Question

In the previous question, obtaining the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Answer 1

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

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

Mass of the electron, m_e = 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

The velocity of the electron is related to the angular frequency as:

v = rω

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

`evB = (mv^2)/r`

`eB = m/r(rω)`

= `m/rr(2piv)`

`v = (Be)/(2pi 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 = (6.5 xx 10^-4 xx 1.6 xx 10^-19)/(2 xx 3.14 xx 9.1 xx 10^-31)`

= `(10.4 xx 10^-23)/(57.148 xx 10^-31)`

= 0.182 × 10⁸

= 18.2 × 10⁶ Hz

≈ 18 MHz

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