# Find the frequency of revolution of an electron?

a) Find the frequency of revolution of an electron with an energy of 100 eV in a uniform magnetic field of 22.0 µT.

in Hz

(b) Calculate the radius of the path of this electron if its velocity is perpendicular to the magnetic field.

in m

### 1 Answer

- MohasaLv 410 years agoFavorite Answer
The electron has energy of 100 eV

So its KE is 0.5mv² = 100 eV...(i)

where m = 9.109 × 10^(-31) kg is the electron mass, and v is its speed

1 eV = 1.6 * 10^(-19) J

So 0.5mv² = 100 * 1.6 * 10^(-19) J

0.5mv² = 1.6 * 10^(-17) J

And v = √[2 * 1.6* 10^(-17) / {9.109 × 10^(-31)} J/kg]

giving v = 5.9271 * 10^6 m/s....(ii)

The magnetic force = the centrifugal force in the magnetic field:

qvB = mv²/r

where q = 1.602 × 10^(-19) coulombs (C) is the electron charge,

v = electron speed, m its mass, B = 22.0 µT is the magnetic field strength

and r is the radius of the circular path.

So qB = mv/r

and v/r = qB/m...(iii)

The period T = 2πr/v

and the frequency f is

f = 1/T = v/(2πr)

or f = qB / (2πm) where we used equation (iii) for v/r

Solve, with B = 22.0 * 10(-6)T

f = 1.602 × 10^(-19) C * 22.0 * 10(-6)T / (2*π*9.109 × 10^(-31) kg)

giving f = 6.158 * 10^5 Hz

or f = 615.8 kHz <=====

b) r = mv/(qB)

r = {9.109 × 10^(-31) kg * 5.9271 * 10^6 m/s} / {1.602 × 10^(-19) C * 22.0 * 10^(-6)T}

r = 1.532 m <====

Hope this helps.