Ground state energy of a finite spherical well with l = 0

[Rahmuss]

Homework Statement

A particle of mass m is placed in a finite spherical well:

$$V(r) = \{^{-V_{0}, if r \leq a;}_{0, if r > a.}$$

Find the ground state, by solving the radial equation with $$l = 0$$. Show that there is no bound state if $$V_{0}a^{2} < \pi^{2}\hbar^{2}/8m$$.

Homework Equations

$$\frac{d}{dr}(r^{2}\frac{dR}{dr}) - \frac{2mr^{2}}{\hbar^{2}}[V(r) - E]R = l(l + 1)R$$.

The Attempt at a Solution

For $$r \leq a$$
$$\frac{d}{dr}(r^{2}\frac{dR}{dr}) - \frac{2mr^{2}}{\hbar^{2}}[V(r) - E]R = 0$$ $$\Rightarrow$$[tex] <br /> $$2r\frac{dR}{dr} + \frac{2mr^{2}}{\hbar^{2}}(V_{0} + E)$$ $$\Rightarrow$$<br /> <br /> $$\frac{dR}{dr} = \frac{-mr}{\hbar^{2}}(V_{0} + E)$$ $$\Rightarrow$$<br /> <br /> $$R = \frac{-m}{\hbar^{2}}\int r(V_{0} + E) dr$$<br /> <br /> But I'm not sure about the second part to show that there is no bound state with the given conditions.[/tex]

 

[Rahmuss]

Ok, so we can say:

$$V_{0} < \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}}$$

When we let $$n = \frac{1}{2}$$ or $$n^{2} = \frac{1}{4}$$

And if there is no bound state, then the ground state of energy must be greater than the potential. And in the ground state we have $$n=1$$ right?

So we have:

$$E_{n0} = \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}}$$.

Which means:

$$V_{0} < E_{n0}$$ $$\Rightarrow$$

$$\frac{\pi^{2} \hbar^{2}}{8ma^{2}} < \frac{\pi^{2} \hbar^{2}}{2ma^{2}}$$.

Is that what they're looking for?