What defines the boundaries of electron standing waves around the nucleus?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 3K views
wil3
Messages
177
Reaction score
1
Hello. I have recently been introduced to the concept of electrons as standing waves around the atomic nucleus. The explanation I read used the simulation of "a particle in a tube" to give a monodimensional interpretation of how the standing wave behaves.

Within an atom, what serves as the "boundaries" for the electron-wave? Is it purely classical electrostatic attraction to the nucleus?

Any accompanying calc-1 level math describing how classical electrostatics shapes these bounds would be very much appreciated.

Thank you in advance for any replies.
 
Physics news on Phys.org
Within an atom, an electron doesn't have sharp bounds. To find the wave solutions to any electron in a ______, you need to know the potential its in. One of the classical examples is an 'infinite square well in 1d.' In this case, there is a region of zero potential, bordered by two regions of infinite potential ---> this leads to a very well bounded electron.

Generally, for an atom, people describe the situation (simplified) by a potential resulting solely from electrostatic interactions, i.e. the potential 'U'
[tex] U \propto \frac{1}{r}[/tex]
and has no clear boundary. The result is that the probability distribution for the electron gradually falls off with increasing distance from the atom's nucleus.
 
yes, the probability distribution of a bound electron looks like a normal distribution curve squished off to the left and peaking at distance = r for that energy level. I think it's zero at the nucleus and probability for radius much below the first energy band...

The probability for such a bound electron fades to zero at infinity while a free actually electron has a finite probability of being found at infinity...but I don't think we have found any there yet! (a poor joke)

I guess we learn about "orbits" with radii like planets, then electron clouds and also electron probability distributions...don't believe ANY of those have been observed...here's a computed graphic: http://en.wikipedia.org/wiki/Electron_cloud
 
Last edited:
Naty1 said:
yes, the probability distribution of a bound electron looks like a normal distribution curve squished off to the left and peaking at distance = r for that energy level. I think it's zero at the nucleus and probability for radius much below the first energy band...

No, the radial probability distribution looks like that. It's not the same thing as the probability distribution, which is |psi|^2 (for a single electron). The radial distribution is what you get if you take the probability at radius r and integrate over the surface of the sphere with that radius. Since a sphere with zero radius has zero surface area, it's zero at r=0. But the 1s (hydrogen ground state) orbital actually has its maximum at r = 0.

In other words, the probability that the electron is in an infinitesimal volume element at (r, theta, omega) should not be confused with the total probability of the electron being at any point with a given radius.
I guess we learn about "orbits" with radii like planets, then electron clouds and also electron probability distributions...don't believe ANY of those have been observed

When we say 'electron cloud' then what we mean is the probability distribution. Which is directly measured all the time in more ways than I could enumerate. X-ray crystallography, STM imaging, even basic chemistry is indirect evidence.