Chapter 4, Problem 18P

To determine

The time taken by the electron of the hydrogen atom to radiate its energy and crash into the nucleus according to classical laws.

Answer to Problem 18P

The time taken by the electron of the hydrogen atom to radiate its energy and crash into the nucleus is $1.56\times {10}^{-11}\text{s}$.

Explanation of Solution

Write the expression for $E_0$.

    $E_0=-\frac{e^2}{8\pi {\epsilon }_0{a}_0}$

Here, $E_0$ is the Energy, $a_0$ is the Bohr’s radius and $e$ is the charge in electron.

Differentiate above equation with respect to time.

    $\frac{d{E}_0}{dt}=-\frac{e^2}{8\pi {\epsilon }_0{r}^2}\frac{dr}{dt}$        (I)

Write the expression for radiated Power in electromagnetic theory.

    $P=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2Q^2}{3c^3}\right){\left(\frac{d^2\overrightarrow{r}}{d{t}^2}\right)}^2$        (II)

Here, $P$ is the power radiated, $Q$  is the charge, $c$ is the speed of light, $t$ is the time and $\overrightarrow{r}$ is the position vector of the electron from the center of the atom.

Power radiated is equal to Energy radiated per unit time.

Compare equation (I) with equation (II).

  $-\frac{e^2}{8\pi {\epsilon }_0{r}^2}\frac{dr}{dt}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2Q^2}{3c^3}\right){\left(\frac{d^2\overrightarrow{r}}{d{t}^2}\right)}^2$

Simplify above equation.

    $-\frac{e^2}{2r^2}\frac{dr}{dt}=\left(\frac{2Q^2}{3c^3}\right){\left(\frac{d^2\overrightarrow{r}}{d{t}^2}\right)}^2$        (III)

In a circular orbit, centripetal acceleration is given by: $\frac{d^{2}r}{d{t}^2}$ .

Write the expression for centripetal acceleration.

    $a=\frac{F}{m}$        (IV)

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

Write the expression for force.

    $F=\frac{e^2}{4\pi {\epsilon }_0{r}^2}$        (V)

Here, $F$ is the force.

Substitute  $\frac{e^2}{4\pi {\epsilon }_0{r}^2}$ for $F$ in equation (IV).

    $a=\frac{e^2}{4\pi {\epsilon }_{0}m{r}^2}$

Substitute $\frac{e^2}{4\pi {\epsilon }_{0}m{r}^2}$ for $\frac{d^2\overrightarrow{r}}{d{t}^2}$ in equation (III).

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

Rearrange above equation.

    $dt=-\frac{{\left(4\pi {\epsilon }_0\right)}^{2}3c^3{m}^2{r}^2}{4e^4}dr$

Integrate above equation.

    $\begin{array}{c}{\displaystyle \int dt}=-\frac{{\left(4\pi {\epsilon }_0\right)}^{2}3c^3{m}^2}{4e^4}{\displaystyle \underset{a_0}{\overset{0}{\int }}{r}^{2}dr}\\ t=-\frac{{\left(4\pi {\epsilon }_0\right)}^{2}3c^3{m}^2}{4e^4}{\left[\frac{r^3}{3}\right]}_{a_0}^0\\ =\frac{{\left(4\pi {\epsilon }_0\right)}^2{c}^3{m}^2{a}_0^3}{4e^4}\end{array}$        (VI)

Conclusion:

Substitute $9\times {10}^9$ for $\frac{1}{4\pi {\epsilon }_0}$ , $1.6\times {10}^{-19}$ for $e$ , $3\times {10}^8\text{m/s}$ for $c$,  $9.1\times {10}^{-31}$ for $m$  and $5.29\times {10}^{-11}\text{m}$ for $a_0$ in

    $\begin{array}{c}t=\frac{{\left(3\times {10}^8\text{m/s}\right)}^3{\left(9.1\times {10}^{-31}\right)}^2{\left(5.29\times {10}^{-11}\text{m}\right)}^3}{{\left(9\times {10}^9\right)}^{2}4{\left(1.6\times {10}^{-19}\right)}^4}\\ =1.56\times {10}^{-11}\text{s}\end{array}$

Thus, the time taken by the electron of the hydrogen atom to radiate its energy and crash into the nucleus is $1.56\times {10}^{-11}\text{s}$.