# In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit.

Question:

In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Solution:

Magnetic field strength, B = 6.5 × 10−4 T

Charge of the electron, e = 1.6 × 10−19 C

Mass of the electron, me = 9.1 × 10−31 kg

Velocity of the electron, v = 4.8 × 106 m/s

Radius of the orbit, r = 4.2 cm = 0.042 m

Frequency of revolution of the electron = ν

Angular frequency of the electron = ω = 2πν

Velocity of the electron is related to the angular frequency as:

$v=r \omega$

In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:

$e v B=\frac{m v^{2}}{r}$

$e B=\frac{m}{r}(r \omega)=\frac{m}{r}(r 2 \pi v)$

$v=\frac{B e}{2 \pi m}$

This expression for frequency is independent of the speed of the electron.

On substituting the known values in this expression, we get the frequency as:

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

$=18.2 \times 10^{6} \mathrm{~Hz}$

$\approx 18 \mathrm{MHz}$

Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.