I'm a little confused about the interpretation of the 1s orbital diagram. I know that the outer boundary is where the wave function of the 1s orbital has a constant value, which means means there's a constant probability of finding an electron on that outer boundary.

The reason I'm confused is because, isn't there an infinite number of distances from the nucleus where there's a constant probability of finding an electron around a cross-sectional slice of the sphere?

What I mean is, if we choose r, meaning distance from the nucleus as 50 pm, you could draw a circle representing that boundary as r = 50pm. You could do the same with an infinite amount of distances from the nucleus and have the same result.

Basically I'm confused as to how/why the boundary is chosen -- does the probability density need to go beneath a certain value for the boundary to be defined? Or is it arbitrary? If r < infinity, there's a finite probability of finding an electron at that distance. So doesn't this mean that the boundary is pretty meaningless?

Graphically what i mean is:

Typical 1s orbital

Why not:

Typical 1s orbital 2


Your help is greatly appreciated :D


1 Answer 1


You are right: the outer boundary is quite arbitrary. There is no any intrinsic threshold; the probability just gradually decreases lower and lower, but never reaches 0. You may draw a sphere so as to have the electron inside with 90% probability, or 95%, or 99%, or any other value as you see fit.

This, in particular, is the reason why atomic radii are somewhat arbitrary too.


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