91Ó°ÊÓ

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.

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

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

Equate equations (1) and (2).

$F_c=F_e$

$$\begin{gathered} \frac{m{v}^2}{r}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{e^2}{r^2} \\ {v}^2=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{e^2}{mr} \\ v=\sqrt{\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{e^2}{mr}}.......\left(3\right) \end{gathered}$$

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

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

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

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

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.

$$\begin{gathered} U=U_{potential}+U_{kinetic} \\ U=-\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{r}+\frac{1}{2}m{v}^2 \end{gathered}$$

Rearrange the above equation,

$$\begin{gathered} U=\frac{1}{2}m{v}^2-\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{r} \\ U=\frac{1}{2}m\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{mr}\right)-\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{r} \\ U=\frac{1}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{r}-\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{r} \\ U=-\frac{1}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{r} \end{gathered}$$

Differentiate the above equation,

$$\begin{gathered} \frac{dU}{dt}=\frac{d}{dt}\left[-\frac{1}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{r}\right] \\ \frac{dU}{dt}=-\frac{q^2}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(-\frac{1}{r^2}\frac{dr}{dt}\right) \\ \frac{dU}{dt}=\frac{q^2}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2}\frac{dr}{dt} \end{gathered}$$

Using the Larmor formula,

$$\begin{gathered} P=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{π}c}{a}^2\left(∴a=\frac{v^2}{r}\right) \\ P=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{π}c}{\left(\frac{v^2}{r}\right)}^2 \\ P=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{π}c{r}^2}{\left(v^2\right)}^2 \end{gathered}$$

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

$$\begin{gathered} P=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{Ï€}c{r}^2}{\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{mr}\right)}^2 \\ P=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{Ï€}c}{\left(\frac{1}{4m{r}^2}\right)}^2 \end{gathered}$$

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

$\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{Ï€}c}{\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q^2}{m{r}^2}\right)}^2=-\frac{q^2}{8\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{1}{r^2}\frac{dr}{dt}$

Here, ${\mathrm{μ}}_0=\frac{1}{c^2{\mathrm{Î\mu }}_0}$.

Hence, the above equation becomes,

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

On further solving, the above equation becomes,

$dt=-3c{\left(\frac{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}mc}{q^2}\right)}^2{r}^{2}dr$

Integrate the above equation,

$$\begin{gathered} t=-3c{\left(\frac{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}mc}{q^2}\right)}^2{∫}_{r_0}^0{r}^{2}dr \\ t=-3c{\left(\frac{2{\mathrm{Ï€}}_{0}mc}{q^2}\right)}^2{\left[\frac{r^3}{3}\right]}_{r_0}^0 \\ t=-3c{\left(\frac{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}mc}{q^2}\right)}^2\left(\frac{r_0^3}{3}\right) \end{gathered}$$

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

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

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