mu=1/m_1+1/m_2$ is the (inverse of the) reduced mass. As you said, this is usually introduced in classical mechanics to ...

4

The approximation that we all started out learning is the linear combination of atomic orbitals (LCAO) approach. The molecular wavefunction, $\Psi$, can be expressed as a sum of some set of basis functions: $$\Psi(\vec{r}) = \sum_n f_n(\vec{r})$$ and a convenient set of basis functions is the atomic orbitals of hydrogen. As a starting point we could take ...

4

With regards your first question: A similar (the same?) question you might reasonably ask is: how can we assume that the proton is stationary, at the centre of the problem, since it is surely going to be attracted by the electron and jiggle about a little? This is a question that would be just as valid directed at a classical system --- say, a planet ...

3

$P(r)dr$ gives you only the probability in an infinitesimal spherical shell around the center. The integration you're expecting is made, when you want to know the probability in a non-infinitesimal shell around the center. For example, you'd like to know what is the probability of finding an electron between $r=1$ and $r=2$ (in whatever coordinates), you'd ...

3

What happens, essentially, is that the S and P wavefunctions get mixed to produce eigenstates that have shifted centres. This means the atom gets an induced electric dipole moment, whose interaction with the external field either lowers or raises the eigenenergy. More specifically, consider the wavefunctions of the states $|200\rangle$ and $|210\rangle$: ...

3

To get it in the momentum representation, one has to do the Fourier transform of this function. This reference can be useful: http://forum.sci.ccny.cuny.edu/Members/lombardi/publications/MOMREP-H-atom.pdf/view At the end, separation of variables after transformation to the momentum space is not trivial, and the mixing of quantum number is presented.

2

A hydrogen atom ion $H^{+}$, with an atomic mass number of A=1, charge number Z=1, is the same as a proton. A hydrogen ion thus usually just refers to a proton. Depending on context, however, you may also have a hydrogen ion which is (a) an ion of a deuterium atom, in which case it is a bound state of a neutron and a proton, with atomic mass number A=2, ...

2

I had a look at the paper, and I think the author means that the energy for the reaction: $$H + H^+ \rightarrow H_2^+$$ is negative i.e. the ground state energy of $H_2^+$ is less than the sum of the ground state energies of $H$ and $H^+$. The reason for this is simply the observation that the $H_2^+$ ion is stable. If the energy of $H_2^+$ were higher ...

2

There is an excellent discussion of this at Pauli principle for particles very far apart from each other. The question isn't a duplicate of yours, so I haven't flagged your question as a duplicate, but Wouter's answer is highly relevant. People have a tendancy to casually throw around the exclusion principle with hand waving arguments such as "when you ...

2

The wavefunction squared (strictly speaking $\psi^*\psi$, but this is equal to $\psi^2$ if the wavefunction is real) gives you the probability density of finding the electron. The probability of finding the electron in some infinitesimal volume $dV$ is given by the probability density times the volume: $$P = \psi^2 dV$$ So if you divide space up into ...

2

Just to make sure we're all on the same page, I'll take a step back before taking two steps forward. The normalized wavefunctions for the $n = 2$ energy level are usually written as \begin{gather} \psi_{2,1,0}(r, \theta, \phi) = \left(\frac{1}{32\pi a_0^3}\right)^{1/2} \frac{r}{a_0} \mathrm{e}^{-r/2a_0} \cos(\theta), \\ \psi_{2,1,\pm1}(r, \theta, \phi) = \mp ...

2

This might not fully answer your question, but maybe it will be a good start. Things to consider Thermal energy received by Jupiter from the sun Thermal energy radiated by Jupiter (hence, net thermal energy) Jupiter's composition Jupiter's temperature Jupiter's gravitation Hydrogen's thermal properties (among other properties) For the first 2 items to ...

2

It's not clear to me exactly why you're unhappy with the answer you get. I would suggest phrasing your expectations in terms of the total charge contained in a sphere of radius $r$, $$4\pi\int_{0^-}^r \rho(r')r'^2\,\text dr'.$$ This should give the positive charge of the nucleus at $r\rightarrow 0^+$ (because the proton is point-sized in this model!) and ...

2

Pulsar and Chris White have given nice explanations of why the dynamics of the early universe would not lead to the formation of one big starlike object. There is also another, much more generic argument against such a process, which boils down to the existence of cosmological horizons. At any given time in the evolution of the universe, there have been ...

2

I'm not sure what you mean by "collapse", but if I interpret that as "no hydrogen is formed" or "the electron is not captured", then 2 things can happen: 1) Elastic electron-proton scattering: the electron and proton just "bounce" off each other under some angle theta. By observing the cross section of the scattering versus the theta angle it was shown that ...

2

It's because there is another vector quantity $A_i$ conserved in addition to the angular momentum $L_i$. Furthermore, the commutation relations of $A_i$'s and $L_i$'s are those of $SO(4)$. See for instance this reference : http://hep.uchicago.edu/~rosner/p342/projs/weinberg.pdf

1

As dmckee commented, angular momentum is barely mentioned in Bohr's revolutionary 1913 paper "On the Constitution of Atoms and Molecules" (Philos. Mag. 26 , 1). Instead, Bohr's bases his argument on Planck's hypothesis that the radiation from a quantum harmonic oscillator "takes place in distinctly separated emissions, the amount of energy radiated out from ...

1

The Strömgren sphere is defined as the sphere bounding a region that, if fully ionized, would undergo recombination at a rate equal to the rate of ionization. Calculating the radius is much simpler than Wikipedia makes it out to be, and it's worthwhile to figure this out on your own rather than look up the formula. Fundamentally, this is just balancing ...

1

The state ket of a system has encoded all the information that can be known about that system. In which way? Well, in its linear decomposition on the eigenstates that form a basis in the Hilbert space you are working at. Your particular ket is represented as a linear combination of 4 kets who have nonzero coefficients. All of those kets represent a ...

1

Suppose you have some wavefunction $\Psi$ that is a linear combination of eigenfunctions, $\psi$: $$\Psi = a_1\psi_1 + a_2\psi_2 + a_3\psi_3 + ...$$ You know the eigenfunctions are orthonormal, so $\langle\psi_i|\psi_j\rangle$ is zero if $i \ne j$ and 1 if $i = j$. Suppose you compute $\langle\psi_i|\Psi\rangle$: \begin{align} ...

1

In your 3-orbital example, you are ignoring the fact that you can get degenerate, linearly independent states. Suppose that you have three orbitals $s_1, s_2,$ and $s_3$. Excluding normalization, you can form the following linearly independent combinations $s_1 + s_2 + s_3$, $s_1 - s_2 - s_3$, and $s_1 - s_2 + s_3$. The last two combinations are degenerate ...

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