Solving for the Ground State Energy of a Finite Spherical Well: Homework Help

[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<br /> <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&#039;m not sure about the second part to show that there is no bound state with the given conditions.

 

[Rahmuss]

Ok, so we can say:

V_{0} &lt; \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} &lt; E_{n0} \Rightarrow

\frac{\pi^{2} \hbar^{2}}{8ma^{2}} &lt; \frac{\pi^{2} \hbar^{2}}{2ma^{2}}.

Is that what they're looking for?