# Thread: how to calculate number of waves in an orbit?

1. hi, i'm not clear about the number of waves in orbits concept. According to the text book number of waves in an orbit is equal to orbit number , but my doubt is each electron is like a wave then number of waves in an orbit should be equal to number of electrons in that orbit. Please explain me this concept....  2.

3. I'm not sure what you mean. Each orbital is a probability distribution of where the electron will be found, these are solutions to the Schrodinger equation, with different solutions having different quantum numbers, the main one being the principle quantum number, n, is this what you mean by orbit number? If so I'm guessing the text book is simplifying things (what level of study are you at?) and describing the n=1 orbital as a standing wave with half a complete wavelength (no nodes), the n=2 as a standing wave with one complete wavelength etc. This is not related to how many electrons an orbital contains any orbital can only contain a maximum of two electrons (read about the Pauli exclusion principle).  4. Originally Posted by chinna hi, i'm not clear about the number of waves in orbits concept. According to the text book number of waves in an orbit is equal to orbit number , but my doubt is each electron is like a wave then number of waves in an orbit should be equal to number of electrons in that orbit. Please explain me this concept....
Yes, you can think of the orbitals as being like standing waves. The most helpful thing, I think, is to think of the NODES, i.e. the points of zero probability in the probability distribution that the "waves" represent. With Principal Quantum Number, n = 1, you have no nodes in the middle. This is just like the fundamental mode of resonant vibration of a violin or piano string, for example. The electron cloud looks like a ball around the nucleus, i.e. it is spherical and all of the same phase (say +ve). With n=2, you still have something that looks like a ball, but if you were to cut a cross section through it you would see there are 2 concentric, spherical layers of probability, with a spherical ring of zero probability inside it. You can think of a layer of +ve phase at distances closer to the nucleus and one of -ve phase at distances further out. So you have one node. So this is like the first harmonic of your violin or piano string, with one node halfway along (and a pitch one octave above the fundamental). With n=3, you have 2 such ring shaped nodes, and so on.

Because electron clouds are in 3D, rather than 1D like a piano string, it gets more complicated. At n=1, you only have the one spherical orbital, but with n=2 you are also allowed nodes in planes at right angles, intersecting the nucleus. These are the nodes of the 3 mutually perpendicular p orbitals, defined by the Azimuthal Quantum Number, l. Ditto for the d orbitals that start at n=3, etc.

The Pauli exclusion principle allows 2 electrons per orbital, provided they have opposite spin, thereby ensuring that no 2 electrons have all the same quantum numbers (which is not allowed for fermions such as electrons). When we say they occupy the same orbital, it means the solution for Schroedinger's equation for each looks identical, which means the shape of the probability distribution they obey is identical. So you don't double anything when you add the second electron, both are in a probability cloud with the same shape and energy, just with opposite spin.  5. what is orbit? motttther.  6.  7. Originally Posted by ASG141 what is orbit? motttther.
Not orbit, orbital.  8. orbit means energy level or shell  9. the number of waves made by an electron in 3rd orbit? i saw this type of questions in some of the books,please help me....  10. Originally Posted by chinna the number of waves made by an electron in 3rd orbit? i saw this type of questions in some of the books,please help me....
Well, by 3rd orbit I assume you mean the orbital (please note they are not "orbits") with Principal Quantum Number, n=3. At n=3, Schroedinger's equation has quite a number of solutions, because the Azimuthal Quantum Number, l, can have integer values from zero up to n-1.

For l=0, you have one orbital, 3s, with 2 "radial"nodes (i.e. the spherical shaped ones I described earlier).

For l=1, you have a set of three 3p orbitals, each of which has a "planar" node in a plane passing through the nucleus, and a further radial node within the dumbell-shape of the orbital, i.e. 2 nodes. P orbitals have a "dipolar" shape, +ve phase in one lobe and -ve phase in the other, reversing at the radial node.

For l=2, you have a set of 5 d orbitals, each of which has a pair of planar nodes at right angles, intersecting the nucleus. Again, 2 nodes. This gives the d orbital a quadrupole shape, with lobes of +ve and -ve phase diagonally opposite each other.

So you have n-1 nodes in all these modes of standing waves. It is just that they are differently distributed in space.  11. Originally Posted by chinna the number of waves made by an electron in 3rd orbit? i saw this type of questions in some of the books,please help me....
Hi chinna. When talking in terms of 'orbits' that was a term used to describe how electons orbited a nucleus with classical periods in an older model proposed by Neils Bohr to explain how the atom worked. This was the model that many of us were taught at school. Though useful in some respects, this model has been replaced by modern quantum mechanics where electrons are described by wavefunctions to more accurately convey the reality of dealing with uncertainty (the Undertainty Principle). Wavefunctions such as standing waves are a great way to understand how a sense of seperation and permanency exists in these wave structures and can replace the notions of classical particles in space. The term orbitals has now subsumed the previous 'orbit' to convey the configuration of electrons and where they are found using probability distributions. The possible position of the electron within these orbitals is smeared out and its occurrence is defined by the probability of finding an electron in this configuration state. More here. Then have another read of post 3 by exchemist. It might make more sense then. :-))  12. ok thanks to all  Bookmarks
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