Question

(a) Calculate$<\left(1/\left|r_1-r_2\right|\right)>$for the state${\psi }_0$(Equation 5.30). Hint: Do$d^3{r}_2$integral

first, using spherical coordinates, and setting the polar axis along$r_1$, so

that

${\psi }_0\left(r_1,r_2\right)=\psi 100\left(r_1\right)\psi 100\left(r_2\right)=\frac{8}{{\mathrm{\pi a}}^3}{e}^{-2\left(r_1+r_2\right)/a}$(5.30).

$\left|{\mathrm{r}}_1-{\mathrm{r}}_2\right|=\sqrt{{\mathrm{r}}_1^2+{\mathrm{r}}_2^2-2{\mathrm{r}}_1{\mathrm{r}}_2{\mathrm{cos\theta }}_2}.$

The${\mathrm{\theta }}_2$integral is easy, but be careful to take the positive root. You’ll have to

break the$r_2$integral into two pieces, one ranging from 0 to$r_1$,the other from$r_{1}to\infty$

Answer: 5/4a.

(b) Use your result in (a) to estimate the electron interaction energy in the ground state of helium. Express your answer in electron volts, and add it to$E_0$(Equation 5.31) to get a corrected estimate of the ground state energy. Compare the experimental value. (Of course, we’re still working with an approximate wave function, so don’t expect perfect agreement.)

$E_0=8\left(-13.6eV\right)=-109eV$(5.31).

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)=\frac{32}{a^4}\left\{a.{\left(\frac{a}{4}\right)}^2-2.2{\left(\frac{a}{8}\right)}^3-a.{\left(\frac{a}{8}\right)}^2\right\}=\frac{32}{a}\left(\frac{1}{16}-\frac{1}{128}-\frac{1}{64}\right)=\frac{5}{4a}. \\ \left(\mathrm{b}\right)V_{ee}\approx \frac{e^2}{4\mathrm{\pi }\infty \text{'}}<\frac{1}{\left|r_1-r_2\right|}>=\frac{5}{4}\frac{e^2}{4\mathrm{\pi }\infty \text{'}}\frac{1}{a}=\frac{5m}{4{\sout{h}}^2}{\left(\frac{e^2}{4\mathrm{\pi }\infty \text{'}}\right)}^2=\frac{5}{2}\left(-E_1\right)=\frac{5}{2}\left(13.6eV\right) \end{gathered}$$

Step-by-step solution

(a) Calculating <1/r1-r2>for the state ψ0

$$\begin{gathered} <\frac{1}{\left|r_1-r_2\right|}>={\left(\frac{8}{{\mathrm{\pi a}}^3}\right)}^2\int \underset{♦}{\underbrace{\left[\int \frac{e^{-4\left(r_1+r_2\right)/a}}{\sqrt{r_2^1+r_2^2-2r_1{r}_2\cos {\theta }_2}}{d}^3{r}_2\right]d^3}}{r}_1. \\ ♦=2\mathrm{\pi }{\int }_0^{\infty }{\mathrm{e}}^{-4\left({\mathrm{r}}_1+{\mathrm{r}}_2\right)/\mathrm{a}}\underset{\stackrel{\circ }{\mathrm{a}}}{\underbrace{\left[{\int }_0^{\mathrm{\pi }}\frac{{\mathrm{sin\theta }}_2}{\sqrt{{\mathrm{r}}_2^1+{\mathrm{r}}_2^2-2{\mathrm{r}}_1{\mathrm{r}}_2{\mathrm{cos\theta }}_2}}{\mathrm{d\theta }}_2\right]{\mathrm{r}}_2^2\mathrm{d}}}{\mathrm{r}}_2.\mathrm{x} \end{gathered}$$

$$\begin{gathered} \stackrel{\circ }{\mathrm{a}=\frac{1}{{\mathrm{r}}_1{\mathrm{r}}_2}}\sqrt{{\mathrm{r}}_1^2+{\mathrm{r}}_2^2-2{\mathrm{r}}_1{\mathrm{r}}_2{\mathrm{cos\theta }}_2}{|}_0^{\mathrm{\pi }}=\frac{1}{{\mathrm{r}}_1{\mathrm{r}}_2}\left[\sqrt{{\mathrm{r}}_1^2+{\mathrm{r}}_2^2+2{\mathrm{r}}_1{\mathrm{r}}_2}-\sqrt{{\mathrm{r}}_1^2+{\mathrm{r}}_2^2+2{\mathrm{r}}_1{\mathrm{r}}_2}\right]. \\ =\frac{1}{{\mathrm{r}}_1{\mathrm{r}}_2}\left[\left({\mathrm{r}}_1+{\mathrm{r}}_2\right)-\left|{\mathrm{r}}_1-{\mathrm{r}}_2\right|\right]=\left\{\begin{array}{l}2/{\mathrm{r}}_1\left({\mathrm{r}}_2<{\mathrm{r}}_1\right)\\ 2/{\mathrm{r}}_2\left({\mathrm{r}}_2>{\mathrm{r}}_1\right)\end{array}\right.. \\ ♦=4{\mathrm{\pi e}}^{-4{\mathrm{r}}_1/\mathrm{a}}\left[\frac{1}{{\mathrm{r}}_1}{\int }_0^{{\mathrm{r}}_1}{\mathrm{r}}_2^2{\mathrm{e}}^{-4{\mathrm{r}}_2/\mathrm{a}}{\mathrm{dr}}_2+{\int }_{{\mathrm{r}}_1}^{\infty }{\mathrm{r}}_2{\mathrm{e}}^{-4{\mathrm{r}}_2/\mathrm{a}}{\mathrm{dr}}_2\right]. \\ \frac{1}{{\mathrm{r}}_1}{\int }_0^{{\mathrm{r}}_1}{\mathrm{r}}_2^2{\mathrm{e}}^{-4{\mathrm{r}}_1/\mathrm{a}}{\mathrm{dr}}_2=\frac{1}{{\mathrm{r}}_1}\left[-\frac{\mathrm{a}}{4}{\mathrm{r}}_2^2{\mathrm{e}}^{-4{\mathrm{r}}_2/\mathrm{a}}+\frac{\mathrm{a}}{2}{\left(\frac{\mathrm{a}}{4}\right)}^2{\mathrm{e}}^{-4{\mathrm{r}}_2/\mathrm{a}}\left(-\frac{4{\mathrm{r}}_2}{\mathrm{a}}-1\right)\right]{|}_0^{{\mathrm{r}}_1} \\ =-\frac{\mathrm{a}}{4{\mathrm{r}}_1}\left[{\mathrm{r}}_1^2{\mathrm{e}}^{-4{\mathrm{r}}_1/\mathrm{a}}+\frac{{\mathrm{ar}}_1}{2}{\mathrm{e}}^{-4{\mathrm{r}}_1/\mathrm{a}}+\frac{{\mathrm{a}}^2}{8}{\mathrm{e}}^{-4{\mathrm{r}}_1/\mathrm{a}}-\frac{{\mathrm{a}}^2}{8}\right]. \\ {\int }_{{\mathrm{r}}_1}^{\infty }{\mathrm{r}}_2{\mathrm{e}}^{-4{\mathrm{r}}_2/\mathrm{a}}{\mathrm{dr}}_2={\left(\frac{\mathrm{a}}{4}\right)}^2{\mathrm{e}}^{-4{\mathrm{r}}_2/\mathrm{a}}\left(-\frac{4{\mathrm{r}}_2}{\mathrm{a}}-1\right){|}_{{\mathrm{r}}_1}^{\infty }=\frac{{\mathrm{ar}}_1}{4}{\mathrm{e}}^{-4{\mathrm{r}}_1/\mathrm{a}}+\frac{{\mathrm{a}}^2}{16}{\mathrm{e}}^{-4{\mathrm{r}}_1/\mathrm{a}} \\ =4\mathrm{\pi }\left\{\frac{{\mathrm{a}}^3}{32{\mathrm{r}}_1}{\mathrm{e}}^{-4{\mathrm{r}}_1/\mathrm{a}}+\left[-\frac{\mathrm{ar}}{4}-\frac{{\mathrm{a}}^2}{8}-\frac{{\mathrm{a}}^3}{32{\mathrm{r}}_1}+\frac{{\mathrm{ar}}_1}{4}+\frac{{\mathrm{a}}^2}{16}\right]{\mathrm{e}}^{-8{\mathrm{r}}_1/\mathrm{a}}\right\} \end{gathered}$$

$=\frac{{\mathrm{\pi a}}^2}{8}\left\{\frac{a}{r_1}{e}^{-4r_1/a}-\left(2+\frac{a}{r_1}\right)e^{-8r_1/a}\right\}.$

$$\begin{gathered} <\frac{1}{\left|r_1-r_2\right|}>=\frac{8}{{\mathrm{\pi a}}^4}.4\mathrm{\pi }{\int }_0^{\infty }\left[\frac{\mathrm{a}}{{\mathrm{r}}_1}{\mathrm{e}}^{-4{\mathrm{r}}_1/\mathrm{a}}-\left(2+\frac{\mathrm{a}}{{\mathrm{r}}_1}\right){\mathrm{e}}^{-8{\mathrm{r}}_1/\mathrm{a}}\right]{\mathrm{r}}_1^2{\mathrm{dr}}_{1.} \\ =\frac{32}{{\mathrm{a}}^4}\left\{\mathrm{a}.{\int }_0^{\infty }{\mathrm{r}}_1{\mathrm{e}}^{-4{\mathrm{r}}_1/\mathrm{a}}{\mathrm{dr}}_1-2{\int }_0^{\infty }{\mathrm{r}}_1^2{\mathrm{e}}^{-8{\mathrm{r}}_1/\mathrm{a}}{\mathrm{dr}}_1-\mathrm{a}{\int }_0^{\infty }{\mathrm{r}}_1{\mathrm{e}}^{-8{\mathrm{r}}_1/\mathrm{a}}{\mathrm{dr}}_1\right\}. \\ =\frac{32}{{\mathrm{a}}^4}\left\{\mathrm{a}.{\left(\frac{\mathrm{a}}{4}\right)}^2-2.2{\left(\frac{\mathrm{a}}{8}\right)}^3-\mathrm{a}.{\left(\frac{\mathrm{a}}{8}\right)}^2\right\}=\frac{32}{\mathrm{a}}\left(\frac{1}{16}-\frac{1}{128}-\frac{1}{64}\right)=\frac{5}{4\mathrm{a}}. \end{gathered}$$

(b) Estimating the electron interaction energy

$$\begin{gathered} V^{ee}\approx \frac{e^2}{4\mathrm{\pi }\infty \text{'}}<\frac{1}{\left|r_1-r_2\right|}>=\frac{5}{4}\frac{e^2}{4\mathrm{\pi }\infty \text{'}}\frac{1}{a}=\frac{5}{4}\frac{m}{\sout{h}}{\left(\frac{e^2}{4\mathrm{\pi }\infty \text{'}}\right)}^2=\frac{5}{2}\left(-E_1\right)=\frac{5}{2}\left(13.6eV\right). \\ {E}_0+V_{ee}=\left(-109+34\right)eV=-75eV,whichisprettyclosetotheexperimentalvalue-79eV). \end{gathered}$$