Chapter 29, Problem 2P

(a)

To determine

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

Answer to Problem 2P

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

Explanation of Solution

Write the equation for acceleration according to Newton’s second law.

  $a=\frac{F}{m}$                                                                                           (I)

Here, $F$ is the force, $a$ is the acceleration and $m$ is the mass.

Write the equation for the force experienced by the particle moving in a uniform circular motion.

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

Here, $k_e$ is the Coulomb constant, $e$ is the electric charge and $r$ is the radius.

Write the acceleration equation using the above equation and substitute $1/4\pi {\epsilon }_0$ for $k_e$.

  $\begin{array}{c}a=\frac{k_e{e}^2}{m_e{r}^2}\\ =\frac{e^2}{4\pi {\epsilon }_0{m}_e{r}^2}\end{array}$                                                                                 (III)

Here, $m_e$ is the mass of the electron and ${\epsilon }_0$ is the permittivity of free space.

Write the equation for centripetal acceleration.

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

Here, $v$ is the velocity of the rotating particle.

Compare equations (III) and (IV) and rearrange to find $m_e{v}^2$.

  $\begin{array}{c}\frac{e^2}{4\pi {\epsilon }_0{m}_e{r}^2}=\frac{v^2}{r}\\ m_e{v}^2=\frac{e^2}{4\pi {\epsilon }_{0}r}\end{array}$                                                                     (V)

Write the equation for the total energy.

  $E=K+U$                                                                               (VI)

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

Write the equation for kinetic energy.

  $K=\frac{1}{2}{m}_e{v}^2$                                                                               (VII)

Write the equation for potential energy.

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

Substitute the equation (VII) and (VIII) in equation (VI) and use equation (V) to find E.

  $\begin{array}{c}E=\frac{1}{2}{m}_e{v}^2-\frac{e^2}{4\pi {\epsilon }_{0}r}\\ =\frac{e^2}{8\pi {\epsilon }_{0}r}-\frac{e^2}{4\pi {\epsilon }_{0}r}\\ =-\frac{e^2}{8\pi {\epsilon }_{0}r}\end{array}$                                                                      (IX)

Conclusion:

Write the time given time derivative equation for energy.

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

Substitute the equation (III) and (IX) in the above equation.

  $\begin{array}{c}\frac{d\left(-\frac{e^2}{8\pi {\epsilon }_{0}r}\right)}{dt}=\frac{-1}{6\pi {\epsilon }_0}\frac{e^2}{c^3}{\left(\frac{e^2}{4\pi {\epsilon }_0{m}_e{r}^2}\right)}^2\\ \frac{e^2}{8\pi {\epsilon }_0{r}^2}\frac{dr}{dt}=\frac{-1}{6\pi {\epsilon }_0}\frac{e^2}{c^3}{\left(\frac{e^2}{4\pi {\epsilon }_0{m}_e{r}^2}\right)}^2\\ \frac{dr}{dt}=-\frac{e^4}{12{\pi }^2{\epsilon }_0^2{m}_e^2{c}^3}\left(\frac{1}{r^2}\right)\end{array}$

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

(b)

To determine

The time interval over which the electron reaches $r=0$.

Answer to Problem 2P

The time interval over which the electron reaches $r=0$ is $0.846\text{ns}$.

Explanation of Solution

Write the equation for the rate at which the electron in classical hydrogen atom spirals into nucleus.

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

Take the integral of the time.

  $T={\displaystyle {\int }_0^{T}dt}$                                                                                         (XI)

Here, $T$ is the time interval over which the electron reaches $r=0$.

Substitute equation (X) in (XI) to find $T$.

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

Conclusion:

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

  $\begin{array}{c}T=\frac{12{\pi }^2{\left(8.85\times {10}^{-12}\text{C}\right)}^2\left(9.11\times {10}^{-31}\text{kg}\right){\left(3.00\times {10}^8\text{m/s}\right)}^3{\left(2.00\times {10}^{-10}\text{m}\right)}^3}{3{\left(1.60\times {10}^{-19}\text{C}\right)}^4}\\ =\left(0.846\times {10}^{-9}\text{s}\right)\left(\frac{{10}^9\text{ns}}{1\text{s}}\right)\\ =0.846\text{ns}\end{array}$

Thus, the time interval over which the electron reaches $r=0$ is $0.846\text{ns}$.