Question

Evaluate $\mathbf{D}$and $\mathbf{X}$(Equations and ). Check your answers against Equations $7.47$and $7.48$.

Short answer

Equations $7.45$and $7.46$matches the two integrals, and the results are the same.

Step-by-step solution

Definition of integral

An integral is a number that is assigned to a function in order to explain displacement, area, volume, and other notions that occur when infinitesimal data is combined.

Integration is the term used to describe the process of locating integrals.

Evaluate  D and X

Calculate two integrals:

$\begin{array}{l}D=a⟨{\psi }_0\left(r_1\right)\left|\frac{1}{r_2}\right|{\psi }_0\left(r_1\right)⟩\\ X=a⟨{\psi }_0\left(r_1\right)\left|\frac{1}{r_1}\right|{\psi }_0\left(r_2\right)⟩\end{array}$

where${\psi }_0\left(r\right)=\frac{e^{-r/a}}{\sqrt{\pi a^3}}$, and$r_1=r\text{\hspace{0.17em}and\hspace{0.17em}}{r}_2=\sqrt{r^2+R^2-2rR\cos \theta }$

The first integral is,

$\begin{array}{c}D=\frac{1}{\pi a^2}\int e^{-r/a}\frac{1}{r_2}{e}^{-r/a}{r}^{2}dr\sin \theta d\theta d\phi \\ =\frac{2}{a^2}{\int }_0^{\infty }{r}^2{e}^{-2r/a}dr{\int }_0^{\pi }\frac{\sin \theta d\theta }{\sqrt{r^2+R^2-2rR\cos \theta }}\end{array}$

As,

$\begin{array}{l}u=r^2+R^2-2rR\cos \theta \\ du=2rR\sin \theta \end{array}$

The limits will be

$0\to {(r-R)}^2\pi \to {(r+R)}^2$

$\begin{array}{c}D=\frac{1}{R{a}^2}{\int }_0^{\infty }r{e}^{-2r/a}dr{\int }_{{(r-R)}^2}^{{(r+R)}^2}\frac{du}{\sqrt{u}}\\ =\frac{2}{R{a}^2}{\int }_0^{\infty }r{e}^{-2r/a}[r+R-|r-R|]dr\\ =\frac{2}{R{a}^2}[{\int }_0^{R}r{e}^{-2r/a}(r+R-R+r)dr+{\int }_R^{\infty }r{e}^{-2r/a}(r+R-r+R)dr]\\ =\frac{2}{R{a}^2}[2{\int }_0^R{r}^2{e}^{-2r/a}dr+2R{\int }_R^{\infty }r{e}^{-2r/a}dr]\end{array}$

Further, evaluate and get,

$\begin{array}{c}D=\frac{4}{R{a}^2}\{\frac{a^3}{8}[e^{-2r/a}(-\frac{4R^2}{a^2}-\frac{4R}{a}-2)+2]+R\frac{a^2}{4}{e}^{-2R/a}(\frac{2R}{a}+1)\}\\ =\frac{4}{R{a}^2}\cdot \frac{a^2}{4}\{\frac{a}{2}{e}^{-2R/a}(-\frac{4R^2}{a^2}-\frac{4R}{a}-2)+a+R{e}^{-2R/a}(\frac{2R}{a}+1)\}\\ =\frac{1}{R}\{-2R{e}^{-2R/a}-a{e}^{-2R/a}+a+R{e}^{-2R/a}\}\\ =\frac{a}{R}-\frac{a}{R}{e}^{-2R/a}-e^{-2R/a}\\ D=\frac{a}{R}-(1+\frac{a}{R})e^{-2R/a}\end{array}$

The Second integral

Now, the second integral,

$\begin{array}{c}X=\frac{1}{\pi a^2}\int e^{-r/a}\frac{1}{r}{e}^{-r_2/a}{r}^{2}dr\sin \theta d\theta d\phi \\ =\frac{2}{a^2}{\int }_0^{\infty }r{e}^{-r/a}dr{\int }_0^{\pi }\exp (-\frac{\sqrt{r^2+R^2-2rR\cos \theta }}{a})\sin \theta d\theta \end{array}$

As,

$\begin{array}{l}u=r^2+R^2-2rR\cos \theta \\ du=2rR\sin \theta \end{array}$

Thus, The limits will be $0\to {(r-R)}^2\pi \to {(r+R)}^2$

Use this in the integral and get,

$\begin{array}{c}X=\frac{1}{R{a}^2}{\int }_0^{\infty }{e}^{-r/a}dr{\int }_{{(r-R)}^2}^{{(r+R)}^2}{e}^{-\sqrt{u}/a}du\\ =\frac{1}{R{a}^2}{\int }_0^{\infty }{e}^{-r/a}2a^2[e^{-|r-R|/a}(\frac{|r-R|}{a}+1)-e^{-R/a}{e}^{-r/a}(\frac{r}{a}+\frac{R}{a}+1)]dr\end{array}$

Separate the integral, which containsthe absolute value

Now separate the integral, which contains the absolute value, into two parts:

$\begin{array}{c}X=\frac{2}{R}\{{\int }_0^{\infty }{e}^{-r/a}{e}^{-|r-R|/a}(\frac{|r-R|}{a}+1)-e^{-R/a}{\int }_0^{\infty }{e}^{-2r/a}(\frac{r}{a}+\frac{R}{a}+1)dr\}\\ =\frac{2}{R}\{{\int }_0^R{e}^{-r/a}{e}^{-R/a}{e}^{r/a}(\frac{R}{a}-\frac{r}{a}+1)dr+{\int }_R^{\infty }{e}^{-r/a}{e}^{R/a}{e}^{-r/a}(-\frac{R}{a}+\frac{r}{a}+1)dr\\ -e^{-R/a}{\int }_0^{\infty }[\frac{r}{a}{e}^{-2r/a}+(\frac{R}{a}+1)e^{-2r/a}]dr\}\\ =\frac{2e^{-R/a}}{R}\{\frac{R^2}{a}-\frac{R^2}{2a}+R+e^{2R/a}{\int }_R^{\infty }{e}^{-2r/a}(-\frac{R}{a}+\frac{r}{a}+1)dr-\frac{1}{a}\frac{a^2}{4}-(\frac{R}{a}+1)\frac{a}{2}\}\\ =\frac{2e^{-R/a}}{R}\{\frac{R^2}{2a}+R+e^{2R/a}\left[\frac{1}{a}\frac{a}{4}{e}^{-2R/a}(a+2R)+(1-\frac{R}{a})\frac{a}{2}{e}^{-2R/a}\right]-\frac{a}{4}-\frac{a}{2}(\frac{R}{a}+1)\}\end{array}$

Further evaluate as:

$\begin{array}{c}=\frac{2e^{-R/a}}{R}\left\{\frac{R^2}{2a}+R+\frac{1}{4}(a+2R)+\frac{a}{2}(1-\frac{R}{a})-\frac{a}{4}-\frac{a}{2}(\frac{R}{a}+1)\right\}\\ =\frac{2e^{-R/a}}{R}(\frac{R^2}{2a}+\frac{R}{2})\\ X=e^{-R/a}(1+\frac{R}{a})\end{array}$

Thus, after calculating the two integrals, it is found that both yield the same outcome as equations $7.45$and $7.46$.