Introduction to the quantum mechanical model of the atom: Thinking about electrons as probabilistic matter waves using the de Broglie wavelength, theSchrödingerequation, and the Heisenberg uncertainty principle. Electron spin and the Stern-Gerlach experiment.

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Keith Stone

8 years agoPosted 8 years ago. Direct link to Keith Stone's post “Why is the spin number ex...”

Why is the spin number expressed as +1/2 or -1/2, rather than just +1 or -1 ?

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(101 votes)

harsh shah

7 years agoPosted 7 years ago. Direct link to harsh shah's post “In the spin quantum numbe...”

In the spin quantum number the electrons are represented either by +1/2 or -1/2, and as shown in the quantum numbers video it is said that the electrons in this type, i.e the spin number can move in two directions ,one towards the left and one towards the right, so as electrons possess like charges(-ve) and because they might be travelling in the opposite directions and finally when they come close to each other they repel, so the electron almost covers 1/2 the circular orbit so probably that is why it is assigned the value +1/2 and -1/2.

(7 votes)

Aryan Sinha

8 years agoPosted 8 years ago. Direct link to Aryan Sinha's post “referring to the image on...”

referring to the image on orbital shapes,

why do the p,d,f orbitals have such "out of the blue" shapes? how were these shapes found out or calculated?•

(15 votes)

Shou Toushiro

8 years agoPosted 8 years ago. Direct link to Shou Toushiro's post “The shapes have been calc...”

The shapes have been calculated from Schrodinger's equation.

Moreover , the shapes represent the region where the probability of finding the electrons is maximum. The orbitals have NO physical BOUNDARY.

(29 votes)

黃岳涵

8 years agoPosted 8 years ago. Direct link to 黃岳涵's post “I've some questions on th...”

I've some questions on this statement:

The square of the wave function represents the probability of finding an electron in a given region within the atom.

Why function square related to probability?•

(16 votes)

Matt B

8 years agoPosted 8 years ago. Direct link to Matt B's post “It's mostly a function in...”

It's mostly a function in mathematics. It would be like asking "what is the average distance of a cosine wave from the center?" - and this might seem easy at first until you realize that the gradient of the wave function keeps changing. Mathematically, the average distance, or in this case the energy or probability of finding the electron, is just the square root of the wave property.

(17 votes)

Akshat Sahay

8 years agoPosted 8 years ago. Direct link to Akshat Sahay's post “So the formula for de Bro...”

So the formula for de Broglie equation says that

lambda= h/mv

and..the v of a stationary object will be 0,

so by mathematical solving it means that the wavelength of a stationary object is infinite?

is this the correct thinking or is there some other theory for a stationary object's wavelength?•

(13 votes)

oscarsimpson1999

8 years agoPosted 8 years ago. Direct link to oscarsimpson1999's post “Isn't to do with the fact...”

Isn't to do with the fact that the velocity is not quite 0? if you know it is exactly 0 then the uncertainty in the position is infinite as well (momentum is a function of velocity, so delta P = 0 -> delta V = 0 -> delta X = inf) therefore it has an equal probability of being anywhere. If you take the infinite wavelength interpretation, then it would be nearly 0 (1/inf) but constant everywhere. The square (probability function) shows that it has an equal chance of being anywhere.

So, if you know with 0 uncertainty what the velocity is, then you have no idea where it is, and all future involvement of the particle is pretty much irrelevant (how is the electron going to diffract around an atom if the electron is in a different galaxy?). As we can't physically measure to perfect accuracy, there is an uncertainty in both measurements of the degree that we know it's probably stationary and it's probably 'over there'. With this uncertainty, the velocity is almost definitely not 0. (1 value in a range of reals is like trying to throw a dart at a dartboard with an infinitely thin wire and hitting the wire). For another layer, you can take the fact that you can never cool anything down to exactly 0K (-273.15C) (although you can get close) and so nothing will ever have 0 velocity.

(22 votes)

clarinsun

8 years agoPosted 8 years ago. Direct link to clarinsun's post “In reference to Shrodinge...”

In reference to Shrodinger's Equation, what's the difference between the V (wave function symbol) on the left side of the equation and the right? There must be some difference, or else they would cancel.

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(5 votes)

Mark Zwald

8 years agoPosted 8 years ago. Direct link to Mark Zwald's post “You can't cancel out the ...”

You can't cancel out the ψ's like that because Hψ is not H times ψ but rather H operating on ψ (that's what the ^ over the H indicates).

The Hamiltonian operator H is actually

H = -ℏ²/2m * d²/dx² + U(x)

so the time independent Schrodinger equation is actually

-ℏ²/2m * d²ψ/dx² + U(x)ψ = Eψ

solving for E gives the allowable energy states of the system.(21 votes)

Guilherme Volpato

8 years agoPosted 8 years ago. Direct link to Guilherme Volpato's post “On Schrödinger's equation...”

On Schrödinger's equation, what exactly is the Hamiltonian operator?

How do we work with it mathematically and what would be its physical interpretation?•

(10 votes)

Samurai Warm

8 years agoPosted 8 years ago. Direct link to Samurai Warm's post “This has to do with Lapla...”

This has to do with Laplacian operator (partial derivative) of a wave function.

Here's the formula http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sch3d.html (I can't type out the partial derivative operator -__-")The h-bar in the equation is defined by h/(2pi)

(8 votes)

Nikitacina

8 years agoPosted 8 years ago. Direct link to Nikitacina's post “I have a really blurry pi...”

I have a really blurry picture of what it means for small particles to behave as both particles and waves. What does it exactly mean when we say that small particle have characteristics of both waves and particle ?

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(3 votes)

Satwik Pasani

8 years agoPosted 8 years ago. Direct link to Satwik Pasani's post “I think I can safely say ...”

I think I can safely say that nobody understands quantum mechanics.

- Feynman

So you are right on track! :)

But to be more pedagogical, you can view them as different behaviors in different situations. When behaving as particles, they show all what we "expect" them to based on our experiences. But under different conditions, they behave as waves, where in these "waves" are equations that determine, in a simple sense, the probability of the particle to be at a particular place at a particular time. Since these equations are not the classic Newtonian equations, their predictions often create the strange outcomes we usually see. (See the double slit experiment to understand the differing behaviors.)(12 votes)

mand4796

6 years agoPosted 6 years ago. Direct link to mand4796's post “Does anyone know exactly ...”

Does anyone know exactly what "spin" is? If a particle has "spin", it doesn't actually physically spin/turn/rotate, right?

In the article, it said that the opposite "spins" of electrons are determined by how the react in a magnetic field. But does this explain what it means for other particles - other leptons, quarks, hadrons - to have spin? What is spin? Please help!•

(2 votes)

Andrew M

6 years agoPosted 6 years ago. Direct link to Andrew M's post “Spin is just a property t...”

Spin is just a property that electrons (and other particles) have. It got named spin back when people were working with the Bohr model and trying to extend it to atoms beyond hydrogen. Just as Bohr imagined that the atoms were little planets revolving around the nucleus as though it were a sun, other scientists tried to extend that idea by imagining that the little planets, just like real planets, had spin. Using this concept actually helped a bit, just like Bohr's imagination of little orbits helped a bit, but again, the electrons are not like little planets: they don't revolve around the nucleus, and they don't rotate on their axes. Still, they have a property that works sort of like an orbit (orbitals) and they have a property that works sort of like rotation about an axis (spin). But there is no point in trying to answer "what is electron spin" by referring to some familiar object like a ball or a planet, because electrons are not balls or planets, they are their own thing, and you just have to accept that they have a property whose effects we understand very well, and we happen to have named it spin. "What is spin" doesn't have any deeper answer than does the question "what is charge". It's just a characteristic that particles have or don't have, and we know what the effect is of having or not having that property.

We have other properties like this for other particles. A quark can be an up quark or a down quark. "Up" and "down" don't have any meaning other than to identify the type of quark. THere's nothing "up-ish" about an up quark. THere's a top quark and a bottom quark. There's a charmed quark and a strange quark. There's nothing unusually strange about the strange quark. It's just the name that we've given a quark that has certain properties.

IN the field of quantum chromodynamics, we say that particles have a color. They don't really have a color, we just use that idea because it's sort of helpful, just like using the idea of positive and negative for charges is helpful.

(10 votes)

Dish*ta

3 years agoPosted 3 years ago. Direct link to Dish*ta's post “Do nodes of the 1s, 2s, 3...”

Do nodes of the 1s, 2s, 3s... subshells overlap?

(wrt to the graph shown above, it seems they do)

If yes, do the nodes of another subshell also overlap,[in fact, if the probability of finding an electron is 0, shouldn't the nodes always overlap(even if multiple subshells /orbitals pass through)?]

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(2 votes)

Charles LaCour

3 years agoPosted 3 years ago. Direct link to Charles LaCour's post “Atomic orbitals are defin...”

Atomic orbitals are defined by energy levels not positions. The electron clouds that are depicted are usually where there is a 90% chance of finding the electron at that energy level. The electron clouds for different energy levels do have overlaps.

(5 votes)

Gerry Dunda

8 years agoPosted 8 years ago. Direct link to Gerry Dunda's post “Can we specify type of th...”

Can we specify type of the wave in schrodinger wave? like transversal or longitudinal

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(3 votes)

Teacher Mackenzie (UK)

8 years agoPosted 8 years ago. Direct link to Teacher Mackenzie (UK)'s post “I never thought about it ...”

I never thought about it before. Its interesting question.

My guess is that it is transverse. though, to honest, there is so much dispute about the wave function and what it actually is, maybe we need to answer that quesiton before we can say tranverse, longitudinal or.... something else?

(3 votes)