What is the radius of the path of an electron (mass $m = 9 \times 10^{-31} \; kg$ and charge $q = 1.6 \times 10^{-19} \; C$) moving at a speed of $v = 3 \times 10^{7} \; m/s$ in a magnetic field of $B = 6 \times 10^{-4} \; T$ perpendicular to it? What is its frequency? Calculate its energy in $keV$. (Given: $1 \; eV = 1.6 \times 10^{-19} \; J$)

(N/A) The radius of the circular path is given by $r = \frac{mv}{qB}$.
Substituting the values: $r = \frac{9 \times 10^{-31} \; kg \times 3 \times 10^{7} \; m/s}{1.6 \times 10^{-19} \; C \times 6 \times 10^{-4} \; T} = \frac{27 \times 10^{-24}}{9.6 \times 10^{-23}} \; m = 0.28125 \; m \approx 28.1 \; cm$.
The frequency of revolution is $f = \frac{v}{2\pi r} = \frac{qB}{2\pi m}$.
$f = \frac{1.6 \times 10^{-19} \times 6 \times 10^{-4}}{2 \times 3.14 \times 9 \times 10^{-31}} \approx 1.7 \times 10^{7} \; Hz = 17 \; MHz$.
The kinetic energy is $E = \frac{1}{2}mv^2$.
$E = 0.5 \times 9 \times 10^{-31} \times (3 \times 10^7)^2 = 4.05 \times 10^{-16} \; J$.
Converting to $keV$: $E = \frac{4.05 \times 10^{-16}}{1.6 \times 10^{-19}} \; eV = 2531.25 \; eV \approx 2.53 \; keV$.