Chapter 7, Problem 7.33P

To determine

The correction to the ground state energy of hydrogen due to the finite size of the nucleus.

Answer to Problem 7.33P

The correction to the ground state energy of hydrogen due to the finite size of the nucleus is ${10}^{-10}{E}_1$.

Explanation of Solution

Using perturbation theory,

$H\text{'}=\frac{e^2}{8\pi {\epsilon }_0}\left(\frac{1}{b}-\frac{1}{r}\right)\left(0<r<b\right)$        (I)

And the energy correction is

$\Delta E=\langle \psi |H\text{'}|\psi \rangle$        (II)

The wave function of the ground state is

$\psi \equiv \frac{1}{\sqrt{\pi a^3}}{e}^{-r/a}$        (III)

Where $a=\frac{4\pi {\epsilon }_0{\hslash }^2}{m{e}^2}$.

Solving equation (II),

$\begin{array}{c}\Delta E=\frac{e^2}{8\pi {\epsilon }_0}\frac{1}{\pi a^3}\left(4\pi \right){\displaystyle \underset{a}{\overset{b}{\int }}\left(\frac{1}{b}-\frac{1}{r}\right)e^{-r/a}{r}^{2}dr}\\ =\frac{e^2}{\pi {\epsilon }_0{a}^3}\left[\frac{1}{b}{\displaystyle \underset{a}{\overset{b}{\int }}{e}^{-2r/a}{r}^{2}dr}-{\displaystyle \underset{a}{\overset{b}{\int }}{e}^{-2r/a}rdr}\right]\\ =\frac{e^2}{\pi {\epsilon }_0{a}^3}{\left[\frac{1}{b}\left[-\frac{a}{2}{r}^2{e}^{-2r/a}+a{\left(\frac{a}{2}\right)}^2{e}^{-2r/a}\left(-\frac{2r}{a}-1\right)\right]-\left[{\left(\frac{a}{2}\right)}^2{e}^{-2r/a}\left(-\frac{2r}{a}-1\right)\right]\right]}_0^b\\ =\frac{e^2}{\pi {\epsilon }_0{a}^3}\left[\left[-\frac{a}{2b}{b}^2{e}^{-2b/a}+\frac{a^3}{4b}{e}^{-2b/a}\left(-\frac{2b}{a}-1\right)\right]-\left[{\left(\frac{a}{2}\right)}^2{e}^{-2b/a}\left(-\frac{2b}{a}-1\right)+\frac{a^3}{4b}-\frac{a^2}{4}\right]\right]\end{array}$

Solving further,

$\begin{array}{c}\Delta E=\frac{e^2}{\pi {\epsilon }_0{a}^3}\left[e^{-2b/a}\left(-\frac{ab}{2}-\frac{a^2}{2}-\frac{a^3}{4b}+\frac{ab}{2}+\frac{a^2}{4}\right)+\frac{a^2}{4}\left(\frac{a}{b}-1\right)\right]\\ =\frac{e^2}{\pi {\epsilon }_0{a}^3}\left[e^{-2b/a}\left(-\frac{a^2}{4}\right)\left(\frac{a}{b}+1\right)+\frac{a^2}{4}\left(\frac{a}{b}-1\right)\right]\\ =\frac{e^2}{4\pi {\epsilon }_{0}a}\left[e^{-2b/a}\left(\frac{a}{b}+1\right)+\left(1-\frac{a}{b}\right)\right]\end{array}$

Let $\frac{2b}{a}=\epsilon \ll 1$, then $e^{-b/a}\approx 1-\epsilon +\frac{{\epsilon }^2}{2}-\frac{{\epsilon }^3}{6}+\mathrm{...}$

Therefore, the above equation becomes,

$\Delta E=\frac{e^2}{4\pi {\epsilon }_{0}a}\left[\left(1-\frac{2}{\epsilon }\right)+\left(a+\frac{2}{\epsilon }\right)\left(1-\epsilon +\frac{{\epsilon }^2}{2}-\frac{{\epsilon }^3}{6}+\mathrm{..}\right)\right]$

Solving the above equation,

$\Delta E=\frac{e^2}{4\pi {\epsilon }_{0}a}\frac{2b^2}{3a^2}$

Since, the ground state energy of the hydrogen is

$E=E_1=\frac{m{e}^4}{2{\left(4\pi {\epsilon }_0\right)}^2{\hslash }^2}$

Where $a=\frac{4\pi {\epsilon }_0{\hslash }^2}{m{e}^2}$.

Therefore,

$E_a=-\frac{e^2}{2\left(4\pi {\epsilon }_0\right)}$

Keeping only the leading term,

$\begin{array}{c}\frac{\Delta E}{E}=\frac{e^2}{4\pi {\epsilon }_0}\left(-\frac{2\left(4\pi {\epsilon }_0\right)}{e^2}\right)\frac{2b^2}{3a^2}\\ =-\frac{4}{3}{\left(\frac{b}{a}\right)}^2\end{array}$

Substitute $a=5\times {10}^{-11}\text{ m}$ and $b={10}^{-15}\text{ m}$ in the above equation,

$\begin{array}{c}\frac{\Delta E}{E}=-\frac{4}{3}{\left(\frac{{10}^{-15}\text{ m}}{5\times {10}^{-11}\text{ m}}\right)}^2\\ =-\frac{16}{3}\times {10}^{-10}\\ \approx -5\times {10}^{-10}\end{array}$

The energy correction for the fine structure is $5\times {10}^{-5}$ and for the hyperfine structure is $3\times {10}^{-8}$.

Therefore, the estimated correction to the ground state energy of hydrogen due to the finite size of the nucleus is very small compared to the fine structure and the hyperfine structure.

Conclusion:

Thus, the correction to the ground state energy of hydrogen due to the finite size of the nucleus is ${10}^{-10}{E}_1$.