91Ó°ÊÓ

1) Radial acceleration,$a=3×{10}^{14}m/s^2$

2) Radius of the circular path,

$$\begin{gathered} r=15\mathrm{cm} \\ =0.15\mathrm{m} \end{gathered}$$

The centripetal acceleration of the object depends on the velocity and radius of the orbit. Using the equation of centripetal acceleration, we can calculate the speed of an electron moving in a circular direction. Using the value of the speed of an electron and the equation of period, we can find the period of motion of an electron.

Formula:

$a=\frac{V^2}{r}$ …(¾±)

$T=\frac{2\mathrm{Ï€}r}{V}$ …(¾±¾±)

From equation (i), we can write,

$a=\frac{V^2}{r}$

By rearranging this equation, we get,

$$\begin{gathered} v=\sqrt{ar} \\ =\sqrt{3×{10}^{14}m/s^2×0.15m} \\ =6.7×{10}^{6}m/s \end{gathered}$$

Therefore, the speed of electron is$6.7×{10}^{6}m/s$ .

From equation (ii), we can write

$T=\frac{2\mathrm{Ï€}r}{V}$

Substitute the value of given radius and speed calculated in part (a), we get

$$\begin{gathered} T=\frac{2\mathrm{π}×0.15m}{6.7×{10}^{6}m/s} \\ =1.4×{10}^{-7}s \end{gathered}$$

Therefore, the time period of rotation of electron is $1.4×{10}^{-7}s$.