# Tag Info

1

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 $T + V = -kZe^2/2r$.

0

It might be possible, but it's uneconomical and probably also uses more electric energy than you get from the power plant. Do you really mean fisson? Fission is splitting a nucleus in two approximately equal parts. This is done with certain isotopes of e.g. uranium 235 which are only barely stable and sometimes fission spontaneously. Other isotopes such as ...

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

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

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 ...

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 ...

0

The Schroedinger model does much more than people here give it credit for. It doesn't just solve the problem of why the ground state doesn't radiate. It solves the problem of why and how the excited states radiate, and it does so by using nothing other than Maxwell's equations. In the Schroedinger model, the states that don't radiate are the states with a ...

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 ...

-5

the magnetic field lines are re aligning as the electron orbits the nucleus. however, even though the electron is accelerating as it orbits the nucleus, it simply does not have enough energy in its orbital to radiate until it reaches the next quantum amount of energy associated with the next orbital.

0

The whole field of Cold Fusion is now viewed by many scientists as pathological science. If such a thing does exist and produces non-trivial amounts of heat then the only way it is going to be validated is if/when units are commercially available for power generation. By this, I do not mean Rossi style demonstrations. I mean units that engineers can take ...

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}\\ ...

0

The problem is that you do two different things with the two methods. Method 1 gives you the (uniteresting) length of the combined mangetic moment vector while Method 2 gived you its expectation value in the quantization direction which is $\vec{J}=\vec{L}+\vec{S}$. $\vec{\mu}_J$ does obviously not point in the same direction as $\vec{J}$, because of the ...

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 ...

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 ...

0

This is an interesting question because you have to start dealing with the atom surrounded by its environment. Atoms with electrons in states other than the lowest possible filled-up states, have lots of opportunities. They can emit photons (and thereby go to a lower energy state) in many possible directions, and that photon can interact with any number of ...

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". ...

0

When "hopping", the particle has enough energy to surmount the potential barrier. Its like water molecules passing from liquid state to gas: only those who happen to have enough kinetic energy KE to escape the average bounding of the other water molecules. This can happen even in room temperature, since their KE follows a Boltzman distribution and it can be ...

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

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 ...

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 ...

0

edit after comments: We have the experimental observation of fixed spectra coming from specific atoms. The same for nuclei, and both are stable in their ground state ( unless energetically disturbed or are unstable isotopes). Quantum mechanical solutions reflect these experimental observations and the spectra of atoms and nuclei have been fitted with ...

0

The human eye, in dark-adapted conditions, has been shown to be capable of detecting a single photon (which is to say, the retina reacts, the signal gets to the brain, and [magic happens] to generate a response in the conscious parts of the brain. Since what you're asking about is the release of one photon, it doesn't matter what the source is. So long ...

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 ...

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 ...

4

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 ...

2

Various physical properties of materials including shear strength are connected to the strength of attractive forces between molecules -inter-molecular forces. The type of inter-molecular force determines the force and energy by which the molecules stick to one another. Inter-molecular force types include ionic bonds, hydrogen bonds, dipole forces and ...

0

This question has another interesting aspect which has more to do with neuroscience than physics: why do we perceive metals with a neutral colour (such as silver) as grey, even why they are shiny and therefore simply reflect the colours of their surroundings? One answer is that such metals always have some roughness and therefore scatter light from a range ...

7

The following passage has been extracted from Bohr's Nobel lecture: While in contradiction to the classical electromagnetic theory no radiation takes place from the atom in the stationary states themselves, a process of transition between stationary states can be accompanied by the emission of electromagnetic radiation, which will have the same ...

18

The sky does not skip over the green range of frequencies. The sky is green. Remove the scattered light from the Sun and the Moon and even the starlight, if you so wish, and you'll be left with something called airglow (check out the link, it's awesome, great pics, and nice explanation). Because the link does such a good job explaining airglow, I'll skip ...

3

Almost but not quite. Qualitatively the spectrum is the same with the $1/n^2$ spacing, but the scale of the spectrum is set by the reduced mass $\mu$, $$\mu = \frac{1}{\frac{1}{m_l}+ \frac{1}{m_p}}$$ where $m_p$ is the proton mass and $m_l$ is the lepton (muon or electron) mass. Since $m_p \approx 2000 m_e$, it is not a large error to take $\mu = m_e$ for an ...

19

Note well: What we perceive as color is bit of a tricky subject. This is a different question, one that has been asked and answered multiple times at this site. Per the typical human eye response, sunlight at the top of the atmosphere is about as "white" as "white" can be. Some of that incoming sunlight is reflected back into space, some is absorbed by the ...

11

You are missing nothing. The Bohr model of the atom is false, and nowadays we replace the idea of the semi-classical "orbit" of Bohr with the fully quantum mechanical notion of orbitals or electron clouds, which give a probability distribution for the position of the electron around the nucleus, but do emphatically not imply that the electron is moving in ...

8

The hand waving explanation in your question is called Rayleigh scattering Rayleigh scattering results from the electric polarizability of the particles. The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency. The particle therefore becomes a small radiating dipole whose radiation ...

5

The sun is technically green because the peak of its black body spectrum is near green wavelengths. When light scatters parallel to the plane of incidence (i.e., during the day time), it is blue-shifted. When light scatters perpendicular to the plane of incidence (i.e., sunset or sunrise), it is red-shifted. The light that is not scattered but makes it ...

3

I think your confusion is cleared up quite simply: you're confusing the terms "orbital" and "electron shell". An orbital is characterized by the three quantum numbers $n,\ell,m_\ell$. This terminology makes sense, because these three numbers together completely determine the spatial component of the wave function. However, this leaves a freedom in the spin ...

2

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 ...

2

Rather than write something unintelligible, I'll quote from a page on cesium clocks. According to quantum theory, atoms can only exist in certain discrete ("quantized") energy states depending on what orbits about their nuclei are occupied by their electrons. Different transitions are possible; those in question refer to a change in the electron and ...

11

It is a very interesting question that allows to point out the differences between a Neutron Star and Nuclei. Although the dedicated article in Wikipedia Neutron Star fully covers the information, it is relevant to summarize here the elements. Nuclei are essentially different to Neutron Stars and some reasons are: Different bounding force: while Nuclei ...

7

The nuclei of heavy elements (lead, gold, ...) approach the asymptotic density of extended nuclear matter (and therefore the density of neutron stars). The lighter elements do not. That said, it would be an error to refer to nuclei as "miniature neutron stars" because the binding force and dynamics are different. Nor are nuclei protected, shielded or held ...

2

If you had a single hydrogen atom, and you watched for a single transition, then yes, you would only see emission at a single frequency. There would be one line in your spectrograph so to speak. But rarely do you have just one atom. And quite often an atom undergoes multiple transitions while you are watching it. When looking at a large ensemble, whether it ...

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