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# explain how the heisenberg principle applies to the current model of the atom

explain how the heisenberg principle applies to the current model of the atom

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Stanton D. | Tutor to Pique Your Sciences InterestTutor to Pique Your Sciences Interest
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So Neveen,
You want to relate the Heisenberg Uncertainty Principle to the current (electron cloud) model of the atom.
First, what did Heisenberg say>
1) .delta.p times .delta.x >= h/2.pi. where .delta. stands for the Greek capital delta letter, and .pi. stands for the mathematical symbol pi (~3.14159).
also:
2) .delta.E times .delta.t >= h/2.pi. This is mathematically equivalent; but we use the first equation for most of the things we think about at the quantum level, unless we're specifically considering energies.

What do these equations mean to you? For the first: whenever you measure, specify, or use values for the momentum (inertia) of any particle, and do the same for values of position (that's the "x") measured, specified, or used at the same time, each of those values has a little experimental uncertainty, that is, you can't measure or know it to infinite precision. You would think that this is just a matter of "self-evident", measurement methods being what they are (example: you're not going to be able to read lengths to a precision of 10^-6 meter from a standard meter stick, with its 10^-3 markings (every mm)); but the Heisenberg equation says something different; it says that NO MATTER HOW GOOD YOUR MEASUREMENT METHODS ARE, you STILL can't measure position and momentum exactly at the same time for any object.
Nature itself prevents you; as you measure you affect the object and its position or momentum then change in a way you can't ever exactly know. You can measure it again, of course; then you know where it is NOW but its momentum has changed yet again, and so forth.
For the nucleus as a whole, it's pretty massive, so even little motions of it at the center of the atom (i.e. wiggles in velocity, which translate into wiggles in momentum = .delta.P) don't prevent getting a pretty good location fix for it (small .delta.position). But for electrons ---- their mass is so small, that their momentum is also small (so, there's not much "wiggle room" or .delta.p, down near zero!) -- and that MUST translate into a high .delta.position -- i.e., the electrons HAVE to take up a large fuzzy cloud as their possible position. That's (more or less directly) the result of the Heisenberg Uncertainty Principle.
I like to use the analogy of cats.
You put an adult cat in a corner, and it washes its face with its paw while it considers the situation.
You put a kitten in a corner, and it starts running out. You can keep putting it in the corner all day, and it will still keep running out. The kitten is like the electron -- you just can't keep it cooped up tightly.
There are a few nifty things that follow from this. Since electrons bop into each other, and thereby keep atoms and molecules from getting too close together (the nuclei are still far apart from each other when two molecules collide - they don't play a part in the collision, other than providing some overall momentum and kinetic energy as a whole to perhaps give the electrons enough force to chemically react), it follows that if a form of matter didn't have electrons (perhaps, it acquired its charge balance directly in the nucleus itself) it might be able to be much denser. Maybe even "dark matter"?
Second, since these electrons are buzzing around in fuzzy clouds, these clouds are overlapping in space. Given that ordinarily electrons like to repel each other electrostatically, what is it about electrons on an atom that permits them to ignore each other? If you haven't started learning yet about orbitals and wavefunctions and their symmetry properties, you will in chemistry class someday. It's great stuff.