In Bohr’s theory of hydrogen, the electron in its ground state was supposed to travel in a circle of radius $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{}\mathbf{m}$, held in orbit by the Coulomb attraction of the proton. According to classical electrodynamics, this electron should radiate, and hence spiral in to the nucleus. Show that$\mathit{v}\mathbf{\ll }\mathit{c}$ for most of the trip (so you can use the Larmor formula), and calculate the lifespan of Bohr’s atom. (Assume each revolution is essentially circular.)

Write the expression for the centripetal force on the electron.

${\mathit{F}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$ …… (1)

Here, m is the mass, v is the velocity of an electron, and r is the radius of the orbit.

Write the expression for the electrostatic force between the nucleus and electron.

${\mathit{F}}_{\mathbf{e}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\Pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{{\mathbf{e}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$ …… (2)

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

Equate equations (1) and (2).

$F_c=F_e$

$$\begin{gathered} \frac{m{v}^2}{r}=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r^2} \\ {v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{mr} \\ v=\sqrt{\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{mr}}.......\left(3\right) \end{gathered}$$

Substitute $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2,\mathrm{e}=1.6\times {10}^{-19}\mathrm{C},\mathrm{m}=9.11\times {10}^{-31}\mathrm{kg}$and$r=5\times {10}^{-11}m$ in the above expression.

$$\begin{gathered} v=\sqrt{\left(9\times {10}^{9}N·m^2/C^2\right)\frac{{\left(1.6\times {10}^{-19}C\right)}^2}{\left(9.11\times {10}^{-31}kg\right)\times \left(5\times {10}^{-11}m\right)}} \\ v=2249039.3\mathrm{m}/\mathrm{s} \end{gathered}$$

Divide the velocity of an electron by the speed of light.

$$\begin{gathered} \frac{v}{c}=\frac{2249039.3\mathrm{m}/\mathrm{s}}{3\times {10}^8\mathrm{m}/\mathrm{s}} \\ \frac{v}{c}=0.007497 \end{gathered}$$

For the radius of one-hundredth of, this v/c is only greater so, for most of the trip, the velocity of safely non-relativistic.

Write the expression for the total power radiated.

$P=\frac{dU}{dt}$ …… (4)

Here, U is the total energy.

Write the expression for the total energy of an orbiting electron.

$$\begin{gathered} U=U_{potential}+U_{kinetic} \\ U=-\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r}+\frac{1}{2}m{v}^2 \end{gathered}$$

Rearrange the above equation,

$$\begin{gathered} U=\frac{1}{2}m{v}^2-\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r} \\ U=\frac{1}{2}m\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{mr}\right)-\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r} \\ U=\frac{1}{8\pi {\epsilon }_0}\frac{q^2}{r}-\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r} \\ U=-\frac{1}{8\pi {\epsilon }_0}\frac{q^2}{r} \end{gathered}$$

Differentiate the above equation,

$$\begin{gathered} \frac{dU}{dt}=\frac{d}{dt}\left[-\frac{1}{8\pi {\epsilon }_0}\frac{q^2}{r}\right] \\ \frac{dU}{dt}=-\frac{q^2}{8\pi {\epsilon }_0}\left(-\frac{1}{r^2}\frac{dr}{dt}\right) \\ \frac{dU}{dt}=\frac{q^2}{8\pi {\epsilon }_0}\frac{1}{r^2}\frac{dr}{dt} \end{gathered}$$

Using the Larmor formula,

$$\begin{gathered} P=\frac{{\mu }_0{q}^2}{6\pi c}{a}^2\left(\therefore a=\frac{v^2}{r}\right) \\ P=\frac{{\mu }_0{q}^2}{6\pi c}{\left(\frac{v^2}{r}\right)}^2 \\ P=\frac{{\mu }_0{q}^2}{6\pi c{r}^2}{\left(v^2\right)}^2 \end{gathered}$$

Substitute the value of equation (3) in the above equation.

$$\begin{gathered} P=\frac{{\mu }_0{q}^2}{6\pi c{r}^2}{\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{mr}\right)}^2 \\ P=\frac{{\mu }_0{q}^2}{6\pi c}{\left(\frac{1}{4m{r}^2}\right)}^2 \end{gathered}$$

Substitute$P=\frac{{\mu }_0{q}^2}{6\pi c}{\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{m{r}^2}\right)}^2$and$\frac{dU}{dt}=\frac{q^2}{8\pi {\epsilon }_0}\frac{1}{r^2}\frac{dr}{dt}$in equation (4).

$\frac{{\mu }_0{q}^2}{6\pi c}{\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{m{r}^2}\right)}^2=-\frac{q^2}{8\pi {\epsilon }_0}\frac{1}{r^2}\frac{dr}{dt}$

Here, ${\mu }_0=\frac{1}{c^2{\epsilon }_0}$.

Hence, the above equation becomes,

$$\begin{gathered} \frac{\left({\displaystyle \frac{1}{c^2{\epsilon }_0}}\right)q^2}{6\pi c}{\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{m{r}^2}\right)}^2=-\frac{q^2}{8\pi {\epsilon }_0}\frac{1}{r^2}\frac{dr}{dt} \\ \frac{1}{6\pi {\epsilon }_{0}c}{\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{cm{r}^2}\right)}^2=-\frac{1}{8\pi {\epsilon }_0}\frac{1}{r^2}\frac{dr}{dt} \\ \frac{dr}{dt}=-\frac{1}{6\pi {\epsilon }_{0}c}{\left(\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{cm{r}^2}\right)}^2\left(8\pi {\epsilon }_0{r}^2\right) \\ \frac{dr}{dt}=-\frac{1}{3c}{\left(\frac{q^2}{2\pi {\epsilon }_{0}mc}\right)}^2\frac{1}{r^2} \end{gathered}$$

On further solving, the above equation becomes,

$dt=-3c{\left(\frac{2\pi {\epsilon }_{0}mc}{q^2}\right)}^2{r}^{2}dr$

Integrate the above equation,

$$\begin{gathered} t=-3c{\left(\frac{2\pi {\epsilon }_{0}mc}{q^2}\right)}^2{\int }_{r_0}^0{r}^{2}dr \\ t=-3c{\left(\frac{2{\pi }_{0}mc}{q^2}\right)}^2{\left[\frac{r^3}{3}\right]}_{r_0}^0 \\ t=-3c{\left(\frac{2\pi {\epsilon }_{0}mc}{q^2}\right)}^2\left(\frac{r_0^3}{3}\right) \end{gathered}$$

Substitute $c=3\times {10}^{8}m/s,{\epsilon }_0=8.85\times {10}^{-12}{C}^2/N·m^2,m=9.11\times {10}^{-31}kg$and$e=1.6\times {10}^{-19}C$in the above expression.

$$\begin{gathered} t=\left[\begin{array}{c}-3\left(3\times {10}^{8}m/s\right){\left(\frac{2\pi \left(8.85\times {10}^{-12}{C}^2/N·m^2\right)\left(9.11\times {10}^{-31}kg\right)\left(3\times {10}^{8}m/s\right)}{{\left(1.6\times {10}^{-19}C\right)}^2}\right)}^2\\ \left(\frac{{\left(5\times {10}^{-11}m\right)}^3}{3}\right)\end{array}\right] \\ t=1.3\times {10}^{-11}s \end{gathered}$$

Therefore, the lifespan of the Bohr atom is $1.3\times {10}^{-11}s$.