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## Homework Statement

Real atomic nuclei are not point charges, but can be approximated as a spherical distribution with radius ##R##, giving the potential

$$ \phi(r) = \begin{cases}

\frac{Ze}{R}(\frac{3}{2}-\frac{1}{2}\frac{r^2}{R^2}) &\quad r<R\\

\frac{Ze}{r} &\quad r>R \\

\end{cases}$$

We can view the modified Hamiltonian as having a perturbation

$$H^\prime = \begin{cases}

-\frac{Ze^2}{R}(\frac{3}{2}-\frac{1}{2}\frac{r^2}{R^2}) + \frac{Ze^2}{r} &\quad r<R\\

0 &\quad r>R\\

\end{cases} $$

Calculate the first order correction to the energy of the ##1s## state.

Given answer: ##\frac{2}{5}\frac{Z^4e^2R^2}{a^3}##, where ##a## is the Bohr radius.

## Homework Equations

Nondegenerate case, first order energy correction to the ##n##'th unperturbed energy $$E_n^1 = H^\prime_{nn}= \langle\psi_n^0 | H^\prime| \psi_n^0 \rangle$$

Radial wavefunction of lowest state of hydrogen/hydrogen like ions

$$\psi(r) = \sqrt{\frac{Z^3}{\pi a^3}}e^{-Zr}$$

*think there may be a typo in the radial wavefunction, as the exponential should have a factor of ##1/a##, making it be ##e^{-Zr/a}##

## The Attempt at a Solution

Tried to evaluate

$$E^1_{nlm=100}=\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} \frac{Z^3}{\pi a^3}e^{-2Zr/a}\times (-\frac{Ze^2}{R}(\frac{3}{2}-\frac{1}{2}\frac{r^2}{R^2}) + \frac{Ze^2}{r})

\times r^2sin\theta \,dr \,d\theta \,d\phi$$

but the integral seemed to fall apart. Wondering 1) if this gives the right answer and 2) is there a better way, e.g. some way to evaluate this without using integration by parts 4 times.***EDIT***

Found the/a solution in another book

They use the approximation ##e^{-2Zr/a} = 1## since ##a \gg R## after which the answer falls out

Feel a bit cheated after having done 3 pages of algebra

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