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## 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{q^{2}}{4\pi\varepsilon_{0}}\frac{b}{r^{2}} & 0<r\leq b\\

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

\end{cases}[/itex]

(a) Find the first order correction for the ground state using the standard integrals

## Homework Equations

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

To find the first order correction - [itex]E_{1}^{(1)}[/itex]=<ψ

^{0}| [itex]\delta\hat{\textrm{H}} [/itex]|ψ

^{0}>

Perturbation - [itex]\delta\hat{\textrm{H}}=-\dfrac{q^{2}}{4\pi\varepsilon_{0}}\left(\dfrac{b}{r^{2}}-\dfrac{1}{r}\right)[/itex] for 0<r≤b

Standard integrals

[itex]\int_{0}^{x}[/itex] e

^{-u}du = 1 - e

^{-x}

-[itex]\int_{0}^{x}[/itex] u e

^{-u}du = 1 - e

^{-x}- xe

^{-x}

## The Attempt at a Solution

Using [itex]E_{1}^{(1)}[/itex]=[itex]\int_{-∞}^{∞}[/itex]ψ

^{(0) *}[itex]\delta\hat{\textrm{H}} [/itex] ψ

^{(0)}

[itex]E_{1}^{(1)}=\dfrac{1}{\pi a_{0}^{3}}\dfrac{q^{2}}{4\pi\varepsilon_{0}}\int_{0}^{b}(e^{\dfrac{-2r}{a_{0}}}) (\dfrac{b}{r^{2}}-\dfrac{1}{r})[/itex] dr

So I've taken the constants out and applied a 'sandwich integral' with the perturbation Hamiltonian... This problem is I have no idea where to start making this easier for me, ive been staring at it for aleast and hour. Can someone help please?

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