91Ó°ÊÓ

• Energy is .$100\text{ }eV=100×1.60×{10}^{-19}\text{ }J$
• The magnitude of the magnetic field is $35.0\text{ }\mathrm{μ}T$.

Here we have to usetheconcept of equilibrium of magnetic force and centrifugal force. From this, we can find radius as well as frequency. To find the radius, first, we need to find velocity usingtheequation of kinetic energy.

Formula:

$F_B=qvB$

$F_c=\frac{m{v}^2}{r}$

$KE=0.5m{v}^2$

We havethe magnetic and centrifugal forces, and in equilibrium, both balance each other. That can be written as:

$\begin{array}{c}qvB=\frac{m{v}^2}{r}\\ qB=\frac{mv}{r}\\ \frac{qB}{m}=\frac{v}{r}\\ \frac{qB}{m}=\mathrm{Ó\neg }\\ 2\mathrm{Ï€}f=\frac{qB}{m}\end{array}$

Substitute the values, and we get,

$\begin{array}{c}f=\frac{qB}{2\mathrm{π}m}\\ =\frac{1.6×{10}^{−10}×35×{10}^{−6}}{2×\mathrm{π}×9.1×{10}^{−31}}\\ =9.79×{10}^5\text{ }Hz\end{array}$

Thus, the frequency of revolution of an electron is $9.79×{10}^5\text{ }Hz$.

Now we can find thevelocity of an electron using the equation of kinetic energy as follows:

$\begin{array}{c}KE=0.5×m{v}^2\\ 100×1.60×{10}^{−19}=0.5×9.1×{10}^{−31}×v^2\\ v=5.93×{10}^6\text{″¾/²õ}\end{array}$

And

$qvB=\frac{m{v}^2}{r}$

$qB=\frac{mv}{r}$

Substitute the values, and we get,

$\begin{array}{c}r=\frac{mv}{qB}\\ =\frac{9.1×{10}^{−31}×5.93×{10}^6}{1.6×{10}^{−19}×35×{10}^{−6}}\\ =0.9641\text{ }m\end{array}$

Thus, the radius of the path of the electron is$0.9641\text{ }m$.