- #1

- 173

- 0

## Homework Statement

Assume that there is a deviation from Coulomb’s law at very small distances, the Coulomb potential energy between an electron and proton is given by

[itex]V_{mod}(r)=\begin{cases}

-\frac{e^{2}}{4\pi\varepsilon_{0}}\frac{b}{r^{2}} & 0<r\leq b\\

-\frac{e^{2}}{4\pi\varepsilon_{0}}\frac{1}{r} & r>b

\end{cases}[/itex]

(a) Specify the perturbation

(b) Find the first order correction for the ground state

(c) Show that the answer in (b) can be approximated by

[itex]E_{1}^{(1)}\approx-\frac{4b^2}{a_0^2}E_R[/itex] where [itex]E_R=\frac{e^{2}}{8\pi\varepsilon_{0}a_0}[/itex] is the Rydberg energy.

## Homework Equations

[itex]\psi_0=\frac{2}{a_0^{3/2}}e^{-r/a_0}[/itex]

[itex]E_{1}^{(1)}=\int_{0}^{b}\left| \psi_0 \right|^2 \delta\hat{\textrm{H}} r^2\,\textrm{d}r[/itex]

[itex]b \ll r[/itex]

## The Attempt at a Solution

(a) [itex]\delta\hat{\textrm{H}}=-\dfrac{e^{2}}{4\pi\varepsilon_{0}}\left(\dfrac{b}{r^{2}}-\dfrac{1}{r}\right)[/itex]

so with [itex]b \ll r[/itex] we have [itex]e^{-2b/a_0}\approx 1[/itex]

(b)

[itex]E_{1}^{(1)}=\dfrac{e^2 b}{2\pi \varepsilon_{0}a_0^2}[/itex]

I think the above is correct, I just can't see how to get part (c).

Last edited: