Inspired by this thread, I was wondering about the radius of an electron orbital. Let’s use a hydrogen atom for simplicity. What is the shape of the probability distribution for location of the electron? I know that it is spherical, but is the maximum probability at the nucleus itself, and then trailing off as one moves further away? Or is the maximum at some spherical shell a distance from the nucleus, and falling off in both directions? At what distance from the nucleus is there a 95% probability that the electron is inside that radius? At, say, 1km, what is the probability of finding the electron at least that far away? Is there a simple formula to determine or describe this probability distribution? And how is this formula complicated by addition of further electrons, or for differently shaped orbitals?
You’re lucky, bryanmcc. The hydrogen atom is one of the few realistic systems for which the time independent Schroedinger equation has an exact, closed form solution.
Here’s the time independent Schroedinger equation for the LaTeX literate:
For the LaTeX illiterate, the TISE is just an eigenvalue equation (A x = lambda x) where the matrix is the Hamiltonian operator (sum of kinetic and potential energy operators), the eigenvector is a stationary state of the hydrogen atom, and the eigenvalue is the energy associated with that state.
To solve the TISE when more than one position coordinate is involved, one would typically separate variables and hope that the resulting ordinary differential equations are tractable. Since the hydrogen atom has a spherically symmetric potential energy function, the natural choice of coordinates is spherical. Thus we assume:
\Psi(\mathbf{r}) = R(r) \Theta( heta) \Phi(\phi)
The resulting ordinary differential equations are indeed tractable; in fact, their solutions were known even before the Schroedinger equation had been published. The solutions for the angular part of the wavefunction are the spherical harmonics, denoted
A table of spherical harmonics can be found in the CRC Standard Mathematical Tables and Formulae. In the 30th edition, they can be found in the chapter on special functions. The quantum numbers l and m uniquely determine a spherical harmonic; they are called “azimuthal quantum number” and “magnetic quantum number”, respectively.
Your question asked specifically about the radial part of the wavefunction. The solution to the ordinary differential equation involving only r is, apart from normalization, the product of e^( - r / (n a ) ) and a Laguerre polynomial of the argument (2 r) / (n a), where “n” is the principal quantum number and “a” is the Bohr radius (5.29 x 10^-11 m). The Laguerre polynomials are also tabulated in the CRC Standard Mathematical Tables and Formulae, but you can derive them independently from the generating formula
For pictures and more readable formulas for some of these wavefunctions, try this site. As you can see, some wavefunctions approach zero at the nucleus (because a factor of r can be pulled out of the radial wavefunction), while others become infinite (because of the e^( - r / ( n a ) ) term).
It depends on which state the electron is in. The solution for a general state would go something like this:
First, write the wavefunction (Psi) for that state. Be sure to normalize it.
Second, compute the squared modulus of Psi. This is typically written as Psi* Psi where the * denotes complex conjugation.
Finally, solve numerically for x:
\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{x} \Psi* \Psi r^{2} \sin heta dr d heta d\phi = 0.95
Here’s the procedure for an arbitrary state.
First, write the wavefunction (Psi) for the state desired. Be sure to normalize it.
Second, compute the squared modulus of the wavefunction (Psi* Psi, as before).
Finally, compute the improper integral
\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{1 km}^{\infty} Psi* Psi r^{2} \sin heta dr d heta d\phi
The value of this integral will give you the probability of finding an electron at least 1 km away from the nucleus.
Sorry, that’s wrong. No wavefunction becomes infinite, particularly not one with that formula. I was just looking at the graph rather than the equation, where the scale made it appear that the function blew up at the origin. In fact, that particular term approaches 1 as r approaches 0. It only blows up as r approaches negative infinity, which is not in the domain of the function anyway.
I wrote the equation without any particular state in mind. You can substitute for psi any wavefunction that satisfies the time independent Schroedinger equation. Some examples you might recall from basic chemistry:
1s (principal quantum number n = 1, azimuthal quantum number l = 0, magnetic quantum number m = 0)
2s (principal quantum number n = 2, azimuthal quantum number l = 0, magnetic quantum number m = 0)
2p (principal quantum number n = 2, azimuthal quantum number l = 1, magnetic quantum number m = -1, 0, or 1)
Some of these stationary states are graphed on the link I provided in an earlier post (or at least the radial parts).
The three quantum numbers n, l, and m are enough to determine the stationary state psi. The stationary state psi is a function of the three spatial coordinates. The probability density, however, is found by computing the squared modulus psi* psi. Integrating the probability density psi* psi over a region in space will give you the probability of finding the electron within that region. I chose the limits on the integrals in my first post to correspond to bryanmcc’s questions.
On preview, I see you’ve provided an expression for the probability density at distance r from the nucleus. I’ll try to compute the same thing using my reference, Griffiths’ Introduction to Quantum Mechanics.
Electron state 1s (n = 1, l = 0, m = 0)
R® = 2 a^( -3 / 2 ) e^( - r / a )
Y(theta, phi) = ( 1 / 4pi )^( 1 / 2 )
psi(r, theta, phi) = R® Y(theta, phi)
psi* psi = a^(-3) e^( -2 r / a ) / pi
psi* psi = u^2 (in your notation)
The probability of the electron being at or in the nucleus must be zero because the electron and the nucleus are repulsing each
other.
Anyways, so is the location of this aforementioned electron unknown, except the location IS known to be limited to somewhere on a certain sphere around the proton? And how far out from the proton is this particular sphere that the electron is on? Like would it be maybe one, two, or three, etc. proton diameters, for example?
And no hiding in equations of calculus and eigenvalues, because I know that you can actually see atoms; to my surprise after reading how you can’t see them I ran across something I can’t find now that mentioned in a subordinate clause that at some point in history we were actually able to see individual atoms. I have to have an actual picture before I can understand anything, though I respect the value of the eigenvalues (a term I looked up several times but can’t figure out what it means). Okay so there is the little proton, as usual (case a:)in the middle of the football field. (Case b:) Or is it at one side of the field?
And the electron is going around the proton somewhere on a sphere whose diameter is (case a:) the length of the football field, or else
(case b:) twice the length of the field. Is that it?
Penultimately, how fast is the electron going around and around in miles per hour, or does its speed change? And finally, is its motion in this sphere always ahead or does it suddenly go backward or take right angles or something? And how many times does it go around the proton per second?
amore ac, yes I relaized you meant any state just after I posted the first post, I’ve never encountered LaTex before so it threw me a bit, so I had to dig out my old quatum mechanics text book.
There reason I used u instead of psi is that the wave function is quite often notated as ulmn, in this case u100 was what I used with Z=1 as I was describing an electron in the 1s state of a hydrogenic atom.
amore ac, yes I relaized you meant any state just after I posted the first post. I’ve never encountered LaTex before so it threw me a bit, so I had to dig out my old quatum mechanics text book.
There reason I used u instead of psi is that the wave function is quite often notated as ulmn, in this case u100 was what I used with Z=1 as I was describing an electron in the 1s state of a hydrogenic atom.
Your explantion was very good and precise, but not having done the maths recently when I saw it it completly puzzled me (I cheated I saw how the textbook had derived the equations rather than deriving them all from the time-independant schrodinger equation).
It is great to see new posters (amore ac studio) come here and manage the level of response given and I think it is a testament to the caliber of people at the SDMB that more than a few other members can follow and add to that explanation.
I also realize that the only way to fully appreciate answers to the OP are via the language of math and (as mentioned in the previous thread bryanmcc linked to ordinary language (ala English) is insufficient by itself to adequately describe these concepts.
With that understanding does anyone care to give an explanation in English at least moderately geared to the layman? This goes along with what don willard was getting at and I second it.
I keep getting hung-up on the shell idea and that an electron must ‘orbit’ (yes…I know that term is not entirely correct) at distinct distances from the nucleus. If the electron changes orbitals it must do so in a quantized fashion…that is it can be in orbital 1 at, say, 1u from the nucleus or orbital 2 at 2u from the nucleus but it can never be at orbital 1.3654u from the nucleus.
The whole probability cloud thing trips me up because it sounds as if the electron is all over the place with regard to distance from the nucleus (I have no problem with it being randomly distributed on the surface of the sphere that represents a given orbital). If the electron were really jumping up and down as the probability cloud seems to suggest I would think you’d be continuously varying its energy state (which I assume without outside interference it should not do but then again maybe quantum fluctuations and uncertainty allow for that). Actually I shouldn’t say the probability cloud concept suggests any such thing…clearly it is my lack of understanding suggesting it but hopefully you guys can clear it up.
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There is a story, possibly apocryphal, in which physicist Richard Feynman is approached by a layman and asked whether there was any way spin could be explained in plain English. Feynman pondered a bit, then told his questioner that he would have to check with his colleagues before giving a definitive answer. After discussing the matter with his fellow quantum physicists, Feynman came to the disturbing conclusion that the essence of spin could not be explained without resorting to the language of mathematics. His response to his questioner could be paraphrased thus: “I believe I can say that nobody really understands spin.” The moral we infer is, that even for a practicing physicist, a thorough understanding of a theory involves language as much as, if not more than, mathematics.
Approaching the hydrogen atom problem with the Schroedinger equation and an arsenal of techniques for solving partial differential equations would not have yielded any knowledge of electron spin. The Schroedinger equation involves the Laplacian operator, a sort of second derivative which acts on the usual three spatial coordinates only. Electron spin is a complicated beast with no dependence on those three spatial coordinates, so it gets totally ignored by the Schroedinger equation. The only way to derive the existence of electron spin is to apply the Dirac version of quantum mechanics, using the algebra of raising and lowering operators to force out eigenstates of the spin angular momentum operators. Even then, the eigenstates do not take the form of functions, and we cannot provide a visual representation of them that would make electron spin accessible to the layman. Without a common analogy that would make the details of electron spin accessible, Feynman was unable to convey just what exactly spin was to his questioner. If the functional form of the electron spin states were known, and the variables upon which they depend were a common part of the human experience (as height, width, and depth are to us Euclidean beings), then Feynman probably could have explained the nature of electron spin to his questioner.
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It’s weird, I’ll be the first to admit. The stationary state is so-called precisely because every time we measure the energy, we’ll get the same (stationary) value. An electron in the state 1s will always be measured as having an energy of -13.6 eV. Even though the location of the electron is smeared over all space, the measurement you record for its energy will remain the same, because the stationary state we calculated was a solution to the time independent Schroedinger equation. Incidentally, two other quantities of the electron in the 1s state are also stationary: the total orbital angular momentum, and the z-component[sup]1[/sup] of the orbital angular momentum. These two quantities are stationary because they are related to the eigenvalues of operators that commute with the Hamiltonian (total energy) operator. So an electron in a stationary state has well-defined energy, orbital angular momentum, and z-component of orbital angular momentum. On the other hand, the electron in a stationary state does not have well-defined position, giving us the pictures of probability density distributions that GargoyleWB linked to.
Footnotes:
One could argue that any of the three Cartesian components of the orbital angular momentum, not just the z-component, could be chosen, since all three operators commute with L^2 and H. However, it is customary to choose the z-component, which can be done in any case by rotating the spatial axes.
An important point, here: Even though you can choose any single component of angular momentum to have a definite value, you cannot choose more than one component. So your atom can have definite energy, total angular momentum, and z component of angular momentum, but it cannot have both definite x component and z component. And no, nobody ever said that this wasn’t weird.
Also, while quantum mechanical spin is not an object in the ordinary space we’re familiar with, it is an object in a space, which can be perfectly well described mathematically. With practice, one can even, to an extent, visualize spin space, but it still can’t be described in words.
First let me state that it is good to have you on-board at the SDMB. While we certainly have other people (notably Chronos but several other people besides him) who can talk on these topics at a very high level I’ve gotten the sense that they get a little tired of dumbing-down complex concepts for the rest of us. That’s probably mostly due to the fact that many of these discussions have been discussed to death over the years yet keep rising again despite their best efforts at answering them. I can hardly blame them for getting tired of it. I suspect Chronos is thrilled to see you here partly to pick-up some of the slack for these incessant questions and mostly to have someone he needn’t dumb things down for (although again there are people here he needn’t dumb anything down for but they are outnumbered by the rest of the unwashed masses…fighting ignorance turns out to be a full-time job as Chronos’ 8,000+ posts can attest to).
Ok…enough with the bootlicking…
Say I have a hydrogen atom with an electron at 1s and take a ‘picture’. I see the fuzzy probability cloud of that electron around the nucleus. Now zap that atom sufficiently to jump the electron up to the 2s orbital and take another picture. Will the probability cloud look larger or do I see the same smeared-out cloud I did in the first picture? (Please don’t get hung-up on whether I can take photos of electron clouds…through the magic of the hypothetical let’s assume I can manage this with the bonus of not affecting the object of the picture in any way.)
The energy of the electron is always measured to be the same at a given orbital but we can’t know the location of the particle (as mentioned in the quote above). That sounds like Heisenberg’s Uncertainty Principle at work. Is that what is at work here or does the electron smear out further than the HUP dictates?
Once again I get hung-up on the probability cloud versus the quantized nature of this creature. You have something smeared out all over the place but it must possess a defined set of energy states. So what I am getting is that an electron can never be between 1s and 2s yet it is all over the place such that it may very well be between 1s and 2s. I suspect I am linking its energy state with its position in a manner that shouldn’t be done but then again I really haven’t a clue so any clarification would be helpful.
I think this has been discussed here before (almost sure of it) but it seems relevant to this discussion and I don’t remember the answer so I’ll ask again. Even though the electron is smeared-out is that merely the universe hiding things from us and the electron actually does have a definite location or is it really all over the place at the same time? I know the double-slit experiment suggests particles really are all over the place simultaneously and I suspect electrons are no different but it’s so counter-intuitive I guess I need reinforcement on that point. I think what stumbles me is the idea that, say, an incoming photon strikes our electron sufficient to knock it into a higher orbital. How can the photon hit the electron if the electron is all over the place?
Actually now that I think about it the photon is all over the place too…
I was hoping someone would provide and english language explanation of the OP question. None appeared, so I will have a go.
In trying to describe this in words obviously the desciption looses accuracy, but hopefuly it will make the whole picture more understandable.
First a definition to help my explanation. I use ‘concentration’ to mean the probability of finding an electron at a point in space. ie psi-star-psi.
Lets start with the simple hydrogen 1s-orbital. In this case we have an exponential decrease in ‘concentration’ of the electron from the nucleus out, in all directions. The most likely spot to find the electron is right in the nucleus, where the potential energy of the electron is lowest. However, the electron does not fall right into the nucleus because the kinetic energy of the electron become rapidly large and it can spread out to regions with less kinetic and more potential energy. Kinetic energy is greatest closest the nucleus.
When you calculate the radius that the electron is most concentrated in a 1s-orbital, it is not going to be right in the centre near the nucleus, because that area of space is small for that radius - thus lower likelihood. Imagine the orbital was superimposed on an onion (a little onion), with the nucleus at the very centre of the onion. The layer of the onion with greatest concentration of electron is not the centre layer, the volume of the layer itself is too small. It is not the outmost layer, which has the greatest volume, because the concentration of the electron drops off exponentially. It is a compromise, a point inbetween has the radius with the greatest probability of finding the electron. The maximum of the differentiated radial distribution function, shown in an earlier post, gives this radius. It is 53 pm for hydrogen. Heavier nuclei suck in the 1s electron even closer - eg 0.6 pm for uranium - giving this electron heaps of kinetic energy.
When the 1s electron jumps up to the 2s-orbital, the electron is now in an obital with an extra buckle radially in 3D space - like a standing wave with an extra node. There is now a shell of extra electron concentration surrounding the nucleus. The orbital has more curvature and thus more kinetic energy. (btw you can say the electron is the orbital). In addition to this the orbital gets bigger, so the electron has more potential energy as well. The same happens with 3s, 4s etc.
There is considerable spatial overlap in the radial distribution functions of these two orbitals. The two wave functions occupy a lot of the same space, but the 2s-orbital is more energetic. The most probable radius that a 2s-electron is found is at 276 pm.
When you introduce the quantum number angular momentum, you get the p-orbitals. The angular momentum gives the electron even more energy. As you increase the angular momentum of the orbital the wavefunction becomes more buckled. It can only buckle in whole increments - like a vibrating string, but in 3D. With each extra buckle, or quantum energy level, the overall energy increases - the kinetic energy of the electron increases. There are three
In a spherical space you can describe an orbital with three coordinates - r the radius, theta the colatitude, and phi the azimuth. The s-orbitals can be described just using a radial dependence, because they do not vary in concentration from a fixed distance around the nucleus. The p-orbitals, d- orbitals etc need to have the theta and phi in the wave function. The p and d etc are more complex and more buckled. The reason - because the orbitals have angular momentum. In these cases the electron is flung away from the nucleus and you dont find it hanging around the nucleus at all. Again more potential energy (bigger orbital) and more kinetic energy (more buckled orbital).
** Whack-a-Mole ** while its true that the probability density for finding the electron in any small volume of the hydrogen atom is highest close to the nucleus, the radial probability density is highest at the Bohr radius of 52.9 pm. And this is the radius at which the electron is most likely to be found. And for higher energy levels the peak radial probability density will be at a still larger radius.
An electron in a higher energy level is not strictly farther from the nucleus than one in a lower one. Sometimes it may even be the other way around. But, on average the one at a higher energy is further out. Therefore its at a higher potential energy and a higher total energy.
E = -13.6 eV / n[sup]2[/sup]
And this is the energy of the photon that will be emitted when an excited electron jumps to a lower orbital.
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I am afraid that I can’t keep these details straight in my mind. I am probably confusing different types of “spin”, but this is what I recall that may be helpful if correct.
(Please note, if you want to learn something, skip this post. This is to the experts who can tell me where I am confused.)
IIRC, isn’t spin related to mirror symmetry? I.E., spin 0 implies that any spatial “rotation” will maintain the symmetry of the particle. Spin 1/2 implies that the particle would have to be rotated 180 degrees to maintain its symmetry. Spin 2 means that the particle would have to be rotated 720 degrees (which is count-intuitive in our macro world) to return to the original symmetry.
Does any of this make sense? If not, never mind. If so, it is perhaps a way to imagine in English what spin is.
