Question related to a 3D finite spherical well

[ClaireBear1596]

I have a particle in a spherical well with the conditions that V(r) = 0 is r < a, and V(r) = V₀ if r ≥ a.
In this problem we are only considering the l=0 in the radial equation.
After solving this I found that in the region 0<r<a, u(r)=Bsin(kr) (k=√2mE/hbar), and in the other region, u(r)=Bexp(-la) (l=√2m(E+V₀/hbar). I was supposed to show that in order to find E I needed to solve a transcendental equation of the form sinθ=±ςθ, however my result was l=ktan(ka), so I am unsure what to do with this.

Also, I was wondering why is it that there is a possibility of there being no bound states if V₀ is too small? And how would I go about finding this minimum potential?

Thanks!

 

[BvU]

u(r)=Bsin(kr)

is a little hard to understand. Could you show the steps you took to find it ?

 

[ClaireBear1596]

is a little hard to understand. Could you show the steps you took to find it ?

Sorry, I actually figured out the problem with that part of my question. I still don't understand however why it is that there will always be a bound state in the 1D case of a (finite) well, whereas there will not always be one with the 3D case?