New answers tagged

1

Electrons and nuclei both have spin. A spinning charged particle has a magnetic dipole moment. When a magnetic dipole is in a magnetic field, it experiences a force. This oversimplified description gives some brief intuition on the origin of the fine and hyperfine splittings. Fine structure is due to the interaction of the electron's spin with the ...


0

This was one of the questions raised during the early stages of QM. If this were to happen atom would not exists as the electron would spiral towards the nucleus. Interference effects at the atomic level are usually assumed to give rise to the stationary states as we know them. The picture of electron actually going round the nucleus has been replaced by the ...


1

The problem with this question is all of the assumptions that go into it. When we're taught physics, we are given analogies that help our understanding, but mislead us when we try to dig deeper. Firstly, charged particles like electrons are always surrounded by an electromagnetic field. Changes in that field propagate through space at the speed of light and ...


0

If you have a Poisson distribution with a probability $$P(n)=\frac{\lambda^n}{n!}e^{-\lambda}$$ that there will be $n$ events per bin, then $\lambda$ is the mean number of events per bin. You can get this via a direct calculation, $$ ⟨n⟩ =\sum_{n=0}^\infty nP(n) =\sum_{n=0}^\infty n \frac{\lambda^n}{n!}e^{-\lambda} =\lambda e^{-\lambda}\sum_{n=1}^\infty \...


3

The atomic states win in a landslide. The reason is that the coulomb force is long ranged while the nuclear force is short ranged. That means that the number od discrete nuclear states is finite while the number of discrete atomic states is infinite. Even for a neutral atom where you might think that cancellation of charges would shorten the interaction ...


3

It's simply a Fourier transform relationship. If an atom is in a metastable state, with the latter coupled to the electromagnetic field, then it is fairly straightforward to show that the probability amplitude for decay to happen in the time interval $[t,\,t+\mathrm{d}t]$ is: $$\psi(t) = \left\{\begin{array}{ll}\frac{1}{\sqrt{\tau}}\,\exp\left(i\,\omega_0\,...


7

There are two separate issues here (not sure which of the two you mean): The first problem is a severe misconception that is similar to Zeno's paradox of Achilles and the Tortoise: Given a hydrogen atom we have (in principle) an infinite number of shells. However, the gap between the shells gets smaller and smaller. If you would jump from shell to shell, ...


36

An animation is worth a million words:


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This will be a purely mathematical treatment. It needs to be combined with some practical playing around to really "get" it. Traveling wave Let's start with the description of a harmonic traveling wave in one-dimension. Here "harmonic" just means the mathematical form of the wave is sinusiodal in both time and space. For concreteness we'll using talk ...


5

how is a standing wave related to the atomic orbit (It is my understanding that the atomic orbit is both a mathematical function that describes the probability of an electron being at a certain place, but it is also the image of this function in terms of real space, i.e. the actual 3 dimensional volume around the nucleus that a particular electron calls "...


2

A standing wave is basically two opposing waves of equal amplitude, as shown in the diagram below (where n is a positive integer): You can see this more clearly if you look at the top line where n=3, and follow it as it goes down, up, down. That's wave one. Then, if you look at the bottom line in the same case as it goes up, down, up, that is the opposing ...


0

Since there are 2$p$ electrons $\ell_1 = 1, \ell_2 = 1 \rightarrow L=0,1,2$ and $s_1=1/2, s_2=1/2 \rightarrow S=0,1$. As you said, ignoring the Pauli principle results in $^3\text{D}, ^1\text{D},^3\text{P},^1\text{P},^3\text{S}$, and $^1\text{S}$ terms. The easiest way to impose the Pauli principle is to make a table as below where the first row and column ...


6

I actually think I have figured it out: if we consider the situation when the principle quantum number for, let's say, excited helium is different, then we can have spin-up electron in $1S_1$ state and spin-up electron in $2S_1$ state. This is going to be a triplet. Please, correct me if I am wrong.


0

Bohr's model is only valid for atoms with a single electron, as it neglects any electron-electron interaction. One might think that the only atom with a single electron is hydrogen. However, ionized helium ($\mathrm{He}^+$) and doubly ionized lithium ($\mathrm{Li}^{2+}$) also fulfill this condition. (Similarly for even heavier atoms...) So in your exercise, ...


0

In reality, you need the quantization field to prepare your atom in a given spin state and have it stay in that state- at zero field, any small perturbation could change it. But in a thought experiment that is not a problem. Still, you have to specify your atom as starting in some initial spin, and if it is starting as usual in an eigenstate of $S_z,$ you ...


1

This is a very good book which covers atomic spectra and Laser Physics, with good Physical intuition and fairly rigorous Mathematical proofs. https://www.amazon.com/Atomic-Physics-Oxford-Master-Optical/dp/0198506961


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What we intuitively think of as "solid objects" are actually electromagnetic force-fields repelling each other. So you are correct; atoms are 'empty' in that they contain no solid objects or things. On the other hand, they are 'full' of basic force field which, in the aggregate, on a macro-scale, creates the illusion of 'solidity' that is what we perceive to ...


0

A good reference for choosing appropriate laser beam for a given type of atom, plz read Michal J Martin's PhD thesis at jila, Prof. Ye Jun's group site. Some moderator would not be happy with links so I delete it. Chapter 2 & 4 all provides condensed materials about laser wavelength.


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Because of the Pauli exclusion principle, it's extremely difficult to compress atomic matter beyond a certain density. It's not impossible, because there are always higher-energy electron states available, but there's a very strong force opposing it (called electron degeneracy pressure). This is what it means for space to be full. If you define "empty space"...


3

A charged particle like electron maybe is point-like (of radius zero), but it is "long-handed" as it is "felt" far away. In this sense it is not so "point-like".


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Although it's commonly said that fundamental particles are point particles you need to be clear what this means. To measure the size of the particle to within some experimental error $d$ requires the use of a probe with a wavelength of $\lambda=d$ or less i.e. with an energy of greater than around $hc/\lambda$. When we say particles are pointlike we mean ...


10

Yes, elementary particles such as electrons and quarks (inside protons) are point-like or at least, their internal structure is incomparably smaller than the size of the atom. So the atom is mostly empty space. However, that doesn't mean that atoms may penetrate each other. Matter is impenetrable because of a combination of the uncertainty principle that ...


5

Regularly spaced light and dark bands far from either anode suggests a simple matter of the electrons picking up the kinetic energy necessary to excite the neutral atoms, giving most of it up when they do so, and then re-accelerating down the tube. We would expect a couple of diagnostic features: The bands nearer the cathode to be more sharply defined ...


2

There are five relevant quadrupole moment operators, and when labeled by the change $\Delta m$ in angular momentum projection they read $$ \begin{array}{c|ccccc} \Delta m & -2 & -1 & 0 & 1 & 2\\ \hat Q_{2m}& (x-iy)^2 & (x-iy)z & x^2+y^2-2z^2 & (x+iy)z & (x+iy)^2 \end{array} $$ These operators arise as (a basis for) all ...


2

The statement of Martin above: Now, can we "see" atoms? This depends, as I already hinted at, what you mean by "see". If you mean "make a picture in visible light", then you can't do that. is actually not quite true. One can take images using visible light that show single atoms. Here is an example: (1) The reason this works is that this is a ...


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This is an image of a Sc2O3 nanocrystal obtained from an abberation corrected scanning transmission electron microscope. The left image is recorded by measuring only electrons that have been bent/deflected by passing through the material (in this case we dont see the oxygen atoms very well) The image on the right measures all the electrons that pass ...


2

In addition to the answer by hsinghal it is worth point out some historical notational quirks. An expression such as $^2P_{3/2}$ is called a Term Symbol. The superscript is the multiplicity of the electron spins, i.e. 2$S$+1 for total spin S. The capital letter, P in this example is the total orbital angular momentum and has letters and values of S=0, P=1, D=...


1

As you know that zeeman splitting is due to the phenomena known as spatial quantization. i.e. if there is a fixed or preferred direction in the space (i.e. symmetry of the space is broken by the electric or magnetic field) then the atom can not assume arbitrary orientation. This orientation depends on the angular momentum of the atomic/spectroscopic state. ...


6

Electron capture Electron capture (K-electron capture, also K-capture, or L-electron capture, L-capture) is a process in which the proton-rich nucleus of an electrically neutral atom absorbs an inner atomic electron, usually from the K or L electron shell. This process thereby changes a nuclear proton to a neutron and simultaneously causes the emission ...


8

The decay of potassium-40 to argon-40 is either a $\beta^+$ decay in which what is emitted is not an electron but a positron $$ {}^{40}{\rm K} \to {}^{40}{\rm Ar} + e^+ + \nu_e $$ or, more frequently (if we have whole atoms), an electron capture that you mentioned in which no charged leptons are emitted at the end! About 11% of the potassium-10 decays ...


2

Please explain by what means electrons extraction can be done. Hot enough plasmas have all the electrons in the plasma leaving the nuclei positive. How person can focus activity on single atom (from precision point of view) to do so? One cannot deal with individual atoms. It is a statistical phenomenon and one can get a beam of ions without any ...



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