what is the radius of the path of an electron (mass=9 x 10^-31 KG and charge 1.6 x 10^-19C ) moving at a speed of 3 x 10⁷ m/s in a magnetic field of 6 x 10^-4 T perpendicular to it ?? what is its frequency ?? calculate its energy in KeV. ( 1 eV = 1.6 x 10^-19J ) .

The force experienced by the electron will be equal to the centripetal force required to keep the electron in its circular path.

So,

evB = mv²/r

=> r = mv/eB

=> r = 0.28125 m

Thus, the circumference of the circular path is = 2 × π × r = 1.76625 m

Time period of revolution = 1.76625/v = 5.8875 × 10^-8 s

So, frequency = 1/(5.8875 × 10^-8) = 1.7 × 10⁷ Hz

Energy of the electron is = ½ mv² = 4.05 × 10^-16 J

So, the energy in eV is = (4.05 × 10^-16)/(1.6 × 10^-19) = 2531.25 eV