# Exam preparation question

• Void123
In summary, when R=\pi/2, there are no bound states and the precise range of bound states is determined by the restriction -V_{0}<E<0.

## Homework Statement

I was reviewing some homework problems and looking at the solutions. There is one problem with a tiny step I just cannot rationalize and I am hoping someone can point me in the right direction.

I have a spherical finite well:

$$V = {- V_{0}: 0 < r < a}$$,

$$= {0: r \geq a}$$

$$- k_{2} = k_{1} cot (k_{1} a)$$ (1)

Refining the notation,

$$\alpha = a \sqrt{(2m(E + V_{0})}/hbar = k_{1} a$$

$$R = a \sqrt{(2m(V_{0})}/hbar$$ and $$k_{2} = \sqrt{(2m(V_{0})}/hbar$$

So (1) may be rewritten as $$\sqrt{R^{2} - \alpha^{2}} = - \alpha cot (\alpha)$$

From part 1.

## The Attempt at a Solution

I don't understand how at $$R = \pi/2$$ there are no bound states.

Also, I am given this restriction: $$-V_{0} < E < 0$$

How is this justified and how is the precise range of bound states determined?

When $R=\pi/2$, the only solutions are at $\alpha=\pm\pi/2$. In these cases, you get k2=0.

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