quantum mechanics - 3D Delta Potential Well

The 1D delta potential well $V(x) = -A\delta(x - a)$ always has exactly one bound state. The same is true for the 3D delta potential well $V(\vec{r}) = -A\delta(\vec{r}-\vec{a})$. I can show this for $\ell = 0$, I don't know how to do the calculations otherwise.

So two questions,

1. 
   

   Can I conclude that there is only one bound state for the 3D potential well for $\ell \not = 0$? I've seen that the energies of the eigenstates for the hydrogen atom depend only on $n$, but I am wondering whether this is an instance of a more general result?

   
   


2. 
   

   When $\vec{a} = 0$ and $\ell=0$, there are no normalizable eigenstates. For $\ell \not = 0$, the effective potential in the radial equation becomes large at the origin, can I use this to conclude that there are no bound states when $\vec{a}=0$?