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

Explanation

From the solution, of exercise 4.11 we have:
` (m_(e)v^(2))/r = evB " " …(i) ["For " theta = 90^(@)]`
If `omega` is the angular velocity of revolution of electron, then, we have
` v = omega r`
Substituting the value in (i) , we get
`(m_(e)(omega r)^(2))/r = e (omega r) B`
` rArr " " omega = (Be)/m_(e) `
The frequency of revolution of electron in its circular orbit is thus given by
`v = omega/(2 pi) = (Be)/(2 pi m_(e))` .....(ii)
` = (6.5 xx 10^(-4) xx 1.6 xx 10^(-19))/(2 xx pi xx 9.1 xx 10^(-31)) `
` = 18.18 xx 10^(6) Hz approx 18 ` MHz
From (ii), we can see that the frequency of revolution is independent of the speed of electron.