Chapter 42, Problem 6P

(a)

To determine

Show that an electron in a classical hydrogen atom spirals into the nucleus at a rate of $\frac{dr}{dt}=-\frac{e^4}{12{\pi }^2{\epsilon }_0^2{m}_e^2{c}^3}\left(\frac{1}{r^2}\right)$.

Answer to Problem 6P

An electron in a classical hydrogen atom spirals into the nucleus at a rate of $\frac{dr}{dt}=-\frac{e^4}{12{\pi }^2{\epsilon }_0^2{m}_e^2{c}^3}\left(\frac{1}{r^2}\right)$.

Explanation of Solution

The uniform circular motion of the electron as a particle about the proton in the hydrogen atom experiences a force which can be expressed as,

    $F=\frac{k_e{e}^2}{r^2}$                                                                                                                  (I)

Here, $k_e$ is the Coulomb constant, $e$ is the charge of electron, $r$ radius of the orbit.

The force can be expressed in terms of Newton’s second law,

    $F=ma$                                                                                                                   (II)

Here, $m$ is the mass of the particle, $a$ is the acceleration.

Use equation (I) in equation (II) and solve for $a$.

    $a=\frac{k_e{e}^2}{m_e{r}^2}$                                                                                                               (III)

Write the expression for the centripetal acceleration.

    $a=\frac{v^2}{r}$                                                                                                        (IV)

Here, $v$ is the speed.

Use equation (II) in equation (IV),

    $\frac{v^2}{r}=\frac{F}{m_e}$                                                                                                             (V)

The value of the $k_e$ is $\frac{1}{4\pi {\epsilon }_0}$, and use equation (I) in equation (V).

    $m_e{v}^2=\frac{e^2}{4\pi {\epsilon }_{0}r}$                                                                                             (VI)

Write the expression for the total energy,

    $E=K+U$                                                                                                 (VII)

Here, $E$ is the total energy, $K$ is the kinetic energy, $U$ is the potential energy.

Write the expression for the kinetic energy.

    $K=\frac{m_e{v}^2}{2}$                                                                                                  (VIII)

Write the expression for the potential energy.

    $U=-\frac{e^2}{4\pi {\epsilon }_{0}r}$                                                                                                     (IX)

Use equation (VIII) and (IX) in equation (VII),

    $E=\frac{m_e{v}^2}{2}-\frac{e^2}{4\pi {\epsilon }_{0}r}$                                                                                          (X)

Use equation (VI) in equation (X),

    $E=-\frac{e^2}{8\pi {\epsilon }_{0}r}$                                                                                               (XI)

The given expression connecting $E\text{and }a$ with $\frac{dE}{dt}$ is given by,

    $\frac{dE}{dt}=-\frac{1}{6\pi {\epsilon }_0}\frac{e^2{a}^2}{c^3}$                                                                                           (XII)

Use equation (XI) and equation (III) in equation (XII),

    $\begin{array}{c}\frac{e^2}{8\pi {\epsilon }_0{r}^2}\frac{dr}{dt}=\frac{-e^2}{6\pi {\epsilon }_0{c}^3}{\left(\frac{e^2}{4\pi {\epsilon }_0{r}^2{m}_e}\right)}^2\\ \frac{dr}{dt}=-\frac{e^4}{12{\pi }^2{\epsilon }_0^2{m}_e^2{c}^3}\left(\frac{1}{r^2}\right)\end{array}$                                                                (XIII)

Conclusion:

Therefore, from equation (XIII) it is shown that an electron in a classical hydrogen atom spirals into the nucleus at a rate as $\frac{dr}{dt}=-\frac{e^4}{12{\pi }^2{\epsilon }_0^2{m}_e^2{c}^3}\left(\frac{1}{r^2}\right)$.

(b)

To determine

The time interval over which the electron reaches $r=0$ starting from $r_0=2.00\times {10}^{-10}\text{m}$.

Answer to Problem 6P

The time interval over which the electron reaches $r=0$ starting from $r_0=2.00\times {10}^{-10}\text{m}$ is $\underset{\_}{0.846\text{ns}}$.

Explanation of Solution

Write the expression for the time interval in terms of $dt$.

    $T={\displaystyle \underset{0}{\overset{T}{\int }}dt}$                                                                                                              (XIV)

Solve equation (XIII) for $dt$ and Use in equation (XIV) and on integrating,

    $\begin{array}{c}T={\displaystyle \underset{0}{\overset{2.00\times {10}^{-10}\text{m}}{\int }}\frac{12{\pi }^2{\epsilon }_0^2{r}^2{m}_e^2{c}^3}{e^4}}dr\\ ={\frac{12{\pi }^2{\epsilon }_0^2{m}_e^2{c}^3}{e^4}\frac{r^3}{3}|}_0^{2.00\times {10}^{-10}\text{m}}\end{array}$                                                                         (XV)

Conclusion:

Substitute $8.85\times {10}^{-12}\text{C}$ for ${\epsilon }_0$ , $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $3.00\times {10}^8\text{m}/\text{s}$ for $c$ and $1.60\times {10}^{-19}\text{C}$ for $e$ in equation (XV) to find $T$.

    $\begin{array}{c}T=\frac{12{\pi }^2\left(8.85\times {10}^{-12}\text{C}\right)\left(9.11\times {10}^{-31}\text{kg}\right){\left(3.00\times {10}^8\text{m}/\text{s}\right)}^3}{\left(1.60\times {10}^{-19}\text{C}\right)}\frac{{\left(2.00\times {10}^{-10}\text{m}\right)}^3}{3}\\ =8.46\times {10}^{-10}\text{s}\times \frac{1\text{ns}}{1\times {10}^{-9}\text{s}}\\ =0.846\text{ns}\\ \end{array}$

Therefore, the time interval over which the electron reaches $r=0$ starting from $r_0=2.00\times {10}^{-10}\text{m}$ is $\underset{\_}{0.846\text{ns}}$.