An electron is released from rest and falls under the influence of gravity. In the first centimeter, what fraction of the potential energy lost is radiated away?

Write the fraction of the potential energy lost is radiated.

$\mathit{f}\mathbf{=}\frac{{\mathbf{U}}_{\mathbf{r}\mathbf{a}\mathbf{d}}}{{\mathbf{U}}_{\mathbf{p}\mathbf{o}\mathbf{t}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{i}\mathbf{a}\mathbf{l}}}$ …… (1)

Here,$U_{rad}$ is the radiated energy and$U_{potential}$ is the potential energy.

Write the expression for the radiated energy.

$U_{rad}=P\times t$ …… (2)

Here, P is the power, and t is the time.

Let the distance travelled by the electron be y, and its initial velocity be $u\left(u=0\right)$.

If an electron travels a distance y in a time t, express the required equation.

$$\begin{gathered} y=\frac{1}{2}g{t}^2 \\ t=\sqrt{\frac{2y}{g}} \end{gathered}$$

Here, g is the gravitational acceleration.

Write the expression for the total radiated power.

$P=\frac{{\mu }_0}{6\pi c}{\left(\ddot{p}\left(t\right)\right)}^2$ …… (3)

Here, p is the dipole moment which is given as:

$$\begin{gathered} p=-ey\hat{y} \\ y=-\frac{p}{e}\hat{y} \\ \frac{1}{2}g{t}^2=-\frac{p}{e}\hat{y} \\ p=\frac{-1}{2}ge{t}^2\hat{y}.......\left(4\right) \end{gathered}$$

Take the double differentiation of equation (4).

$$\begin{gathered} p=\frac{-1}{2}ge\left(2t\right)\hat{y} \\ \ddot{p}=-ge\hat{y} \\ \left|\ddot{p}\left(t\right)\right|=ge \end{gathered}$$

Substitute$\left|\ddot{p}\left(t\right)\right|=ge$ in equation (3).

$P=\frac{{\mu }_0}{6\pi c}{\left(ge\right)}^2$

Substitute $P=\frac{{\mu }_0}{6\pi c}{\left(ge\right)}^2$and $t=\sqrt{\frac{2y}{g}}$in equation (2).

$$\begin{gathered} U_{rad}=\left(\frac{{\mu }_0}{6\pi c}{\left(ge\right)}^2\right)\times \left(\sqrt{\frac{2y}{g}}\right) \\ {U}_{rad}=\frac{{\mu }_0{\left(ge\right)}^2}{6\pi c}\left(\sqrt{\frac{2y}{g}}\right) \end{gathered}$$

Write the expression for the loss in potential energy of an electron falling a distance.

$U_{potential}=mgy$

Substitute$U_{rad}=\frac{{\mu }_0{\left(ge\right)}^2}{6\pi c}\left(\sqrt{\frac{2y}{g}}\right)$ and$U_{potential}=mgy$ in equation (1).

$$\begin{gathered} f=\frac{{\displaystyle \frac{{\mu }_0{\left(ge\right)}^2}{6\pi c}}\left(\sqrt{{\displaystyle \frac{2y}{g}}}\right)}{mgy} \\ f=\frac{{\mu }_0{g}^2{e}^2}{6\pi cmg}\left(\sqrt{\frac{2y}{g}}\right) \\ f=\frac{{\mu }_0{e}^2}{6\pi cm}\sqrt{\frac{2g}{y}} \end{gathered}$$

Here, m is the mass of an electron $\left(m=9.11\times {10}^{-31}\mathrm{kg}\right)$.

Substitute ${\mu }_0=4\pi \times {10}^{-7}\mathrm{H}/\mathrm{m},\mathrm{e}=1.6\times {10}^{-19}\mathrm{C},\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s},$$m=9.11\times {10}^{-31}\mathrm{kg},g=9.81\mathrm{m}/{\mathrm{s}}^2$and$y=1cm$ in the above expression.

$$\begin{gathered} f=\frac{\left(4\pi \times {10}^{-7}\mathrm{H}/\mathrm{m}\right){\left(1.6\times {10}^{-19}C\right)}^2}{6\pi \left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(9.11\times {10}^{-31}\mathrm{kg}\right)}\sqrt{\frac{2\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}{\left(1\mathrm{cm}\times {\displaystyle \frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}}\right)}} \\ f=2.76\times {10}^{-22} \end{gathered}$$

Therefore, the fraction of the potential energy lost is$2.76\times {10}^{-22}$ .