91Ó°ÊÓ

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.

$U_{rad}=P×t$ …… (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.

$P=\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}{\left(\stackrel{¨}{p}\left(t\right)\right)}^2$ …… (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).

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

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

$P=\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}{\left(ge\right)}^2$

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

$$\begin{gathered} U_{rad}=\left(\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}{\left(ge\right)}^2\right)×\left(\sqrt{\frac{2y}{g}}\right) \\ {U}_{rad}=\frac{{\mathrm{μ}}_0{\left(ge\right)}^2}{6\mathrm{π}c}\left(\sqrt{\frac{2y}{g}}\right) \end{gathered}$$

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

$U_{potential}=mgy$

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

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

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

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

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

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