91Ó°ÊÓ

An electron of charge -e is orbiting around a positive charge e in a circular orbit of radius r.

The total energy of the electron is

$E_{\text{orbit}}=-\frac{e^2}{8\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}.....\left(I\right)$

Here ${\mathrm{Î\mu }}_0$is the permittivity of free space.

The power of electromagnetic radiation radiated by an electron with acceleration a is

$P=e^2{a}^2/6{\mathrm{Î\mu }}_0{c}^3.....\left(II\right)$

Here c is the speed of light in vacuum.

The Coulomb force between two charge particles of charge q and distance r is

$\mathit{F}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{r}}$ .....(III)

The centripetal force of a particle of mass m moving in circular orbit of radius r and angular velocity $\mathrm{Ó\neg }$is

$\mathit{F}\mathbf{=}\mathit{m}{\mathit{Ó\neg }}^{\mathbf{2}}\mathit{r}\mathbf{}\mathbf{}.....\left(IV\right)$

The angular acceleration of a particle undergoing uniform circular motion of radius r and angular velocity $\mathrm{Ó\neg }$ is

$\mathit{a}\mathbf{=}{\mathit{Ó\neg }}^{\mathbf{2}}\mathit{r}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}.....\left(V\right)$

Equate equations (III) for $q=e$and (IV) to get

$$\begin{gathered} \frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}r}=m{\mathrm{Ó\neg }}^{2}r \\ {\mathrm{Ó\neg }}^2=\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}m{r}^3} \\ \mathrm{Ó\neg }=\frac{e}{\sqrt{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}m}{r}^{3/2}} \end{gathered}$$

Substitute equation (V) in (II) and use form of angular velocity to get

$$\begin{gathered} P=\frac{e^2{r}^2{\mathrm{Ó\neg }}^4}{6{\mathrm{Î\mu }}_0{c}^3} \\ =\frac{e^2{r}^2}{6{\mathrm{Î\mu }}_0{c}^3}×\frac{e^4}{16{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0^2{m}^2{r}^6} \\ =\frac{e^6}{96{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0^3{m}^2{c}^3{r}^4} \end{gathered}$$

To get the energy lost per orbit multiply this by the time period $T=2\mathrm{Ï€}/\mathrm{Ó\neg }$

$$\begin{gathered} \mathrm{Δ}E=P×T \\ =\frac{e^6}{96{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0^3{m}^2{c}^3{r}^4}×\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }} \\ =\frac{e^6}{48\mathrm{Ï€}{\mathrm{Î\mu }}_0^3{m}^2{c}^3{r}^4}\frac{\sqrt{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}m}{r}^{3/2}}{e} \\ =\frac{e^5}{24\sqrt{\mathrm{Ï€}}{\mathrm{Î\mu }}_0^{5/2}{m}^{3/2}{c}^3{r}^{5/2}} \end{gathered}$$

Differentiate equation (I) with respect to r and get

$$\begin{gathered} \frac{d{E}_{\text{orbit}}}{dr}=\frac{e^2}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2} \\ dr=\frac{8\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2×d{E}_{\text{orbit}}}{e^2} \end{gathered}$$

Substitute expression of change in energy per orbit from the previous section to get

$$\begin{gathered} â–\mu r=\frac{8\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2×\mathrm{Δ}{E}_{\text{orbit}}}{e^2} \\ =\frac{8\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}{e^2}×\frac{e^5}{24\sqrt{\mathrm{Ï€}}{\mathrm{Î\mu }}_0^{5/2}{m}^{3/2}{c}^3{r}^{5/2}} \\ =\frac{\sqrt{\mathrm{Ï€}}{e}^3}{3{\mathrm{Î\mu }}_0^{3/2}{m}^{3/2}{c}^3\sqrt{r}} \end{gathered}$$

Substitute the values

$$\begin{gathered} e=1.6×{10}^{-19}\text{C} \\ {\mathrm{Î\mu }}_0=8.85×{10}^{-12}{\text{C}}^{\text{2}}\text{/N}·{\text{m}}^{\text{2}} \\ m=9.1×{10}^{-31}\text{kg} \\ r={10}^{-10}\text{m} \\ c=3×{10}^8\text{m/s} \\ andget \end{gathered}$$

$$\begin{gathered} â–\mu r=\frac{\sqrt{3.14}×{\left(1.6×{10}^{-19}\text{C}\right)}^3}{3×{\left(8.85×{10}^{-12}{\text{C}}^{\text{2}}\text{/N}·{\text{m}}^{\text{2}}\right)}^{3/2}×{\left(9.1×{10}^{-31}\text{kg}\right)}^{3/2}×{\left(3×{10}^8\text{m/s}\right)}^3×\sqrt{{10}^{-10}\text{m}}} \\ =0.0039×{10}^{-8}×\left(1{\text{C}}^3\right)·\left(\frac{1}{1{\text{C}}^3}\right)·\left(1{\text{N}}^{3/2}×\frac{1{\text{kg}}^{3/2}·{\text{m}}^{3/2}/{\text{s}}^3}{1{\text{N}}^{3/2}}\right)·\left(1{\text{m}}^3\right)·\left(\frac{1}{1{\text{kg}}^{3/2}}\right)·\left(\frac{1}{1{\text{m}}^3/s^3}\right)·\left(1{\text{m}}^{-1/2}\right) \\ =0.39×{10}^{-10}\text{m} \end{gathered}$$

Thus the change in radius per orbit is $0.39×{10}^{-10}\text{m}$.

$d{E}_{\text{orbit}}/dr$is the change in energy per unit change in radius and power is the change in energy per unit time. Dividing the two will give the inverse of change in radius of time. Thus multiplying with dr will give the time taken to change radius by dr .

The expression will be

$$\begin{gathered} dt=\frac{1}{P}\frac{d{E}_{\text{orbit}}}{dr}dr \\ =\frac{96{\mathrm{Ï€}}^2{\mathrm{Î\mu }}_0^3{m}^2{c}^3{r}^4}{e^6}\frac{e^2}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}dr \\ =\frac{12\mathrm{Ï€}{\mathrm{Î\mu }}_0^2{m}^2{c}^3}{e^4}{r}^{2}dr \end{gathered}$$

Integrate this from $r_{\text{initial}}={10}^{-10}\text{m}$to 0 to get

$$\begin{gathered} \mathrm{Δ}t=\frac{12\mathrm{Ï€}{\mathrm{Î\mu }}_0^2{m}^2{c}^3}{e^4}\left|\underset{r_{\text{initial}}}{\overset{0}{∫}}{r}^{2}dr\right| \\ =\frac{12\mathrm{Ï€}{\mathrm{Î\mu }}_0^2{m}^2{c}^3{r}_{\text{initial}}^3}{3e^4} \\ =\frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0^2{m}^2{c}^3{r}_{\text{initial}}^3}{e^4} \end{gathered}$$

Substitute the values to get

$$\begin{gathered} \mathrm{Δ}t=\frac{4\mathrm{π}{\left(8.85×{10}^{-12}{\text{C}}^{\text{2}}\text{/N}·{\text{m}}^{\text{2}}\right)}^2{\left(9.1×{10}^{-31}\text{kg}\right)}^2{\left(3×{10}^8\text{m/s}\right)}^3{\left({10}^{-10}\text{m}\right)}^3}{{\left(1.6×{10}^{-19}\text{C}\right)}^4} \\ =3.36×{10}^{-11}·\left(1{\text{C}}^4\right)·\left(\frac{1}{1{\text{N}}^2}×\frac{1{\text{N}}^{\text{2}}{\text{s}}^4}{1{\text{kg}}^{\text{2}}{\text{m}}^2}\right)·\left(\frac{1}{1{\text{m}}^4}\right)·\left(1{\text{kg}}^2\right)·\left(1{\text{m}}^{\text{3}}{\text{/s}}^3\right)·\left(1{\text{m}}^3\right)·\left(\frac{1}{1{\text{C}}^4}\right) \\ =3.36×{10}^{-11}\text{s} \end{gathered}$$

Thus the time taken is $3.36×{10}^{-11}\text{s}$.