Question related to a 3D finite spherical well

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    3d Finite Spherical
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SUMMARY

The discussion centers on the quantum mechanics of a particle in a 3D finite spherical well, defined by the potential V(r) = 0 for r < a and V(r) = V0 for r ≥ a. The radial wave function is derived as u(r) = Bsin(kr) for 0 < r < a and u(r) = Bexp(-la) for r ≥ a, with k = √(2mE/ħ) and l = √(2m(E + V0)/ħ). The participant encountered a transcendental equation sin(θ) = ±ςθ while attempting to find energy E, resulting in confusion over the relationship l = k*tan(ka). The discussion also highlights the condition for the existence of bound states, noting that a minimum potential V0 is required for bound states in 3D, unlike in 1D cases.

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ClaireBear1596
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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!
 
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ClaireBear1596 said:
u(r)=Bsin(kr)
is a little hard to understand. Could you show the steps you took to find it ?
 
BvU said:
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?
 

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