Given,

Magnetic field strength (B) = 6.5 G = $6.5 \times {10^{ – 4}}T$

Speed of the electron (v) = $4.8 \times {10^6}m/s$

Charge on the electron (e) = $1.6 \times {10^{ – 19}}C$

Mass of the electron (m_e) = $9.1 \times {10^{ – 31}}kg$

Radius of the orbit, r = 0.042 m

Frequency of revolution of the electron = v

Angular frequency of the electron = $\omega $ = $2\pi \nu $

Velocity of the electron is related to the angular frequency as: $\nu $= rω

The magnetic pull on the electron is balanced by the centripetal force in a circular orbit.

Hence, we can write:

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

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

$eB = \frac{{m(r\omega )}}{r}$

$eB = \frac{{m(r.2\pi v)}}{r}$

This frequency expression is independent of the electron’s speed. We get the frequency by inserting the known values in this expression:

$v = \frac{{6.5 \times {{10}^{ – 4}} \times 1.6 \times {{10}^{ – 19}}}}{{2 \times 3.14 \times 9.1 \times {{10}^{ – 31}}}}$

$v = 1.82 \times {10^6}Hz$

$v \approx 18MHz$

As a result, the electron’s frequency is around 18 MHz and is independent of the electron’s speed.