Meaning of the 1s orbital diagram

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](https://i.stack.imgur.com/fOOc0.png)](https://i.stack.imgur.com/fOOc0.png)


Why not:


[![Typical 1s orbital 2](https://i.stack.imgur.com/bvlAH.png)](https://i.stack.imgur.com/bvlAH.png)


?


Your help is greatly appreciated :D




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](https://en.wikipedia.org/wiki/Van_der_Waals_radius) are somewhat arbitrary too.


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