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25

The Sommerfeld model, and the Bohr model from which it is derived, are toy models developed in an attempt to describe spectral lines in the era before modern quantum mechanics. You might be interested to look at the question Is it possible to recover the old Bohr-Sommerfeld model from the QM description of the atom by turning off some parameters? for more on ...


21

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 ...


13

Further to John Rennie's answer, which I wholeheartedly agree with (insofar that the planetary orbit picture is outdated by nearly a century), you might like find the following interesting. If the electron really were like a planetary orbit, then it would generally be an ellipse rather than a circle. The force field has the same functional form as the ...


9

The number of atoms (or molecules) in a body is given by Avogadro's constant, or $6.022 \times 10^{23}$ per mole. A mole is the amount of material, in grams, equal to the atomic or molecular mass of the substance in question. For example, for water ($H_2O$), 1 mole equals 18 grams. To get this number, remember that hydrogen ($H$) has an atomic mass of ...


5

Why do electrons in an atom occupy only the stationary states? This isn't true. An electron in an atom can be in any superposition of states. This is one of the basic postulates of quantum mechanics: linearity. For example, say an atom has a ground state 1 and an excited state 2, and let's say we're able to prepare it in a pure state 2. It will decay ...


5

Firstly the Bohr model involves fixed orbits, not probablity distributions. Also, the Bohr model does not consider sublevels (such as 2s versus 2p). The concept of probabilty distribution is an interpretation of the Schrodinger equation wavefunction solutions. The 1s orbital has zero nodes, 2s has one node, 3s two nodes, etc. The 2p subshell has a ...


5

The OP acknowledges that $m_e/m_p$ is constant and correctly points out that the fine sturcture energy levels of the hydrogen atom are proportional to $m_e$ (as opposed to $m_e/m_p$). It is also true that the energy levels of the Lamb Shift are proportional to $m_e$. However, the hyperfine structure of the hydrogen atom is proportional to $m_e/m_p$. ...


4

I would guess the book is using the virial theorem, which states that for a stable system: $$ 2T = -V $$ This immediately gives us $T = kZe^2/2r$ and therefore the total energy is $T + V = -kZe^2/2r$.


4

The simple answer to find the average number of atoms/molecules per unit volume is.... N/V (average atoms or molecules/$m^3$ ) = density ($kg/m^3$) * 1000 / atomic(or molecular) mass * $N_a$ where $N_a$ is Avogadro's number (~$6 \times 10^{23}$) In general in solid or liquid the distance between the nuclei of atoms is approximately 1 Angstrom = $10^{-10}$ ...


3

From Maxwells equations we know that accelerated charged particles emit em waves. This can be seen from the electromagnetic wave equation, where a second derivative of the electric field is involved. If the second time derivative of the charge density is nonzero, also the second time derivative of the electric field is non vanishing and electromagnetic ...


3

That depends on what is meant by "solving" the atom. What Feynman probably is referring to is the usual atomic Hamiltonian, which is already an approximation from the field theoretic point of view (no strong forces, etc.). The main problem is electron-electron-interactions. If you have an atom with more than one electron, the interaction term between the ...


2

If you assume separability of the wave function, i.e., $\psi(\mathbf x)=u(x)v(y)w(z)$, you can solve the individual components separately: \begin{align} -\frac{\hbar^2}{2\mu}\frac{d^2u(x)}{dx^2}+V_1(x)u(x)&=E_1u(x)\\ -\frac{\hbar^2}{2\mu}\frac{d^2v(y)}{dy^2}+V_2(y)v(y)&=E_2v(y)\tag{1}\\ ...


2

Hopping and tunneling are often used as synonyms, but they are really very different terms with a fundamentally different basis. Tunneling is an inherently quantum-mechanical feature which means that a particle wave-function tends to overlap into it's energetically disallowed area which leads to a non-zero probability of finding it "where it should not be". ...


2

One reason we focus on energy eigenstates is that atoms spend almost all of their time in an energy eigenstate, and their spectrum is a result of transitions between them. Another reason is pedagogical: to peel back the onion one layer at a time. But before too long, many courses do include examples of systems that are not in an energy eigenstate. One ...


2

I think something like this really works.* Here's how mainstream physicists would describe it. Let's use particle-physics natural units ($c=\hbar=1$). Then we have just one dimensionful unit, usually the eV. Nothing is stopping you from using a different unit, the cV (short for "crazy eV"), where the conversion is 1cV = 1eV in this galaxy, but 1cV=0.99eV in ...


2

Technically the electron and proton are both orbiting the barycenter of the system, both in classical and quantum mechanics, just as in gravitational systems. You find the same dynamics for the system if you assume the proton and electron are moving independently about the barycenter, or if you convert to a one-body problem of a single "particle" with the ...


1

The short answer is: protons are much more (1800 times) massive than electrons. That makes them (approximately) the center of mass of the system, that's why electrons are the ones orbiting protons and not vice versa. The term 'orbiting', however, means something essentially quantum. It is the reason of the stability of the atom (electrons don't radiate ...


1

You ask "How can an electron change orbitals staying at same positions, if it can then it must be able to do it all the time, why not only at the time of deexcitation?" Why do you think the electron needs to change position to jump from one orbital to another? Orbitals are energy states of an electron not "position" states. An electron simply needs to ...


1

I will stress what the other answers also include. The Bohr model does not have orbitals, but orbits, similar to the planet orbits around the sun. It is the first successful effort to describe the experimental behavior, i.e. the photon spectra, that excited atoms emitted. It postulated steady orbits. A postulate is like an axiom in a theoretical model that ...


1

Interesting question. I would have thought that if you were aware of the exact number of energy states and the populations thereof, you could apply boltzmann statistics to each of the levels in order to fit an appropriate temperature to the population in each state. This temperature, if comparable amongst the included levels, would therefore require that the ...


1

One of the problems with Bohr's theory to describe the hydrogen atom, was that the electron orbiting around the nucleus has an acceleration. Therefore it radiates and loses energy, until it would collapse with the nucleus. This is a common error in physicist's history of physics. The problem was not with Bohr's model, but (as Bohr thought) with ...



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