schroedinger equation - Why doesn't a quantum particle in an attractive 1D potential accumulate at the center?

I have two questions regarding (possibly counter intuitive results) Schrodinger equation and its application to two (strictly hypothetical) scenarios.

1. 
   

   Consider the 1D potential $V(x) = - \frac{\alpha}{|x|}$, which is an attractive one. As it is attractive, after a sufficiently large amount of time I'd expect the probability density of finding the particle at $x=0$ (center) to be higher than that anywhere else. I hope it is reasonable to expect such a thing.

   
   

   But the solution according to Schrodinger equation is (as given in this answer) is $$u_n(x,t) \sim \lvert x\rvert e^{-\lvert x\rvert/na} ~L_{n -1}^1\biggl(\frac{2\lvert x\rvert }{na}\biggr) e^{-E_nt/\hbar}$$ Consider the amplitude at $x=0$ $$\lvert u_n(0,t)\rvert = 0$$ which I think is contradictory to our reasonable expectations based on intuition. I'd appreciate some comments on why it is counter intuitive.

   
   



2. 
   

   Consider the particle in box problem and according to Schrodinger's equation, in almost all energy states the probability of finding the particle close to the boundaries is zero. Now instead of intuition (which is not predicting anything) I'd give a practical example where it is quite contradictory. (Pardon me if this example is not suitable, I am learning and I'd love to know why it isn't intuitively not suitable). Consider a metal conductor, I guess its an example for particle in a box as the potential inside is zero and we know that the charge is usually accumulated at the edges of the conductor.

   
   

EDIT

An interesting thing to add. The probability density vanishing at center does not seem to arise in 3-D hydrogen atom. Check the wave function of 1s orbital of Hydrogen atom, it of the form $\psi_{1s}(r) = a e^{-kr}$, where $k$ and $a$ are some constants and $r$ is the radial distance from nucleus.