17

Now that's a mildly non-trivial observation. Why would they be equal, really? Let's say a particle with mass $m$, charge $q$, and initial velocity $v$ enters an area of length $L$ where an electric field $E$ starts to deflect it sideways. This is a clear example of uniformly accelerated motion, and its laws are well known: $x=vt,\;y={at^2\over2}$, where the ...


15

This is an example of terminology which should be taken with a grain of salt. The term "effective nuclear charge" is often casually symbolized with $Z_\mathrm{eff}.$ This is a universally accepted simplification, but you should keep in mind that the effective nuclear charge is, strictly speaking, $Z_\mathrm{eff}e,$ where $e$ is the elementary ...


13

This is a very fundamental question and for really understanding the "why" some advanced physics is involved. I will describe the process rather superficially. As you might know, the level energies of atoms and molecules can be calculated (in principle) using quantum mechanics. The simplest system is the hydrogen atom as it consists of a single ...


10

I'm not aware of the Russell–Saunders effect, but the Russell–Saunders coupling scheme is definitely a thing. As you noted, the Wikipedia page on "spin-orbit interaction" doesn't talk about it, but a different Wikipedia page does, and basically tells you the same thing as I will. The answer is... yes and no. The word "coupling" refers to ...


10

It is very tempting (and often also very useful!) to picture electron spin as an angular momentum vector, similar to a spinning top. Using this analogy, there are two properties (or numbers) of this angular momentum vector that we need in order to describe the electron spin. The first one is the spin itself and this is often designated the symbol $s$. The ...


9

An orbital of current quantum models of atoms and molecules has 3 major interpretations: orbital(1) as a wave function, being a solution of the quantum wave equation describing behaviour of electrons around an atomic kernels. Functions do not have energy, even if they describe it. orbital(2) is the 3D region describing probability of occurance of an ...


9

I'd like to copy the answer by John Rennie to a similar question at Physics.SE, since it's much better than current answers here IMHO. Although it mostly speaks about binding energy, the same principle applies to the quanta of excitation energy. The mass of a hydrogen atom is $1.67353270 \times 10^{-27}$ kg. If you add the masses of a proton and electron ...


7

transported electrons in $\ce{NADH}$ have a higher energy than those in $\ce{FADH2}.$ This is jargon describing the redox potential of the electron carrier $\ce{NADH}/\ce{NAD+}$ vs the electron carrier $\ce{FADH2}/\ce{FAD}.$ So if you compare the two reduction half reactions $$ \begin{align} \ce{NAD+ + H+ + 2e- &-> NADH} &\quad E^{\circ '}_\...


6

Here's what I believe they're trying to say. First, note that the discussion is mostly limited to high spin complexes (3 unpaired electrons out of the 7 d electrons), so the figure you reproduced is only representative of high spin states. Second, recall that when D > 0 (ie left side of figure), the five d orbital energy levels split so that we have three ...


6

There are no spatial barriers, much like there are no sky-high blue walls along all meridians and parallels, even though the map of the world might have convinced you otherwise. There is, however, the energy barrier. You can't skip that by tunneling. The tunneling is about getting to the state of equal energy when getting there seems to be topologically ...


6

Does electron mass decrease when it changes its orbit? Essentially yes. If you add the mass of a free proton and a free electron you'll get a greater mass than that of a hydrogen atom. The mass difference will be equivalent to 13.6 eV which is the ionization energy of hydrogen. Now for any "practical" chemistry experiment the assumption is that ...


5

Be gentle with me, I am going to try to explain orbitals in a very simple way using many similes. I do this with primary school kids so many of the words I will use aren't literally accurate but I hope the overall picture is as accurate as possible (without knowing quantum math). Please fell free to make suggestions where the overall picture deviates from &...


5

In QM, properties that can be measured (such as energy, spin, position, etc) are each associated with an "operator". Every observation will always return an eigenvalue for the relevant operator. Eigen values are values for $\lambda$ for which the equation $$\hat{O}\Psi = \lambda\Psi$$ is true, where $\hat{O}$ is the operator of interest and $\Psi$ ...


5

Very nice question. I also had the same question. These orbitals are nothing more than functions (wavefunctions). They are solutions to Schrodinger's equation, a second order differential equation. So you solve the equation and the solutions you get are determined by $4$ numbers also called quantum numbers $n, l, m_l, m_s$. A shell consists of all orbitals ...


5

An electron is not a wave nor a particle. An electron is usually described as a quantum object with some wave-like properties and some particle-like properties. Some of them, like particularly a spin, do not have direct counterpart in our familiar macro world. It is a kind of a mysterious, specifically quantum property of all elementary particles and atomic ...


5

Single electron systems I believe that particularly high values of n[V] are unlikely because as you pack more and more electrons in the same region, their repulsion would start to overpower their attraction to the nuclei, meaning the electron density would likely spread out more. The main constraint on high electron density is not the repulsion between ...


5

Short Answer: There is a limit to the maximum value of electron density $\rho_e(\vec{r})$. Long Answer: It is true that electrons repel each other, so as the electron density increases, the interelectronic repulsion also increases. This means that you can find high electron density only if the nucleus-electron attraction is stronger than the e-e repulsion. ...


5

Just to be sure: Note that orbitals(*) themselves have neither spin, neither charge. They are not real objects, but theoretical constructs of quantum atom models, that fit well the observations. Every atom has theoretically infinite number of orbitals ( 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f etc), all but some unoccupied. For most scenarious, it is practical ...


5

Pauli's exclusion principle is a consequence of electrons being indistinguishable fermions. Fundamental particles are indistinguishable in that two of the same type differ only in a few properties but otherwise behave identically (compare that say to two apples - they are always in principle distinguishable because they differ in so many ways). Fermions ...


5

Many things are only stable in their lowest energy state: electrons are no different Hold a ball in your hand. It is, in effect, in an excited state. Open your hand and the ball falls to the floor, without much effort or any push. Set the ball on the floor and it doesn't move. It is in its lowest energy state and won't move around unless given a push. Many ...


4

If this is a serious question and you wish to get a conceptual grasp of it, there is a lovely thin book called "Hydrogen" by Rigden (from Harvard). It walks you through the conceptual development. Briefly, the need of better and better model of hydrogen atom arose because of the need to completely rationalize the experimental hydrogen spectrum. By ...


4

I think part of your confusion stems from neglecting the fact that the electrons in a single molecule are all entangled with each other, so while an individual electron can occupy many states simultaneously, in each of those states it is entangled with all the other electrons such that the sum of states remains the same. To start with a simple example, ...


4

Thomson described the mechanical effect produced by cathodic rays on the first edition of its book Conduction of Electricity Through Gases (1903, pp. 501–502) and again later on the second edition (1906, pp. 629-630). As you mentioned it refers to the Crookes experiment with the paddle wheel in which the electron beam collides with the mill coausing it to ...


4

My question is, since are looking at the equivalence point in a titration, why can't we use $N_\text{acid} \times \text{Volume}_\text{acid} = N_\text{base} \times \text{Volume}_\text{base}$, since equivalents of acids and bases are the same at the equivalence point? If we can, how would we do it? First of all, I would suggest that you stick to the original ...


4

I can't present you a movie of electron interactions, but maybe a fuzzy picture. Imagine a metal body (start with copper): the atoms are held in a solid structure - in a sea of electrons! Those loose electrons can be pushed around easily, so we can make wires out of copper and transmit electricity thru them. If you put two different metals together, the sea &...


4

@KarstenTheis made it clear by using E0 values of each half reaction. Here I describe what they actually mean by using the phrase "high energy electrons" in biochemistry texts. Consider Ethyne and Ethane. Therefore Ethynide has a high energy electron compared to Ethanide Another Example : H- , OH- , SH- . H- is stronger than OH- and OH- is ...


4

The Bohr Model tried, quite successfully for its time, to model the energy states of an electron. This model has turned out inadequate, as it cannot answer question like yours. There have been more refined models (where your question isn't possible), but they all have their drawbacks. Not an answer to what you were asking, but some kind matching your ...


3

Slater's rules are an attempt to lump the effect of all other electrons on the wavefunction, and thereby other properties such as energy, of an electron (described by a hydrogen-like wavefunction). The effect of electron-electron repulsion is modeled indirectly by saying that inner shell electrons effectively screen the attractive nuclear charge sensed by ...


3

The half-equations are not correctly written. There is no Oxygen atom released, and no H atom emitted, as the author proposes. And the cathode does not get 4 electrons, as he or she states. The correct half-equations should be, first at the anode : $$\ce{Pb + SO_4^{2-} -> PbSO4 + 2 e^-}$$ And at the cathode it is : $$\ce{PbO_2 + 4H+ + SO_4^{2-} + 2 e^- ...


3

Isolated in a vacuum, in the absence of external fields, the first configuration is correct - $[\ce{Ar}]\mathrm{(3d)^6(4s)^1}$ is the ground electronic state of the iron(I) cation $\ce{Fe^{+}}$. More specifically, the ground state also has the term symbol $\mathrm{^6D_{9/2}}$. NIST's Atomic Spectra Database has compiled the ground state electronic ...


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