Chapter 23, Problem 77P

(a)

To determine

The value of ${\rho }_o$ that is needed for the hydrogen atom to be neutral

Answer to Problem 77P

The value of ${\rho }_o$ that is needed for the hydrogen atom to be neutral ${\rho }_o=-3.56\times {10}^8\text{C}/{\text{m}}^{\text{3}}$ .

Explanation of Solution

Given:

The distribution of charge density is $\rho \left(r\right)=-{\rho }_o{e}^{\frac{-2r}{a}}$ .

The value of most probable distance of electron from the proton is $a=0.\text{523}\text{nm}$ .

Formula used:

The expression for charge in a spherical shell is given by,

  $dq=\rho dV$

Calculation:

The value of ${\rho }_o$ that is needed for the hydrogen atom to be neutral is calculated as:

  $\begin{array}{c}dq=\rho dV\\ =\left(-{\rho }_o{e}^{\frac{-2r}{a}}\right)\times \left(4\pi r^{2}dr\right)\\ q={\displaystyle {\int }_0^{\infty }\left(-{\rho }_o{e}^{\frac{-2r}{a}}\right)\times \left(4\pi r^{2}dr\right)}\\ =-4\pi {\rho }_o{\displaystyle {\int }_0^{\infty }{r}^2{e}^{\frac{-2r}{a}}dr}\end{array}$

Using the integral table,

  ${\displaystyle {\int }_0^{\infty }{x}^2}{e}^{bx}dx=\frac{e^{bx}}{b^3}\left(b^2{x}^2-2bx+2\right)$

On comparing and solving further,

  ${\displaystyle {\int }_0^{\infty }{r}^2{e}^{\frac{-2r}{a}}dr}=\frac{a^3}{4}$

  $\begin{array}{c}q=-4\pi {\rho }_o\left(\frac{a^3}{4}\right)\\ 1.602\times {10}^{-19}\text{C}=-\pi {\rho }_o{\left(\text{0}\text{.523}\text{nm×}\frac{\text{1}\text{m}}{{\text{10}}^{\text{9}}\text{nm}}\right)}^3\\ {\rho }_o=-3.56\times {10}^8\text{C}/{\text{m}}^{\text{3}}\end{array}$

Conclusion:

Therefore, the value of ${\rho }_o$ that is needed for the hydrogen atom to be neutral is ${\rho }_o=-3.56\times {10}^8\text{C}/{\text{m}}^{\text{3}}$ .

(b)

To determine

The electrostatic potential as a function of the distance $r$ from the proton

Answer to Problem 77P

The electrostatic potential as a function of the distance $r$ from the proton is $ke\left(\frac{1}{a}+\frac{1}{r}\right)e^{\frac{-2r}{a}}$

Explanation of Solution

Formula used:

The expression for electrostatic potential of the proton is given by,

  $V_1=\frac{ke}{r}+\frac{k{Q}_1}{r}$

The expression for the charge of the proton is given by:

  $Q_1=\pi {\rho }_o{\displaystyle {\int }_0^r{\left(r_1\right)}^2{e}^{\frac{-2r_1}{a}}}d{r}_1$

The expression for electrostatic potential of the electron is given by,

  $V_1={\displaystyle {\int }_0^{\infty }\frac{k\rho \left(r_1\right)4\pi {\left(r_1\right)}^{2}d{r}_1}{r_1}}$

The total electrostatic potential of the system is given by:

  $V=V_1+V_2$

Calculation:

The charge of the proton is calculated as

  $Q_1=\pi {\rho }_o{\displaystyle {\int }_0^r{\left(r_1\right)}^2{e}^{\frac{-2r_1}{a}}}d{r}_1$

Using the integral table,

  ${\displaystyle {\int }_0^{\infty }{x}^2}{e}^{bx}dx=\frac{e^{bx}}{b^3}\left(b^2{x}^2-2bx+2\right)$

  $\begin{array}{c}{\displaystyle {\int }_0^r{\left(r_1\right)}^2{e}^{\frac{-2r_1}{a}}}d{r}_1=-{\frac{a^3{e}^{\frac{-2x}{a}}}{8}\left(\frac{4x^2}{a^2}+2\left(\frac{2}{a}\right)x+2\right)|}_0^r\\ =-\frac{a^3{e}^{\frac{-2r}{a}}}{8}\left(\frac{4r^2}{a^2}+2\left(\frac{2}{a}\right)r+2\right)-\left(-\frac{a^3}{8}\left(2\right)\right)\\ =-\frac{a^3{e}^{\frac{-2r}{a}}}{8}\left(\frac{4r^2}{a^2}+\left(\frac{4}{a}\right)r+2\right)+\left(\frac{a^3}{4}\right)\\ Q_1=-4\pi {\rho }_o\frac{a^3{e}^{\frac{-2r}{a}}}{8}\left(\frac{4r^2}{a^2}+\left(\frac{4}{a}\right)r+2\right)\left(\frac{a^3}{4}\right)\end{array}$

The electrostatic potential of the proton is calculated as,

  $\begin{array}{c}{V}_1=\frac{ke}{r}+\frac{k\left(-4\pi \left(-\frac{e}{\pi a^3}\right)\frac{a^3{e}^{\frac{-2r}{a}}}{8}\left(\frac{4r^2}{a^2}+\left(\frac{4}{a}\right)r+2\right)\left(\frac{a^3}{4}\right)\right)}{r}\\ =\frac{ke}{r}{e}^{\frac{-2r}{a}}\left(\frac{2r^2}{a^2}+\left(\frac{2}{a}\right)r+1\right)\end{array}$

Similarly,

  $\begin{array}{c}{V}_2={\displaystyle {\int }_0^{\infty }\frac{k{\rho }_o{e}^{\frac{-2r}{a}}4\pi {\left(r_1\right)}^{2}d{r}_1}{r_1}}\\ =4\pi k{\rho }_o{\displaystyle {\int }_0^{\infty }{e}^{\frac{-2r}{a}}\left(r_1\right)d{r}_1}\end{array}$

Using the integral table,

  ${\displaystyle {\int }_0^{\infty }{x}^2}{e}^{bx}dx=\frac{e^{bx}}{b^3}\left(bx-1\right)$

  ${\displaystyle {\int }_r^{\infty }{e}^{\frac{-2r}{a}}\left(r_1\right)d{r}_1}=\frac{-a^2}{4}{e}^{\frac{-2r}{a}}\left(\frac{2}{a}r+1\right)$

  $\begin{array}{c}{V}_2={\displaystyle {\int }_0^{\infty }\frac{k\left(\frac{-e}{\pi a^3}\right)e^{\frac{-2r}{a}}4\pi {\left(r_1\right)}^{2}d{r}_1}{r_1}}\\ =4\pi k\left(\frac{-e}{\pi a^3}\right)\frac{-a^2}{4}{e}^{\frac{-2r}{a}}\left(\frac{2}{a}r+1\right)\\ =ke\left(\frac{1}{a}\right)e^{\frac{-2r}{a}}\left(\frac{2}{a}r+1\right)\end{array}$

The net electrostatic potential is calculated as:

  $\begin{array}{c}V=V_1+V_2\\ =\frac{ke}{r}{e}^{\frac{-2r}{a}}\left(\frac{2r^2}{a^2}+\left(\frac{2}{a}\right)r+1\right)+ke\left(\frac{1}{a}\right)e^{\frac{-2r}{a}}\left(\frac{2}{a}r+1\right)\\ =ke\left(\frac{1}{a}+\frac{1}{r}\right)e^{\frac{-2r}{a}}\end{array}$

Conclusion:

Therefore,the electrostatic potential as a function of the distance $r$ from the proton is $ke\left(\frac{1}{a}+\frac{1}{r}\right)e^{\frac{-2r}{a}}$ .