Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Explanation

When a particle of mass m and charge q performs uniform circular motion with constant speed v in a plane perpendicular to uniform magnetic field B, radius of its circular path is,
`r=(mv)/(Bq)`
`thereforer=(m(romega))/(Bq)" "(becausev=romega)`
`thereforeomega=(Bq)/m` (Where `omega` = angular frequency)
`therefore2pif=(Bq)/m`
(Where f = frequency of revolution = no. of revolutions in unit time)
`thereforef=(Bq)/(2pim)`
`thereforef=((6.5xx10^(-4))(1.6xx10^(-19)))/((2)(3.14)(9.1xx10^(-31)))`
`thereforef=18.2xx10^(6)Hz` (= 18.2 MHz)
`thereforef=1.82xx10^(7)Hz`