91Ó°ÊÓ

• Kinetic energyis$K=1.20\text{ k±ð³Õ}=1.20×{10}^3×1.6×{10}^{-19}\text{ J}$
• Orbit radius is$r=25.0\text{ }\text{cm}=0.25\text{ }\text{m}$

We use the formula for kinetic energy to find speed. Then, we can find the frequency and period from speed and radius. To find the magnetic field, we have to use the formula for magnetic force and centrifugal force.

The formula for kinetic energy is as follows:

$K=0.5×m{v}^2$

Substituting the values in the above expression, and we get,

$\begin{array}{c}1.20×{10}^3×1.6×{10}^{−19}=0.5×9.11×{10}^{−31}×v^2\\ v=2.05×{10}^7\text{″¾/²õ}\end{array}$

Speed of electron is $2.05×{10}^7\text{″¾/²õ}$.

Now the magnetic force is balanced by centrifugal force so:

$\begin{array}{c}{F}_m=F_c\\ qvB=\frac{m{v}^2}{r}\\ B=\frac{mv}{qr}\end{array}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}B=\frac{9.11×{10}^{-31}×2.05×{10}^7}{1.6×{10}^{-19}×0.25}\\ =4.67×{10}^{-4}\text{ }T\end{array}$

The magnitude of the magnetic field is$4.67×{10}^{-4}\text{ }T$ .

Now the frequency is as follows:

$\begin{array}{l}v=r\mathrm{Ó\neg }\\ v=r×2\mathrm{Ï€}×f\\ f=\frac{v}{2\mathrm{Ï€}r}\end{array}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}f=\frac{2.05×{10}^7}{2\mathrm{Ï€}×0.25}\\ =1.31×{10}^7\text{ H³ú}\end{array}$

Circling frequency is $1.31×{10}^{7}Hz$.

The period of rotation can be calculated as:

$T=\frac{1}{f}$

Substituting the values in the above expression, and we get,

$\begin{array}{c}T=\frac{1}{1.31×{10}^7}\\ =7.63×{10}^{-8}\text{ }s\end{array}$

The period of the motion is $7.63×{10}^{-8}s$.